The digits in a measured quantity which are reliable and confidence
in our measurement + the digit which is uncertain.
MASS
•Unified atomic mass unit(amu) is used to measure
mass of atoms & molecules
•1amu =(1/12)th mass of
one C12 atom
•1amu = 1.66×10-27 kg
•Electron mass- 10-30 kg
2. A zero becomes significant figure if it appears between two
non-zero digits. For example, 5.03 has three significant figures;
5.604 has four significant figures; and 4.004 has four significant
figures.
•Earth mass : 1025 kg
•Observable Universe 1055 kg
TIME
•SI unit is second (based on caesium clock with an
uncertainity less than 1 part in 10-13
ie,3μs loss every year)
•Timespan of unstable particle: 10
s
•Age of universe: 10 s
17
•Parallax angle=
BASIS
DISTANCE
b
=x
x
•1 =1.745 x 10 rad
•1‛=2.91×10­4 rad.
•1"=4.85×10­6 rad.
O
x
b
3. If the digit to be dropped is 5 followed by digits other than zero,
then the preceding digit is raised by one. For example, x = 16.351 is
rounded off to 16.4 and x = 6.758 is rounded off to 6.8.
•For very small sizes, optical microscope,
tunneling microscope, electron microscope
were used.
4. If the digit to be dropped is 5 or 5 followed by zeros, then the
preceding digit, if it is even, is left unchanged. For example,
x = 3.250 becomes 3.2 on rounding off and x = 12.650 becomes 12.6
on rounding off.
•1 AU = 1.496×1011 m
•1 ly = 9.46 × 1015 m
•1parsec= 3.08 x 1016 m
5. If the digit to be dropped is 5 or 5 followed by zeros, then the
preceding digit, if it is odd, is raised by one. For example,
x = 3.750 is rounded off to 3.8, again x = 16.150 is rounded off
to 16.2.
•Size of proton: 10-15 m
•Radius Of Earth: 107m
•Distance to Boundary Of
Observable Universe
: 1026 m
RULES FOR ROUNDING OF A MEASUREMENT
ADDITION & SUBTRACTION
In addition or subtraction, the final result should be reported
to the same number of decimal Places as that of the original
number with minimum number of decimal places
SI SYSTEM
7 Base units and 2 supplementary units
1
Length
2
Mass
Time
3
4
5
6
7
NO.
Base Units
Unit
Quantity
Temperature
Electric current
Luminous intensity
Amount of
substance
Symbol
meter
m
kilogram
second
kelvin
ampere
candela
mole
kg
s
K
A
cd
mol
Quantity
Unit
Symbol
Plane angle
radian
2
Solid angle
steradian
rad
sr
(has two decimal places)
(Answer should be reported to two decimal
places after rounding off)
MULTIPLICATION & DIVISION
When numbers are multiplied or divided, the number of
significant figures in the answer equals the smallest number
of significant figures in any of the original numbers
51.028
x 1.31
66.84668
(Three significant figures)
(Answer should have three significant figures
after rounding off)
Answer = 66.8
Q. Unit of permittivity of free space
(a)
(b)
(c)
(d)
coloumb/newton-metre
newton-metre2 /coloumb²
coloumb²/newton-metre2
coloumb2/(newton-metre)2
ε0 is
If L=2.331cm, B= 2.1cm,then L+B = ?
Instrumental
1) Pressure=stress=Young‛s modulus=ML T
2) Work=Energy=Torque=M L2 T-2
3)
4)
5)
6)
7)
Power P=M L2 T-3
Gravitational constant G=M-1 L3 T-2
Force constant=Spring constant=M T-2
Coefficient of viscosity=M L-1 T-1
Latent heat L=L2 T-2
8
9
ε
0
Least Count:
Smallest quantity an instrument can
measure
mm scale
↓
1mm
vernier scale
↓
0.1mm
screw gauge
↓
0.01mm
VERNIER CALIPERS
1VSD = n-1 MSD
n
n-1
Least Count = 1MSD - n MSD = 1MSD
n
Total Reading = Main Scale Reading + coinciding
Vernier Scale division x least count
Time period
L
R
l
g
α
m
k
R
g
α
= RC = LC
In a vernier calipers, one main scale division is x cm
& n division of vernier scale coincide with n-1 divisions
of the main scale. the least count (in cm) of the
calipers is.
n-1
nx
x
x
a) ( n ) x
b)
c)
d) n
(n-1)
(n-1)
SCREW GAUGE
DIMENSIONLESS
QUANTITIES
Main Scale Reading
Pitch =
No.of rotations
1) Strain
2) Refractive index
3) Relative density
4) Plane angle
5) Solid angle
Least Count =
pitch
Total no.of divisions on
circlular scale
Total Reading = Linear Scale Reading + circular scale
reading x least count
In SI Units, the dimensions of ε
0
μ0
(a) 4.431 cm
(b) 4.43 cm
a)A-1 T M L3
b)A T2 M-1L-1
(c) 4.4 cm
(d) 4 cm
c)A T-3 M L3/2
d)A2 T3 M-1 L-2
is:
The least count of the main scale of a screw gauge
is 1mm the minimum no.of divisions on its circular
scale required to measure 5μm diameter of wire is
a) 200 b) 50
c) 400 d) 100
Δamean
amean
x 100
COMBINATION OF ERRORS
Operations
Absolute
error Δ Z
Formula Z
Relative
error ΔZ/Z
Percentage error
100 x Δ Z / Z
Sum
A+B
ΔA+ ΔB
ΔA+ΔB
A+B
ΔA+ΔB
A+B
x 100
Difference
A-B
ΔA+ ΔB
ΔA+ΔB
A-B
ΔA+ΔB
A-B
x 100
AxB
AΔB+ BΔA
Division
A
B
BΔA+ AΔB
Power
An
Multiplication
A
Root
1/n
B2
ΔA
A
+
ΔA
A
+
ΔB
B
ΔB
B
( A + B ( x 100
( AA+ BB(x 100
n
1 A 1/n-1 ΔA
n
1 ΔA
n A
Δ
Δ
ΔA
n A n - 1 ΔA
ΔB
ΔA
A
n
ΔA x 100
A
1 ΔA x 100
n A
General rule:
If Z = APBq
Cr
Tα
Δamean
amean
• Percentage Error:-
If n VSD Coincides with (n-1)
MSD,
then (n-1) MSD= n VSD
10) Capacitance=M-1 L-2 T-4 A2
11) Permittivity ε0=M-1 L-3 T4 A2
12) Angular momentum = planck‛s constant
=M1 L2 T-1
a1+a2+a3+ ....+an
n
; Δamean= Δa1+Δa2+Δa3+ ....+Δan
n
• Absolute Error :- Δa = |ai-amean|, amean=
Least Count = 1 MSD - 1VSD
I
=M L2 T-3 A-2
μ0
-2
Personal
Due to individual
bias,Lack of proper
setting of apparatus
Limitations in
experimental
technique
• Least count error is the smallest value that can be measured by
instrument (occurs with random & systematic errors)
INSTRUMENTS
-1
Irregular and at random
in magnitude & direction
Experimental
Due to inbuilt defect
of measuring instrument
• Relative Error:-
Answer = 3.47
Supplementary Units
1
3.1421
0.241
+0.09
3.4731
Deducing relation
among physical
quantity
M A L B T C
n1= n2 [ 2 ] [ 2 ] [ 2 ]
M1
L1
T1
DIMENSIONAL FORMULA
Random Errors
Errors which tend to occur
only in one direction,
either positive or negative
conversion of one system
of unit into another
n1u1=n2u2
Eg: n1[M1A L1B T1C] = n2[M2A L2B T2C]
checking the correctness of
various formulae
Eg: If Z=A+B,[Z]=[A]=[B]
4. Trailing zeros or the zeros placed to the right of the number are
significant. For example, 4.330 has four significant figures; 433.00
has five significant figures; and 343.000 has six significant figures.
2. If the digit to be dropped is more than 5, then the preceding digit
is raised by one. For example, x = 6.87 is rounded off to 6.9 and
x = 12.78 is rounded off to 12.8.
-2
NO.
3. Leading zeros or the zeros placed to the left of the number are
never significant. For example,0.543 has three significant figures;
0.045 has two significant figures; and 0.006 has one significant
figure.
1. If the digit to be dropped is less than 5, then the preceding digit is
left unchanged. For example,x = 7.82 is rounded off to 7.8 and
x = 3.94 is rounded off to 3.9.
Systematic Errors
APPLICATIONS
RULES FOR ROUNDING OF A MEASUREMENT
p
•Large distance is measured by
parallax method
Difference between true value
& measured value of a quantity
Dimensions of a physical quantity are power to which units of base quantity
are raised. Eg: [M]a [L]b [T]c [A]d [K]e
5. In exponential notation, the numerical portion gives the number of
significant figures. For example,1.32 x 10-² has three significant
figures and 1.32 x 104 has three significant figures.
MEASUREMENT OF LENGTH
ERRORS IN MEASUREMENT
Dimensional Analysis
RULES FOR SIGNIFICANT FIGURES
1. All non-zero digits are significant. For example, 42.3 has three
significant figures; 243.4 has four significant figures; and 24.123 has
five significant figures.
-24
UNITS & MEASUREMENTS
SIGNIFICANT FIGURES
MEASUREMENT OF MASS & TIME
,Then the maximum fractional relative
error in Z will be:
ΔZ =p ΔA +q ΔB
Z
A
B
+r ΔC
C
In an expirement four quantities a,b,c
and d are measured with percentage
error1%, 2%, 3% and 4% respectievely.
Quantity P is calculated as follows:
P=
a2b2
cd
then percentage error in P is
(a) 14% (b) 10%
(c) 7% (d) 4%
Motion with constant acceleration: Equations of motion
(i) v=u+at
1 at2
(ii) S =ut+2
A Person travels from A to B covers unequal distances in equal
interval of time with constant acceleration a
then
Distance = Length of actual path
Displacement = Length of
shortest path
initial velocity U=
Distance > |displacement|
Acceleration a =
(
Ratio of Displacement to Distance = 2Sin θ
2
Time t =Rθ
θ
θ
(
Average Acceleration = 2U2Sin θ
2
Rθ
U
2t
S2-S1
t
t
t
A
The number of planks required to stop the bullet
v= u2+v2
Mid
2
u
Calculation of stoping distance
Displacement = velocity x time
u2
2a
s=
=
s1 + s2 + s3 + ....+s n v 1t1 +v 2t 2 + v 3 t3 + ......
=
t1 + t2 + t3 + ......
t1 + t2 + t3 + ....+ tn
(Arithmetic mean of speeds)
Total distance covered s1 + s2 + s3 +.....+ sn
s + s + s3 + .....+ s n
=
v av =
= 1 2
s
s
s
s
Total time elapsed
t1 + t2 +t3 + ....+ tn
1
+ 2 + 3 +....
.. + n
v1 v2 v 3
vn
If s1= s2= s 3= ...... sn s
2V1V2
then
n
=
(Harmonic mean of speeds)
v av =
1 1 1
1
V1+V2
.+
+
+ + .....
v1 v2 v 3
vn
Instantaneous Acceleration
∆ x =∫vdt
dv
a=
dt
∆ v =∫adt
Case 3
V or x = f(t)
V= f(x)
dV
a= V
dx
t=f(x)
then
a=-(double diff. of t w.r.to x)
Displacement
Differentiation
Integration
Velocity
Differentiation
Integration
3. Uniformly accelerated with
u = 0 at t=0
X
V3
Acceleration
6. Uniformly retarded then
accelerated in opposite direction
A car accelerates from rest at a constant rate α for some time, after which it decelerates at
a constant rate β, to come to rest. If the total time elapsed is t, the maximum velocity
attained
αβ
αβ
t
Vmax =
A
v
Total Distance = 1
t2
α+β
2 α+β
(
=
v
s
vt
s=
s =½ at
2
s
t
v
s=ut+½at2
t
v
s
t
a
u+
v=
u
u
s
at
s=s0+ut+½at2
s
s=ut-½at2
u-
at
t0
t
s
t0
t
+ve = upward motion
-ve = downward motion
(ii) Acceleration
t0
u=0
h
(iii) Ratio of distance covered at the end of time t:2t:3t:....=12:22:32....
(i) Maximum height attained H =
H
t
Maximum height H =
t1
H
t2
h
1
g(t1 +t2) 2
8
A body is thrown upward, downward & horizontally with same speed
takes time t1, t2 & t3 respectively to reach the ground then
t3 = t t
1 2
u
u2
2g
(ii) Time of ascent = time of descent u
g
2u
(iii) Total time of flight = g
(iv) Velocity of fall at the point of projection = u (downwards)
height of point h=½ gt1t2
t1+t2= 2u
g
t0 t
v
u
(i) initial velocity
At any point on its path the body will have same speed for upward journey and
downward journey. If a body thrown upwards crosses a point in time t1 & t2
respectively then
t
v
v=
Sign Convension
If a body is thrown vertically up with a velocity u in the uniform gravitational field (neglecting
air resistance) then
t
u+
v=
u
MOTION UNDER GRAVITY
(i) Ratio of displacement in equal interval of time S1:S2:S3....=1:3:5....
(ii) Ratio of time of covering equal distance
t1:(t2-t1):(t3-t2):.......:(tn-tn-1)= 1: ( 2- 1):( 3- 2):...:( n- n-1
t
at
t
t2
Object is dropped from top of a tower
s-t graph
v=
B
t1
+ve = final position is above initial position
-ve = final position is below initial position
Zero = final position & initial position are at same level
t
v
O
(iii) Displacement
u+v
2
v=constant
max
Always -ve
t
B
v-t graph
2. Uniformly accelerated motion
with u =0 at t=0
5. Uniformly retarded motion till
velocity becomes zero
Case 2
dv d2x
=
dt dt2
S3
S2
t
vavg
Different Cases
4. Uniformly accelerated motion
with u=0 and s=s0 at t=0
Case 1
a=
S1
t
for uniform accelerated motion
1. Uniform motion
Distance average speed
dx
dt
0
s
.........
A
v1 + v2 + v 3 +.....+V n v1+v2
= 2
n
Instantaneous Velocity v =
u
S1
Time average speed
v av =
v
v
S1:S2:S3 = 1:3:5
If t1 = t2 =t3 = .....= t n
then
u
a
a (2n-1)
(iv) sn =u+ _
2
Ratio of distance travelled in equal interval of time in a uniformly
accelerated motion from rest
Average speed = |average velocity|=|instantaneous velocity|
Total distance covered
Total time elapsed
a-t curve area gives change in velocity.
The two ends of a train moving with constant acceleration pass a certain
point with velocities u and v. The velocity with which the middle point of
the train passes the same point is
For uniform motion
v av =
B
0
2
v-t curve area gives displacement.
Slope of velocity-time curve = instantaneous
acceleration
u2
N= 2 2
u -v
(
(
θ
Average Velocity = 2USin
2
(
U
S2
[ [
[∆ x=∫vdt[
dv
[a = dt [
[ ∆ v =∫adt[
(iii) v2=u2+2a.s
U
(
(
(
S1
Instantaneous velocity is the slope of
position time curve
MOTION
ALONG A
STRAIGHT
LINE
dx
v=
dt
(
A particle moves from A to B in a circular path of
radius R covering an angle θ with uniform speed U
θ
Distance = Rθ
Displacement = 2RSin
2
3S1-S2
Important point about graphical
analysis of motion
& height from where the particle was throw is h= ½ gt1 t2
v
1
v
2
v
3
VECTORS
Magnitude Of Vectors

ˆ ˆ ˆ
r=xi+yj+zk

|r|= x +y +z
2
2
2
ˆ ˆ ˆ
r=xi+yj+zk

-ˆ
ˆ
i East
-ˆ
j
South
A2+B2+2A Bcos




(x2y2z2)

ˆ
ˆ
ˆ
∆ r=(x2-x1)i+(y
-y1) j+(z
-z1)k
2
2

(x2-x1) +(y2-y1) +(z2-z1)
|∆r|=
2
2
R=Rcos θ ˆi+Rsin θjˆ
R
)θ
2


Rmax=A+B



tan = B sin θ
Rcos θ
A+B cos θ
Vector product
5m north
3m south
2m east
=
R=
Parallel Vectors
b
ˆ
= 2j+2i

a=a1ˆi+a2ˆj+a3ˆk


ˆ ˆˆ
5j-3j+2i


a
k Vertically downward (Zaxis)
-ˆ
ˆ
ˆ
b=b1ˆi+b2j+b
k
3


β
a=mb
a1 a2 a3
= = =m
b1 b2 b 3
γ
)
2
cos
+ cos
sin
2
sin
+ cos
=1
+ sin
=2
2β
+
2β
2γ
2γ


C=A B= AB sin θ
ˆ
ˆ
i ˆj
k
ˆ
ˆ
ˆ
A B= Ax Ay Az =i[A
B -ByAz]-j[A
B -AzBx]+k[A
B -AyBx]
y z
x z
x y


+
o
3
North
i
West
(x1y1z1)


+
ˆ
j

|r|or r= x2+y2+z2

ˆ ˆ ˆ
xi+yj+zk
unit vector=ˆr= r =
|r|
x2+y2+z2

r
ˆ ˆ ˆ
i-j+k
ˆr=
Vertically
upward (Zaxis)
k
Displacement Vector
|r|= 12+12+12= 3
ˆj
ˆ
p

ˆk
Addition Of Vectors
Components of Vector

ˆ ˆ ˆ
r=i-j+k
Rsi
nθ
)
1,1
(1,
-
)
i
Bx By Bz
Dot product
 
MOTION
IN A PLANE
x=A.B= AB cos θ
Projectile motion
PROJECTILE MOTION
Same range for θ and (90-θ)
Horizontal component = Ucos θ
Vertical component = Usin θ
Uy2
2
2
2
sin
U
(Usin
)
θ
θ
H=
=
=
2g
2g
2g
2
θ = 2Usin θ Ucosθ =
R= U sin2
g
g
KE at maximum height =Kcos θ
R=4 H1.H2
θ and (90-θ)
eg;
Equation of Velocity
2
T2
H2
45+ θ,45- θ
y
y
Momentum
u sin θ -gt
θ
300 & 600
2UxUy
g
) )θ
0
shortcut
T+
T2= 2R
1
g
4H=Rtan θ =5T2
0
0
0
30
45
0
60
T12+T22=
Ucos θ
4 Rmax
g
Pf=m(ucosθ ˆi+(usin θ-gt) ˆj)
H1
=tan2θ
H2
T1
=tan θ
T2
15
u
u cos θ
H1
(9
(u cos θ)2 +(u sin θ-gt)2
=
T1
0
θ)
0-
Vx2+Vy2
Vnet=
u
2
H+
H2= R
1
16
R
H1+H2= max
2
x
Net displacement
θ
x
x=Ucos θ t
ˆ
θ ) jˆ
Pi=m(ucos θ )i+(usin
y=Usin θ t- 1 gt2
2
ˆ
Maximum range
Δ p=-mgt j
From the relation,
=-mg+ u sinθ ˆj
g
4H=Rtan θ =5T2
For θ = 45o
R
2
4H=R tan45 H = max = U
4
4g
4H=Rmax
2
Rmax= U
g
r= x2+y2
r
=-mu sinθ ĵ
Rmax = U2
g
y
=tan-1(y/x)
x
RELATIVE MOTION
Relative Motion
VA
1) Velocity of A with
respect to B VAB=VA-VB
2) VA/B=VA-VB
=VA-(-VB)=VA+VB
VB
3)VA/Tree=VA-VTree=60-0=60
VB/Tree=VB-VTree=-40
VA
Stopping distance
Relative Motion in one dimension overtaking & chasing
Tree
VA=60km/hr
VB=40km/hr
1) t=
VB
2) t=
d+L1+L2
V1+V2
d+L1+L2
V1-V2
d
L1
V2
V1
L2
L1
V1
L1
urel=u1-u2
3) d+L1+L2=(u1-u2)t+ 1 (a1-a2)t2
2
V1
d
V2
L2
V2
d
L2
0=u2-2as
0= u2rel-2arels
arel=a1-a2
u2
S= rel
2arel
u1
a1
d
u2
a2
RELATIVE MOTION
tcross=
2) Down stream

1) VMR or VM/Still water= velocity due to
effort of man
[V]/ velocity in still water.
Shortest time
Xdrift=(VMRcos θ +VR)+ t
VM/R
VM=VM/R+VR
VMR
=VMRcos θ t
+
+
2) VR= velocity of River
3) Vm= Resultant velocity of man
with respect to ground
VR
1) Upstream
VMRsin θ
VM=VM/R-VR
VM/R
VMR
VMRsin θ
VR
Additional
θ =90
sin θ =1
d
VMRsin θ
VMRsin θ = VR
= sin θ =
Xdrift=VR+ t
θ
VMRcosθ
Escalator
VR
t 3=
VMRcos θ =Vm
VR
VMR
VR
tt
d
d
=
= 1 2
VE+VM/E d + d
t1+t2
t1 t2
t1=Time taken by a man to move distance d
on a stationary escalator
= tcross= d
VM
Vm = (VMR)2+(VR)2
2) Vllel =VMRcos θ +VR
VMRcos θ
Condition for no drifting
tmin= d
VMR
1) V =VMRsin θ
Vm
Vm
θ
t=
Due to effort
of man
Swimming across the river
VMR
Vm
VRt
VR

Shortest Path
= Vm = VMR2-VR2
t2=Time taken by a man to move distance d
along a moving escalator
= Drift=0
t3=Time taken by a man to move distance d
while walking along a moving escalator
VR
VE=Velocity of escalator
VM/E=Velocity of man w.r.t escalator
MAN RAIN PROBLEM
Method
VA
VB
VA
B
A
VB/A=VB-VA ,
VB
VB w.r.t A
=tan-1
VR/m=
=
2
t
V =
sec
2
60
+2
(
=
2
T
v
2
=
hour
12 x 60 x 60
min
hour
12
= 1
v
at = r
at =
Equation of angular motion
2) Constant angular acceleration
ac
= constant
VRsin θ -Vm
VRcos θ
VR/m=
VR
VRcos θ
VR/m
V
tan = m
VR
VR/m=VRsin θ
)
v
u
VR
VR/m= V 2+V 2
R
m
Vm=VRsin θ
(VRcos θ)2+ (Vm-VRsin θ)2
x
d
B
shortest path,x=
Time taken to reach
shortest path
=
ac
ac
at
v
ac
at
ac
Tangential Force
vl
llll
In direction of tangent
=d
at= dv = r
dt
dt
Ft= mat
Resultant acceleration
ar=
ac2 + at2
= Constant
2
- ac = vr ,r
v
Velocity is tangent
to the circle
2
Non-uniform Circular Motion
- Speed not Constant
- Velocity changes in direction
and magnitude
- ac = Centripetal acceleration
- at = tangential acceleration
-
dv
dt
T1=m1l1
dt
0
m2
T2
l1
2
+m2(l1+l2)
0
m3
T3
l2
Banking of Road
d V
V2+U2
d U
V2+U2
+m3(l1+l2+l3)
2
+m3(l1+l2+l3)
Vopt =
- Speed Constant
- Direction of velocity changes
T =
3gR
Rg
s
mv2 + mg
r
of the circle =
5gR
At Top
a) Tmin =
m
mv2
= mr
r
)
b) min velocity at bottom
T
l
tan
At Bottom
Flat circular track
Horizontal circular motion
Uniform Circular Motion
rg(tan
1 +
gR
Vmax =
)
2
2
2
+
tan
rg tan
a) Tmax =
angular
acceleration
Circular Motion
rg(tan
1 -
l3
2
Vmax =
Vmin =
T2=m2(l1+l2)
T3=m3(l1+l2+l3)
=d
= Changes
0
m1
T1
- at = 0
=
ac= v2 = ac= r 2
R
→→
→
→
ac
Power= F V = 0
v
→→
→
→
Work= F S = 0
s
Fc
K=0
vll
t
-
ac
-change the direction of velocity
x r
1) Constant angular velocity :-
ac
va
lll
)
v
ac
Tangential acceleration
t2
Directed towards centre
Angular acceleration
d
dt
=
t +1/2
2
o
VR/m
Vm>VRsin θ
VRsin θ -Vm
VRcos θ
VRsin θ
A
Vm
CIRCULAR MOTION
Not a constant vector
2
=
min
60 x 60
=
o
(VRsin θ -Vm)2 +(VRcos θ)2
t
Centripetal acceleration
x r
=
+
o
=
2
T
tan =
Case 4
Vm
VRcosθ
VRsin θ >Vm
Vm
10 m/s
Vm-VRsin θ
VRcosθ
Shortest path
Case 3
VR/m
VR cosθ
20 m/s
Angular velocity
VRsin θ -Vm
VR
VA/B= VB-VA , VA w.r.t B
10 m/s
Case 2
θ
VR/m= Velocity of Rain w.r.t man
1o= 180
=
VR sinθ
VR = Velocity of rain w.r.t
stationary man
Vm = Velocity of man
20 m/s
Case 1
x
Inorder to find the
relative velocity of B
with respect to A
we have to reverse
the direction of
vector A and add it
with vector B
=
Vm
Terms
)
Man-rain problem
x
MOTION IN A PLANE - 02
d
= d
VMRcos θ
V
T

T
Man-river problem
mv2
- mg
r
b) min velocity at top to
2
= ml
complete the circle =
2
5gR
gR
INERTIA
A body cannot change its state of rest or
uniform motion along a straight line. This
property is called inertia.Inertia has no
unit and no dimension.
INERTIA
Inertia
Inertia
Inertia of
of Rest
of Motion
Direction
- Inertia of Rest
Inability to change state of rest by itself.
- Inertia of Motion
Inability of a body to change its state of
uniform motion by itself.
- Inertia of Direction
Inability of a body to change direction of
motion by itself.
Newton‛s Second Law
Fnet= Rate of change of linear momentum.
M
T
M2
a
F N|
a=
T
M2g
a
M1
N|=F|
T =
M1g
Case 3: Accelerated upward at
a rate of ‘a‛
M1g
M1 + M2
M1M2
M1 + M2
R = m(g+a) = Wapparent
g
T
F
INCLINED PLANE + PULLEY
Restoring force developed
when a longitudnal force is
applied on a body
Wapparent > Wactual → Feels over weight
Accelerated upward at a rate of ‘g‛
R -mg = mg , R = 2mg , W = 2 x W
app
act
Case 4: Accelerated downward at
a rate of ‘a‛
mg - R = ma , R = m(g-a) = Wapp , W < W
app
act
F|
Accelerated downward at a rate of ‘g‛ [ Freefall]
mg - R = mg , R = mg - mg =0 , W = 0
app
If a > g : body looses contact with the weighing
Lift moves from ground floor to first floor
Ideal Rope
*Massless
*Tension is same everywhere
T
*Can support only elongation
*Cannot support compressive force
F
*On compression it becomes slack.
*Tension always acts away from the object.
v decrease
A
Rod
Graph
F
B
x
Q
N - Mg = Ma (Ma=0)
Fwall = 2ρAu2cosθ
Change in momentum=-2mu cos θ
N
N = Mg
Along x-axis
F - 0 = Ma
a=
F
M
a
F
M
Mg
M
L
x
Note :
given
total mass
x
Mass of given length=
length
total length
mass
= constant
length
Along y-axis a=0 ,
LAWS OF MOTION
T=
m2F
M
C
V
1
2
t
t
Retardation,a2=tanθ2
M
L
Mass of x length of rope =
SINGLE BLOCK
A
2
1
Acceleration,a1=tanθ1
L
Mass per unit length=
Acceleration is along x-axis only
ucos
F
V
)
-a 2
(g
THICK ROPE
Tension will be different at different points.
a=0
=m
N2
Compressive
)
T
1
F
Horizontal Force
u
Ground floor
L
+a
T
When liquid jet strikes obliquely
usin
N=m(g+a)
a
N=mg (true weight)
u
ucos
a (retardation)
v Increase
MOTION OF BLOCKS CONNECTED BY MASSLESS STRING
can support both elongation and compression
usin
first floor
B
-ve
2
N=m(g-a)
)
-a 2
(g
Fjet = -2ρAu2
N=mg (true weight)
v=0, a=0
L
C
u
Δt
Fjet = -2ρAu2cosθ
Case 2: Lift moving up or down with
constant velocity
R = mg , W
= Wactual = mg
apparent
N
N=F
Elongation
+ve
-2mu
u
M1g
(g
mx0-mu
Fwall = 2ρAu
Fnet
M1
=m
N2
Δt
F1
Apparent weight ,(Wapparent) = Reaction
force (R)
Case 1: Lift is at rest
R = mg , Wapparent = Wactual = mg
T
M2g
machine and R becomes zero
When liquid jet bounce back
=
M
g
=m
v=0
-mu
=
=-ρAu2
=
Δt
Δt
Δt
= ρAu2 (m=ρAΔx, Δx = u)
Δt
Pf-Pi
a=
F2
T
M2
M1 + M2
Reading of weighing machine = reaction
force exerted by weighing machine (N)
a
)
wall
m
x
Fjet=
F1 - F2 = Ma
F = Ma
a
1
u
Fwall
2M1M2
T =
Apparent weight of body in a lift
T
g+
a
Area=A
Fjet=
F = M1- M2
g
M
M1 + M2
a=
a
F1 > F2
T
T
LIQUID JETS
Pf-Pi
If, Fsinθ >Mg
the block leaves contact with ground and
it begins to accelerate.
M1 > M2
W
Tension
T
P=mv
-It is a vector quantity having direction
same as that of velocity
-Unit is kg m/s.
Mg
If, Fsinθ =Mg
block just leaves
contact with ground
LIFT PROBLEMS
Ideal
1
MOMENTUM
N
Fcos
PULLEY
N
p pf-pi
=
t
t
Fav=
B
N
M
F
If, Fsinθ <Mg
block remains in
contact with ground
MOTION OF CONNECTED BODIES
Action
Normal Reaction
- Occurs when two surfaces are in contact with
each other
Always perpendicular to the surface
and is self-adjusting.
N=W
a
=m
(
dp
dt
Reaction A
FAction= -FReaction
Fsin
N
1
F=
Average
* Force exerted on body A by body B (action )
* force exerted on body B by body A (reaction)
Inclined Forces
N
Instantaneous
NEWTON‛S THIRD LAW
-To every action, there is always an equal
(in magnitude ) and opposite (in direction) reaction.
- Forces in nature always occurs in pairs.
- A single isolated force is not possible.
- Counter force experienced by a body- reaction
- Action and reaction never act on the same body
M
(L-x) = m2
L
M
x= m1
L
x
L
F
FRAME OF REFERENCE & PSEUDO FORCE
Frame of Reference
A frame in which and observer is situated
and makes his observation
Inertial frame of
reference
Non-Inertial frame
of reference
At rest or moving with
uniform velocity along
straight line.
i.e unaccelerated
Accelerated frame
of reference.
Holds Newton‛s law
of motion
Newton‛s law of motion
not applicable.
u
a
M0g
F1
Static friction
F3
F
F2
F1
F = F1 + F2 , F = F12 + F22 , tan =
F2
MAN-CAGE PROBLEM
T
N
m
M
N =
F1
g
m-M
2
g
t
Pf-Pi = Fip dt = area under F-t graph
o
I = Impulse = Pf - Pi =
mg
t
o
Fip dt = area of
F-t graph
fs max
N
s
=
fs
N
I = Pf - Pi
= mV2 -(-mV1)
h2
v2 = 2gh2
I = m (V1+V2)
impulse
P
= t
to
o
h
h
m(v+v) = J
= m(2v)
v
v
F
Case-2
Impulse
N
Fcosθ ≥ µs(mg-Fsinθ)
Fcos
M
Fs
s
m
J = 2mv
k
m
F
µsmg
cosθ - µssinθ
aT>a0
fk
mgcos
o
mgsin
o
VERTICAL TRUCK BOX
o
fsmax
mg
.
s
As angle of inclined plane increases, R remains
constant and when sliding starts R starts
decreasing.
ma0
M
k
N
mg
a0
As angle of inclined plane increases, angle of
friction will also increases and as sliding
N < R < N 1+µs2
a=0
N
m
µs
then box
does not slip
Fmin is the minimum force required to
make the block stationary if the
angle of inclined plane is
greater than angle of repose
fs
tan θ = µs
Fext
N
s
Fmax is the force required to
move the block along the inclined
plane if the angle of inclined plane
is greater than angle of repose
fs
Fmax = mg ( sinθ + µscosθ)
Fpulling =
g
Fmin = mg ( sinθ - µscosθ)
iii) fext = fsmax
Rmax
If a0 >
Minimum & Maximum force (applied parallel to
inclined plane)
µsmg
cosθ + µssinθ
cosθ + µssinθ
Fcosθ ≥ µs(mg+Fsinθ)
M
m
starts its value becomes constant and tanθ=µk
Fcos
F≥
Fsin
µsmg
= aT- gµ
Box Slips on the truck
arel
Variation of angle of friction.
max
F≥
Average impulsive
force
2mv
P
=
Favg=
to
t
ii) 0 < Fext < fsmax
PULLING FORCE & PUSHING FORCE
Fsin
v=0
tanθ0 = µs
Variation of R
mg
Fsmax=
v1 = 2gh1
u=0
Fext = 0
depends only on µs and
is independent of mass.
arel
k
Average impulsive
force
Favg=
fsmax
N
in
Impulse
atruck> a0=gµs
ANGLE OF REPOSE
µN
fsmax = µkN
Box does not slip on the truck
Case-2
fs = 0
kinetic region
Static region
fs = MaT
mg
gs
v=0
h1
fk =
aT
fs
N
N = mg
R
a0 = gµs
M
Fext
m
u=0
F
f
Case-1
k
µs is Coefficient of friction
Rmax = N 1+µs
µ
Mg
tan θ = µs
F k=
0 < tanθ < µs
If the applied force is increased further and
sets the body in motion, the friction opposing
the motion is called kinetic friction.
dt
fs =fs max R =
mg
Kinetic friction
dp
Fip =
m
Fs
fs
Case-1
atruck< a0=gµs
N
fsmax = µsN
F
IMPULSE
Large force acting for short period of
time, As a result of impulse there will
be a sudden change in momentum
m+M
T =
2
T
F2
N
R
- It is a self adjusting force.
- The opposing force that comes into play,
when object tends to move over the surface
of other object, but the actual motion has not
yet started.
- As applied force increases static friction
also increases.
- If the applied force is increased than the
force of static friction also increases.
- The body doesn‛t move untill maximum
value of static friction is attained
- The value is called limiting friction or fsmax
PARALLELOGRAM LAW
When
R
i) Fext = 0
FRICTION
M0
EQUILLIBRIUM & LAMI‛S
THEOREM
F1
F2
F3
=
=
sin
sin
sin
=vtan
velocity of B towards A = vsin
velocity of A away from B = ucos
Mo = M+m
cot - 1
F
x(-v) , u
R is resultant
fs
tan θ =
aT
fs max
in
m1
a =
g
m2
m
M
y
x
M
θ = Angle of kinetic friction
ax
u =-
ucos
1+µk2
m
x
A usin
u
N
m
m1
R = N2 + fs2
Minimum mass such that
there is no relative slipping
m2
M
, a = gtan
R =
F
Minimum acceleration
“M” must be pushed such
that there is no relative
slipping.
vcos
Angle made by resultant of normal (N)
& frictional force(fs) with normal
vsin
HORIZONTAL TRUCK BOX
iv) f > f
ext
smax
m
gs
F = (m+M) gtan
M a
B
ANGLE OF FRICTION
F
Minimum force required to push the inclined plane
such that “m” does not slip with respect to “M”
m
F
ROD SLIDING ON A WALL
in
RELATIVE SLIPPING
µsmg
Fpushing =
cosθ - µssinθ
mg(sinθ -
µscosθ) < F < mg (sinθ + µ cosθ)
s
LAWS OF
MOTION
WORK
Work is said to be done
when a force applied on the
body displaces the body
through a certain distance in
the direction of force
WORK DONE BY
CONSTANT FORCE
Fsin
WORK DONE BY
CONSERVATIVE & NON
CONSERVATIVE FORCE
WORK DONE IN PULLING
THE CHAIN
L
B
(1/n)
3
WA
S
W=Fcos x S
=FScos =F.S
WA
M
(Path 1)= WA
B
B
(Path 2)=
B
(Path 3)
(for conservative force)
NATURE OF WORK DONE
1) Positive work (0o< <90o)
F
Direction
of motion
S
Direction
of motion
S
2) Zero work
Work done becomes 0 for
three conditions
1.Force is perpendicular to
displacement
2.if there is no displacement
3.if there is no force acting
on the body
WA
WA
B
B
(Path 1)= WA
B
(Path 3)
W= F.ds = Fds cos
in terms of rectagular
components
F=Fx i+Fy j+FZ k
ds=dx i+dy j+dz k
W= Fxdx+
dW=F.dx
WORK DONE BY DIFFERENT
FORCES
B
B
B
h1
A
A
A
force
xi
= F.dx
xf
Displacement
W= Area under curve
Relation between different units
h
l
0
work done by kinetic friction
-ve
Wfk=fk.S=fK S cos 180 , = -fkS
work done by spring force
magnitude of spring force=-kx
Fs
s
Fext
xf
xi
dx
x
xf
Ws= Fs.ds =- kxdx
=- 1 K xf2-xi2
2
xi
Fext
P
E
1
m
If particle displaced from
equillibrium position force
acting on it tries to displace
further away from equillibrium
position
If particle is slightly
displaced from equillibrium,
then it doesn,t experience
a force or continues in
equillibrium
Potential energy is
minimum
Potential energy is
maximum
Potential energy is
constant
F= -dU =0
dx
d 2U
=positive
dx2
F= -dU =0
dx
F= -dU =0
dx
d 2U
=negative
dx2
d2U
=0
dx2
P
m
P constant,& E
P
- Defined only for conservative force
- Energy possessd by a body by
virtue of its position
- Can either be positive, negative or
zero according to point of reference
- Body always move from higher
potential to lower potential
Identifying forces with potential
energy
1) Attractive force:On increasing x, if U increases
Capacity of doing work
Scalar quantity
Dimension ML2T-2
work done by static friction
Fs
WORK
ENERGY&
POWER
ENERGY
=mgh
W3=mgh1+0+mgh2+0+mgh3+0+mgh4
W= dW
xi
h
A
W1=mgh = mgh
W2=mgxl sin =mgxl
E
Work done in pulling, the hanging
portion on the table W= MgL
2n2
h4
h2
p2
m constant, E
NUETRAL
POTENTIAL ENERGY
Note :
Work done for a complete cycle
by a conservative force is zero
l
V
Mass of chain
(for non conservative force)
h3
v2
UNSTABLE
If particle displaced from
equillibrium position force
acting will try to bring
back to the initial position
1eV=1.6 10-19Joules
1kWh=3.6 106Joules
1calorie= 4.18Joules
1 Joule=107erg
Kinetic Energy
Energy possessed by virtue of
motion Expression
K.E= 1 mv2
2
Always positive
Depends on frame of reference
Work Energy Theorm
Change in kinetic Energy
mv2
Δ K.E= 1
2
Change in kinetic energy of a body
is equal to work done on the body


K2-K1=- F.dr
dU
=positive
dx
(BC portion of graph)
dU
=negative
dx
(AB portion of graph)
3) Zero force:On increasing x, if U doesnot change
dU
=0
dx
C
A
B
B,C points on the graph
For an isolated system for a body in
presence of conservative forces,the
sum of kinetic and potential energies at
any point remains constant throughout
the motion
K.E+P.E=constant
POWER
2) Repulsive force:On increasing x, if U decreases
U(x)
CONSERVATION OF ENERGY
D
x
Types of Potential Energy
-Elastic Potential Energy
-Electric Potential Energy
-Gravitational Potential Energy
Types of equilibrium
If net force acting on a particle
is zero it is said to be in equillibrium
Rate at which body can do work
W
Average power(Pav)= Δ t
Δ
W
Instanteneous power(Pinst)= dd t
 

F.ds = F.v
=
dt
Relation between units:
1 watt=1joule/sec =107 erg/sec
1 HP=746watt,1MW =106 watt
1 KW=103 watt
If work done by two bodies
1
is same then power
time
Unit of power multiplied by time
always give work
1 KWh=3.6 106 Joules
Slope of work-time curve gives
instanteneous power
+
xf
Part of length hanging
(Path 2)=
Fydy+ Fzdz
Graphical representation of
work done
m constant, E
+
dW=F.ds
Total length
th
+
WORK DONE BY VARIABLE
FORCE
E
E
P constant,& E
2) Negative work (90o< <180o
F
L/n
Variation of graph of kinetic Energy
1
A
Fcos
STABLE
Linear momentum:- P= 2m E
Conservative: work done doesnot
depend on path followed
Non-conservative: work depends
on the path followed
2
F
RELATION OF KINETIC
ENERGY WITH OTHER Quantities
E
P.E=max
P.E=max
K.E=min
K.E=min
P.E=min
P.E
K.E=max
K.E
work
)
θ
tan θ = d W
dt
=Pinst
time
Area under power time graph
gives work done
P= d W
dt
Power
=> dW=Pdt
W= Pdt
time
W=Area under P-t graph
Position and velocity in
terms of power:-
[
1/2
Line of impact
Line passing through common normal to surfaces
in contact during impact
1/2
[
1) Velocity,V=[2Pt
m
2) Position,S= [ 8P
9m
t3/2
B
Power delivered by an elevator
Line of motion
of ball B
a=0, T=(M+m) g
  
Power of a water drawing pump
[
V2
Power,P= d W = d m gh+ 2
dt
dt
h=height of water level
d m => mass flow rate of pump
dt
[
velocity of the water outlet
Power required to just lift
water, V=0
dm
P=gh
dt
Efficiency of pump
(
(
Oblique collision
m1
Line of impact
Output Power
μ=
Input Power
Line of motion of ball A
Coefficient of restitution (e)
e =
=
Velocity of separation along the line of impact
Velocity of approach along the line of impact
U1
b
m2
V1=U1
V2=U2
Oblique
Perfectly
inelastic
Head on/one
dimensional
Perfectly elastic collision
K.E before and after collision is same
Inelastic collision
WORK
ENERGY&
POWER
K.E after collision is not equal to K.E before
collision then it is said to be inelastic collision
Head on collision / One dimensional collision
m1
U1
m2
U2
V2
V1
m2
After collision
U2
=
m1-m2
m1+m2
m2-m1
m1+m2
+
+
m1+m2
(1+e) m1U1
V2 =
m1+m2
1-e
1+e
+
Loss in kinetic energy
+
(m1-em2)U1
m1+m2
(m2-em1)U2
K=
m1+m2
m1m2
1
2
m1+m2
(1-e2) (U1-U2)2
First height of rebound
h1 = e2ho
ho
V1
m1
m2
m2
m1+m2
2m1U1
m1+m2
Special cases:
1) Projectile and target having same mass m1=m2
v1=u2,v2=u1, the velocities get interchanged
2) If massive projectile collide with a light target
ie m1>>>m2 v1=u1,v2=-u2+2u1
3) If the light projectile collides with a very heavy
target, m1<<m2 v1=-u1+2u2,v2=u2
h1
ro
V2
2m2U2
h2
V1
to
2t1
V2
h3
2t2
h4
V3
2t3
V4
t
2t4
Total distance travelled by the ball before it stops bouncing
H =ho
1+e2
1-e2
Total time taken by the by to stop bouncing
T =
Perfectly inelastic collision
1+e
1-e
2ho
g
colliding bodies are moving in the same direction
m1
U1
m2
Loss in kinetic energy
V
U2
k=
1
2
m1m2
m1+m2
V =
m1U1+m2U2
m1+m2
(U1-U2)2
Colliding bodies are moving in the opposite direction
V =
m1U1-m2U2
m1+m2
Change in kinetic enargy K= 1
2
m1m2
m1+m2
(U1+U2)2
Energy transfer from projectile to target
1) Fractional decrease in kinetic energy
(If target is at rest)
4m1m2
K
=
K
(m1-m2)2+4m1m2
V2
Impact porameter b=0
(1+e) m2U2
Rebounding of ball
Greater the difference in masses, less will be
transfer of K.E and vice versa
Before collision
m1
V1
Velocity after collision
Based on direction of
colliding bodies
Relative velocity of approach
Ratio of velocities
After collision
m1
Classification
Relative velocity of seperation
U2
U1
Conditions
1.For elastic collision: e=1
2. For inelastic collision: e<1
3. For perfectly elastic collision: e=0
Inelastic
U1-U2
Perfectly elastic Head on collision
Relative velocity before collision along the line of impact
Based on conservation of
kinetic energy
V2-V1
Velocity after collision V1 =
V2
- Particle collision is glancing
- Direction of motion after collision are not
along initial line of motion
- If they collide in same plane , collision is 2
dimensional otherwise 3 dimensional
- Impact parameter 0<b<(r1+r2) r1,r2 are radii
of colliding bodies
Relative velocity after collision along the line of impact
Perfectly
elastic
e =
m1
m2
A
(M+m)g
Inelastic collision
V1
Before collision
P=T.V
V
=TV
Power, P=(M+m)gV
V
COLLISION
Event in which impulsive force acts between two
or more bodies which result in change of their
velocities.
If m2=nm1
K = 4n
K
(1+n)2
Collision in two dimension
If the initial velocities of two colliding bodies are not along the line
of impact, then the collision is said to be oblique collision or collision
in two dimension.
F
m1
F
m2
r
F =
Gm1m2
r2
6.67x10-11 Nm2Kg-2 (SI or MKS)
6.67x10-8 dyne cm2g-2 (CGS)
r1
v1
T2 = 4
r
Time period
or
T2
Dimensional formula [G]
M-1L3T-2
IMPORTANT POINTS ABOUT
GRAVITATIONAL FORCE
1. Gravitational force
* Always attractive in nature
* Independent of the nature of medium
between masses
* Independent of presence or absence
of other bodies
2. Are central forces, acts along the centre
of gravity of two bodies.
3. Conservative force
4. Force between any two masses Gravitational force
Force between earth and any other body Force of gravity
F21 force on 2 due to 1
y
m1
r1
r12 vector from m1 to m2
m2
G m1m2
F21 =
r2
x
r122
G m1m2
or -
r123
Similarly
V
F21 =
r122
r12
or
Clearly
Newtons third law F21
G m1m2
r123
G(m1+m2)
V1 =
l
l
R
c
r
3
l
V
GM
l
=
M
R 3
Fractional decrease =
GM
(2 2+1)
2 2 l
1
V =
2
l
R
T
GM
R
l
2
3
Here R =
GRAVITY
Acceleration due to gravity
GMe
On the surface of earth g =
m
Re2
M - mass of earth
Re
R - Radius of earth
M
Re
Gravitational force is a two body interaction.
Force between two particles does not depend
on the presence or absence of other particles.
The principle of superposition is valid here.
“Force on a particle due to a no. of particles
is the resultant of forces due to individual
particles.”
l
2
g = g [ 1-
d
R
“If density is mentioned use the
above equation”
gl = g -
= 0o
2
e
r
r = R
gr
R
1
r2
1
1
r1
r2
GMm
W = E 2 - E1 =
2
WORK DONE IN MOVING OBJECT
FROM SURFACE OF EARTH TO
HEIGHT h ABOVE SURFACE
m
Orbit at a height ‘h‛ from the surface
Vo =
Re
M
GM
(R+H)
GM
R
=
=
If orbit is closer to
earth‛s surface( neglect ‘h‛)
R
gR2
(R+H)
gR
Vo =
or
= 8 km/s or
GM
R
GM
R
GM
R
=
gR
= 8 km/s = 8x 103 m/s
KE, PE OR TE FOR AN ORBITING
SATELLITE
2r
GMm
r
GMm
TE =2r
U =-
W = (U f - Ui
Or,
W = mgh
=
-GMm
R
-
-GMm
R+h
R
R+h
Work done to move object
to a height h = R
W =
Work done to move object
to a height h = R/2
= 64 x 106
GMm
R
R
Note - for easy calculations
gR
h
m
(called minimum orbit, velocity-first cosmic velocity)
KE =
gl
CONCEPT - WORK DONE BY
EXTERNAL AGENT = CHANGE IN
MECHANICAL ENERGY
ORBITAL VELOCITY
m,Vo
gl
r2
R
l
W =
mgR
2
mgR
3
WORK DONE IN MOVING OBJECT
FROM SURFACE TO CIRCULAR ORBIT
h
r
Vo
m
Re
h
M
Relation KE, U & TE
U = 2 x T.E
K.E = - T.E
r
for r < R, gl =
R
Graph
gR2
r2
g IN TERMS OF DENSITY OF EARTH
Re
gl = g
When a body of mass m is moved from equator to
the poles, weight increases by an amount
2
R
m (g - g ) = m
r
GRAVITATION
GM
r2
R = 0.034
o
Vo =
g
V02 =
r1
The effect of rotational motion of the earth on
the value of g at the equator is maximum.
]
dg
g
l
= g-g =
R
g
l
g-g
d
g
=
=
Fractional decrease =
R
g
g
d
g
Percentage decrease =
x100
x100 =
R
g
Very imp graph
The graphical representation of change in
the value of g‛ with height and depth
for r < R, gl =
g
For equator
GM
r1
V01 =
R cos2
2
There is no effect of rotational motion of the
earth on the value of g at poles.
Re
[Put GMe = g Re2 to solve
problems easily]
g=4 G Re
= 90
For poles
h
l
gl
2
WORK DONE IN MOVING OBJECT
FROM ONE ORBIT TO ANOTHER
m,Vo
Absolute decrease =
(2 2+1)
Note = value of
p
g
g-gl
2h x 100
=
x 100 =
g
g
R
g
r
M
FOUR EQUAL MASSES UNDER MUTUAL
GRAVITATIONAL FORCE
l
g
=
g
Variation due to depth ‘d‛
R
V =
2hg
R
g-gl
2h
=
R
g
g = g - gl =
Percentage decrease =
d
r12
= - F12
h<<<<<<R (h < 100 km)
2h
use, gl= g 1R
Note the point
If h<<<<R, then decrease in the value
of g with height
Absolute decrease =
Latitude Angle which the line
joining the point to
the centre of earth
makes with the
equatorial plane
gl = g -
Approximate equation
GM
Time period
T2
l3
V
M
l
r12
Re
Fc
mgl
mg
General Equation
M
M
-r12
F12 force on 1 due to 2
G m1m2
r
GM
(R + h)2
gR2
g =
(R + h)2
g
r
P
r
g =
3
THREE MASSES(EQUAL) REVOLVING
UNDER MUTUAL GRAVITATIONAL FORCE
l
VECTOR FORM
2
g
h
(m1+m2) r
r2
COM
m
Gm12
V2 =
Value of G
Variation due to height ‘h‛
(m1+m2) r
v2
G - Universal gravitational constant
r12
Gm22
V1 =
Variation of g due to rotation of earth
VARIATION IN THE VALUE OF
ACCELERATION DUE TO GRAVITY
(
ROTATION OF 2 MASSES UNDER MUTUAL
GRAVITATIONAL FORCE OF ATTRACTION
NEWTON‛S LAW OF
GRAVITATION
W = Ef - Ei
energy
W = Etotal - Ui
R
W =
-GMm
2(R+h)
+
GMm
R
KE
TE
distance
U
PHYSICS
WALLAH
LONGITUDNAL STRESS
m
Tensile stress causes increase in
length
M
Tensile stress
m
a
T1
T.S1 =
T2
2m
F
Tensile stress =
A
T.S2 =
T3
unit N/m2
3m
T.S3 =
T1
A
T2
A
T3
A
=
=
=
Longitudnal stress= Δ L
L
Change in length
Original length
MECHANICS
OF SOLIDS
Same as pressure
Causes change in volume
m(g+a)
A
T.S =
m
Fm
(M+m)/A
T.S =
F
01
VOLUME STRESS
PHYSICS
WALLAH
volume stress =
F
A
= pressure
STRAIN
F = normal force/thrust
6 mg
02
Change in volume
Volumetric strain= ΔV
Original volume
V
SHEARING STRESS
A
Causes change in shape
5 mg
PHYSICS
WALLAH
shearing stress =
A
3 mg
Ft
Ft
A
Ft = tangential force
A
03
Shearing strain= Φ =
Δx
L
RATIO OF DENSITY OF BODY TO
THAT OF LIQUID IN WHICH BODY
IS IMMERSED
BREAKING STRESS
HOOKE‛ S LAW
E=
STRESS STRAIN CURVE
A1
Stress
Stress
al
tion
(E= modulus of elasticity)
or
Prop
1. For rigid body E = infinity
2. Steel is more elastic than rubber
3. Depends on :(a) Nature of metal
Young‛s modulus
(b) Temperature
Longitudnal strain
E
Fluid region
FL
Ax
F=
AYx
=kx
L
ok
E.PE =
=
POISSON‛S RATIO
=
Lateral strain
D
l
Longitudinal strain
ΔD
r
- Δr
- D
(-dr/r)
=
=
=
Δl
Δl
( dLL (
l
l
=
F
1
2
1
2
1
2
kx2 =
YA
L
x
F
2k
2
x
=
1
2
D - ΔD
INCREASE IN LENGTH DUE
TO IT‛S OWN WEIGHT
< 0.5
< 0.5
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=
1
2
x
F = kx
YA
Mg =
x
L
2
mgL
x=
2YA
Fx
2
Stress
Strain
x volume x strain
Δl
Theoretical value :- -1 <
Practical value :- 0 <
E - Fracture occur
F
Stress
x volume x strain
2
F2
A2
P.E =
=
2 k
YA
2
L
2
B.S
Lmax =
ρg
RATIO OF EXTENSION
=
1
M2 g2 L
4
YA
ρ
lb
F1
k1
F2
k2
where l=
=
Y1 x A1
=
ρ
Lmax
=
la-lb
la
=
la
la-lb
BULK MODULUS
P
Bulk modulus,B = ΔΔV
K= 1 =compressibility
B
2mg
Bisothermal = P
3mg x L1
Badibatic
-
2
Y2 x A2
L1
Y1
D1
y =
d =
L2
Y2
D2
Fl
η= AF0 = Ax
V
= P
MODULUS OF RIGIDITY
3l
2mg x L2 2yd
b
a
ρ
3mg
2m
x1
=
x2
F
M2 g2
B.S A = Vx ρxg = A Lmax ρ g
mg
2
( =Kl
(1- ( = l
(B.S)1 = (B.S)2
m
L/2
ρ
B.S A T= mg
Plastic reigion is large for
ductile materials and smaller
for brittle materials
beyond D - added strain is produced even
for a small applied force.
x
=
BREAKING OF WIRE UNDER IT‛S OWN WEIGHT
D - Ultimate stress point
ELASTIC POTENTIAL ENERGY
AY
L
A1
B to D - Body doesn‛t regain it‛s original
dimension. Beyond B is plastic
region.
Doesn‛t obey
Hooke‛s Law
FL
AΔ L
k=
F1
B - Yield point
Elastomer
Eg:-Rubber
W(1-
F
AB - Not proportional but body regains
it‛s original shape and size when
load is removed
Comparing with a spring of force constant K
Y=
OA - Hooke‛s law obeyed
W=Kla
L2
A2
A - Proportional limit
Yield point
Modulus of rigidity
=
Fracture point
risid body
YOUNG‛S MODULUS
Longitudnal stress
A BC
D
E
Bulk Modulus
Y=
c
sti
Pla gion
i
re
Ho
Strain
e‛s
La
w
tan = modulus of elasticity
limit
A
L1
F
gth
Ultimate Tensile stren
Strain
4. Independent of dimensions
B.F = breaking stress x area, B.F
Liquid of density ρ
- Density (ρ)=
Pressure
mass (m)
volume (v)
Bubble rising up
at constant temperature
Normal force or thrust exerted by liquid
at rest per unit area
m
ρ v
P=
- unit= kg/m
- density of water=1000 kg / m3=1g/cc
ρgh = patm [n3-1]
3
- for same mass
ρ1v1=ρ2v2
Pressure depth relation
P= h ρ g
Patm
R= n r
h1
h1d1 = h2d2
h2
h1
h1d1 = h2d2 + h3d3
L1
d1
L2
d2
Conditions for equal forces on wall
and bottom in a cylinder
Hydrostatic paradox
if ρ1>ρ2
v1<v2
Patm
radius ‘r‛ becomes ‘R‛ when bubble rises
in liquid from bottom to the surface
F
A
4)
1) U-Tube manometer
Whatever the shape or width the pressure
at any particular depth is same
h2
h3
L3 d3
2) U-Tube type
PHYSICS
WALLAH
5)
h1d1 = h2 d2
Mixing of liquid
A
B
C
f2
D
PA=PB=PC=PD
Calculation of resultant/final density
1) Volumes are equal
d=
- Gauge pressure = P-Patm
= hρg
2) Masses are equal
2d1d2
3d1d2d3
Inclined barometer
d1d2+d2d3+d1d3
Total volume
m1+m2
= v +v
1
2
m =ρ1v1 & m =ρ2v2
2
1
v1 =
m1
ρ1
& v2 = ρ
2
1) Relative density of a body
ds
wa
=
(R.D)s=
dω wa-ww
2) Relative density of liquid
dL
(R.D)L=
dω
wa - w L
=
wa - w w
3) Relative density of a solid
to that of liquid
(R.D)L
=
wa
wa-wL
if θ=angle with horizontal
Sinθ =
m2
Relative density (R.D)/Specific gravity
(R.D)S
h2
L1
d2
h2
h1
Ll =
sp. gravity = (R.D) =
L2
Ll =
h
Ll
(hl)
h
(R.D) =
dL
dw
h
sinθ
L
h
Cosθ
L1
h1,d1
=
h3
hL
hw
L3 d3
hL
6)
L2
h
θ
Ll
ρm
h2,d2
L2
Special case: U - tube rotating
h
h2
L
tan
L
h2 d2
d3 h3
h = h1 - h2
h1
h1d1 + h3d3 = h2d2
L3
h
Px=Po+ hρmg
hρmg = Px- Po
L1
d1 h1
L3.
h
FLUID
MECHANICS 01
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h1d1 = h2d2 + h3d3
hw
U - tube accelerating
l
h2
L2
d2
h1d1 = h2d2
Px
Ll
h
h1 = h2 + h3
L1
d1
If one of the liquid H2O,
then, F1=F2
Patm
θ
if θ=angle with vertical
Cosθ =
h1
Note:
Manometer
d1 + d2
Total mass
d2
h1
L=R
=
3) The third liquid is in level with other
- Patm= 1.01325 x 105 Pa
3) Masses and volumes are different
d=
d1
R
2
For 3- liquid⟹ d=
f1
If
d1 + d2
For 2- liquid⟹ d=
d1
L
=
w
a
h =
g
=
h
L
=
w2l2
2g
a
g
l
h2
h1
x1
PHYSICS
WALLAH
W
x2
h1 =
w2x12
2g
h2 =
w2x22
2g
h = h2 - h1 =
w2 (x22 - x12)
2g
Archimedes principle
If gravity effect is neglected, the pressure at
every point of liquid in static equilibrium is same
Unit of Coefficient of viscosity
upthrust=weight of the liquid displaced= Vρ g
Apparent weight=Actual weight-upthrust
1) The CGS Unit of η is dyne s cm-2 and is called poise.
2) The SI unit of η is Nsm-2 or decapoise or poiseuille
Wapp=Wair-U
Fc
FB.Cos θ
Ab
Ac
FB
=Wair [1- _
ρ
θ
1 poiseuille = 10 poise
net force acting upward=V+ ρL+ g
)
)
θ
[
A
PHYSICS
WALLAH
Pascals Law
Poiseuille‛s formula
πPr4
Q= _
8 ηl
PHYSICS
WALLAH
FB.Sin θ
Law of floatation
W
AA
B
FA
C
W
Stoke’s law
F=6 ηπrv
W
Fnet=Apparent weight-viscous force
U
U
Application
U
Hydraulic Lift
(A) W>U
F
A
=
F
a
If the cylinders are connected
F1
R12
F1
R1
2
F1
D12
F1
D1
=
=
=
F2
R22
F2
R2
2
F2
D22
F2
D2
FLUID
MECHANICS 02
F >> f
f
a
A
ρb= ρL
ρb > ρL
As A>> a therefore
F
(B) W=U
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Terminal velocity
(C) W=U
2 r2 (ρ- ) g
Vt= _
9η
1) If ρ > ,the body will attain terminal velocity in the downward direction.
ρb < ρL
2) If ρ < the terminal velocity will be negative and the body will move in the
upward direction.
ρ
3) = ,the body remain suspended in the fluid.
W ⇒ Weight
U ⇒ Upthrust
Critical velocity
Fractional submerged volume
Displaced
volume(Vd) _
_
= ρ (submerged fraction)
Total volume(V)
Ve
Significances of Reynold number.
Exposed volume(Vl)
_ = 1- _ (Exposed fraction)
ρ
Total volume(V)
If Re lies between 0 and 2000 the folow is stream lined or laminar.
Vd
ρb
wa
weight of solid in air
_
_
Relative density of a solid= _ = w -w = ρW
a
w
Loss of weight in water
-wL _
ρL
Loss of weight in liquid w
_
a
Relative density of a liquid= _ = w -w = ρW
a
w
Loss of weight in water
Reynold number
ρvD
Re = _
η
If Re>3000,the liquid in turbulent.
σ
If Re lies between 2000 & 3000 the flow of liquid is unstable.It may
change from laminar to turbulent and vice versa.
P
Velocity gradient
dv
Velocity gradient= _
dx
F
d
dv
_
>F=- ηA _
A dv =
dx
x
_
F/A
F/A
_
F =
F/A
η= _
>coefficent of viscosity =
> η= _
dv/dx= v/l =d
Adv/dx
(x/l)
dt
shearing stress
=
>η= _
strain rate
Equation of continuity
V1A1 Δ t ρ1= V2A2 Δ t ρ2
v2
A2
v 2dt
since the liquid is incompressable ρ1= ρ2
V1A1=V2A2
Av=constant.
dv
=Q=
Av=
>Volume rate of flow
dt
B
A
A2
v1dt
v1
Energy of fluid in a study flow
SURFACE TENSION
VENTURIMETER
Device to measure the flow of speed of incompressible fluid
Surface tension T =
BERNOULLI’S PRINCIPLE
1
kinetic Energy =
mv2
2
v1 =
P₁V₁ - P₂V₂ = 1 m (v₂²-v₁²) +mg (h₂-h₁)
2
kinetic energy per unit mass = 1 v2
2
2hg
(A12/A22) -1
Unit in SI system =
1
P +
2
Potential energy per unit mass = gh
Potential energy per unit volume = ρgh
ρv + ρgh = constant
P
v
ρg + 2g + h = constant
2
Pressure energy = PV
P = pressure head
ρg
Pressure energy per unit mass = ρP
v2
2g = velocity head
Pressure energy per unit volume = P
h = Gravitational head
D
Soap film
T
Pi - Po =
Then
Pi - Po =
P
h
v= 2gh
H
B
C
Pa
T
p
T
x
p<p1
Pressure on concave side> pressure on convex side
But F = 2TL
2T
R
Pconcave - Pconvex =
FLUID 03
MECHANICS
4T
R
p>p1
p=p1
Pinside - Poutside =
2T
R
Pinside - Poutside =
4T
R
One surface
[ Liquid drop or air bubble ]
[ Soup bubble ]
Two surfaces
Capillarity
Meniscus
< 90
Meniscus
o
h<0
Water
)
R
Fa =
SL
SV
Glass
2(H-h)/g
LV
vx = Horizontal component of velocity
R is max. when h= H
2
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Mercury
> 90
o
Consider the equilibrium of at line of contact
O
O
l
Range R = vx xt
T
p
A
B1
Shape of liquid meniscus
Fa
R = 2 h (H-h)
A
T
F
Work done W = F x x
2T
R
SV
= 2gh x
A
p
h>0
(H-h)
Time of fall, t = 2(H-h)
g
P1
P1
T
Excess pressure inside a soap bubble
If tank is open, P = Pa
m
F
Excess pressure inside a liquid drop
Pa
R
F
l
= T (2LX)=T∆A
v= 2(P-Pa) + 2gh
ρ
Q
S
P1
A1
A
Energy of the additional surface = W = 2TLx
Torricelli‛s Law of Efflux
N
B
C
P
PRESSURE DIFFERENCE ACROSS A CURVED LIQUID SURFACE
W = 2TL x x
APPLICATIONS OF BERNOULLI’S PRINCIPLE
l
PHYSICS
WALLAH
SURFACE ENERGY
2
F
=
A
PHYSICS
WALLAH
Potential Energy = mgh
Length
Unit in CGS system = dyne / cm
(P₁ - P₂) V = 1 m (v22 - v1²) + mg (h₂-h₁)
2
mg
1 m
(P₁ - P₂) V =
(v22 - v1²) +
(h -h )
V ₂ ₁
2 V
kinetic energy per unit volume = 1 ρv2
2
Force
cos
LV
=
SL
=
Ascent formula:
sin
+
SV
LV
-
cos
SL
h =
2T
Rρg
LV
Water
= Angle of contact.
h =
2Tcos
rρg
h > 0
)
< 90
h < 0
)
> 90
o
)
o
)
(1) Δ U =ΔQv =nC ΔT=n f RΔT
v
2
ΔU
(PV)
= nγRΔT= Δγ
-1
-1
P V -P V
ΔQv
= f γf i i
-1
(Internal Energy is a function of temperature)
Work done from P-V Graph
Compression
Adiabatic
p
v
Area under P-V diagram gives
work done by the gas
First law of T.D=
>ΔQ=ΔU+ΔW
ΔQ,ΔW
=
>path functions
Δ U=
>state function
1
state A
state B
Δ U1=Δ U2=UB-UA
P1
p
A
D
>
D
>
ΔU=UA-UA=0
P2
P1
ΔW=PΔV=0
C
V2
V1
(P -P )(V -V )
W= 1
2 2 1 2 1
p
W= (P2-P1)(V2-V1)
Work done:
path function
V
ΔW=F.dx= Pdv
1cal=4.2 Joule
p A
W
System
-ve
(compressive
workdone on the gas)
+ve
(expansion)
workdone
by the gas
Q
System
p
A
-ve
p
A
V
Expansion
Adiabatic process
B
W2
Q
Heat
Work
V
B
2
V
Isobaric
Heat: path function
CP
ΔQ
=
> unit:calorie/Joule
Cv
Adiabatic
Adiabatic process =
>Δ Q=0 [No heat transfer]
At constant volume =
>ΔQV=Δ U=nCV ΔT
Isothermal
p
Isochoric
V
01
W1
Adiabatic
Wisobaric>Wisothermal>Wadiabatic>WIsochoric
At constant presence =
>ΔQp= Δ U+ΔW=nCPΔT
W1=
/ W2
ΔQ=0 [no exchange of heat]
W1 > W2
Rapid or spontaneous
process/insulated vessel
FLOT=
> Δ Q= Δ U+ΔW
Δ Q= 0
Δ U=-ΔW
Compression
ΔW= - ve ΔU= + ve
>Temperature
ΔU =
=
> Pressure
Expansion
ΔW=-ve Δ U=+ve
>Temperature
ΔU =
=
>Pressure
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High T
02
Isothermal process
Source
T1
Q1
equation of states =
> P1V1=P2V2
Workdone by the gas
V
W=2.303 nRT log 2
V1
( (
P
W=2.303 nRT log ( (
P
1
Slope =
>
Slope of adiabatic process
2
=
heat released
Sink
(surrounding)
efficiency(η)
Q -Q
Q
η = W = 1 2 =1- 2
Q1
Q1
Q1
η=
β=
T1-T2
T
=1- 2
T1
T1
Relationship between η& β
η
β = 1η
Cascaded engine
T1
03
D
E1
W2
T= T T
1 3
A
Isothermal
expansion
T
V1
= 1
T2
V2
specific heat of gas
f
R
=
>Cp= 1+ 2 R= -1
T1+T2
2
T2=T1T3
T -T
η= 1 2
T1
ΔP=0
FLOT=
> Δ Q= Δ U+ΔW
T=
T
1- T =1- 3
T
T1
Ideal engine
Isobaric process
ΔW=PΔ V= P(V2-V1) = nR(T2-T1)
W1
2T=T1+T3
Same efficiency
T
Carnot Engine
Workdone by the gas
W1=W2
T1-T=T+T3
(not possible)
C=
equation of states =
>V T
Same work output
ηmax=
>Where Q2=0 or T2=0K
slope of isothermal process
specific heat
T2
T1-T2
W=Q1-Q2
Q2
Low T
T2 (Low T)
Coefficient of performance (β)
Q2
Q
β= 2 =
W Q1-Q2
heat absorbed
E
T2
Sink
+
W
+ve
Thermodynamic process
B
1
‘Devices that convert
heat into work‛
W
Q2
Heat Engine
FLOT=
> Δ Q= Δ U+ΔW
Δ Q=ΔW
V2 V
E
f
R
specific heat of gas =
>CV= 2 R= -1
PiVi-PfVf
-1
T1 (High T)
Q1
ΔW=0
=> ΔT=0 =>ΔU=0
eg:- perfectly conducting
slow process
V1
Sign Convension
V2
P
B
>
C
V2
T P1 = T1
P2
T2
S
C
I
S
Y
H
P
H
A
L
L
A
W
P1
W= π (P2-P1)(V2-V1)
4
Isochoric process
>
B
V1
Δ U=0
ΔV=0
Refrigerator
Source
Workdone by the gas
Slope of adiabatic process
= slope of isothermal process
specific heat of gas =
>C=0
Q
Δ
C=0
C= Δt
ΔQ =0
P2
P2
V1
A
[
-1
ΔW=-ΔU=nCv(T1-T2)
=n f R(T1-Tf)
2
nR
ΔW = -1 (T1-Tf)
V
ΔW=P(V2-V1)
A
[
Isochoric process
ΔV=0 or V=constant
equation of states =
>P
Workdone by the gas
Isobaric
Isobaric process
2
Cyclic process
P T
ΔW=
W1=
/ W2
Q1 =
/ Q2
Isothermal process =
>ΔT=0
Isochoric
Isothermal
p
04
Equation of state
PV =a constant
-1
TV =a constant
Pressure
Internal Energy(U)
T3
B
Adiabatic
expansion
D
Adiabatic
compression
C
Isothermal compression
Volume
THERMODYNAMICS
BIMETALLIC STRIP
TEMPERATURE SCALE
C F-32 K-273
=
=
5
9
5
Result
Cu
(celcius-fahrenheitkelvin conversion)
any scale conversion formula
Reading on any scale - lower fixed point
=
Upper fixed point - lower fixed point
>
cu
= constant
l
1. Δ.l= l Δ θ
So when temperature increases
Δl of Cu > Δl of Fe
strip with higher value of
will be on
convex side
Fe
2
l2‛,
1
Δl
2
APPLICATIONS OF LINEAR EXPANSION
Pendulum clock
Fact
When temperature increases, time period
increases, clock runs slow
When temperature decreases, time period
decreases, clock runs fast
1) Loss of time in any given time interval t,
Δt= 1 Δθ t
2
2) Time lost by clock in a day
Δt= 1 Δθ t = 1 Δ θ 86400=43200 Δ θ
2
2
CUBICAL EXPANSION/VOLUME EXPANSION
1. ΔV=V Δ θ
2. V =V(1+ Δθ)
3. = Δ V
unit
VΔθ
k, dimension-[K-1]
4.
/ 0c or /K
Variation of density with temperature
1
Density
Volume
V = V(1+ Δθ)
then
Δθ
l
Thermal Stress=Y Δ θ
Thermal Force=YA
Δθ
Result
(1) At θ‘>θ True value> Scale reading
At θ‘<θ True value< Scale reading
True value= Scale reading (1+ Δ θ)
HEAT CAPACITY
Heat capacity=mass+ specific heat capacity
Unit= cal
=>SI unit J
0
K
C
The mass of water that will absorb
or lose as same quantity of heat as
substance for the same changes in
temperature
msT1 mLf
ms(T2-T1) mLv
= t =
= t
Δ 34
Δ t1
Δ 12
Δ t23
T1
T2
ΔQ = ∫ msdT
T1
PRINCIPLE OF CALORIMETRY
Cu glass
Vessel
Fe
initial
Liquid
Hot iron
phere
m1
m2
m3
s1
s2
s3
T1
T2
T3
Heat lost by the hotter body = Heat gained by colder bodies
Q3=Q1+Q2
REAL AND APPARENT EXPANSION OF LIQUID
1. Δ Vapparent=V0
2. Δ Vapparent=V0 )
apparent
l
-
3. Δ Vapparent=V0 )
l
-3
4. apparent= l -3
s
s
Real expansion of liquid Expansion of solid in which liquid
is contained
Δθ
) Δθ
s
) Δθ
-Real expansion of liquid
l
s
00C water
Δ Q3=m2Swater θ 2
Δ Q3=,>,or< Δ Q1+ Δ Q2
check
Δ Q3< Δ Q1+ Δ Q2
1. Whole ice melts into water
1. Only m‛ g of ice melts
2. Additional heat is used
to increase the temperature
of system from 0 0C
2. Mass of ice melts can be
found by
3. Final temperature can
be found out by
3. Final temperature is 00C
m Lf=Q
Δ Q = MtotalSwaterT
where
ΔQ
additional heat
-coefficent of linear expansion
of solid
Final equilibrium temperature,
Teq=
m1s1T1+m2s2T2+m3s3T3 ∑ msT
= ∑
m1s1+m2s2+m3s3
ms
Facts :
Calorimeter A device in which
the measurement
of heat can be
done.
CONVERSION OF MECHANICAL ENERGY TO HEAT ENERGY
1. Potential energy to Heat energy
Δ U=mgh=
>Δ Q=m Lf
w=water
b=body
mWsW=mbsb
T2
water
2. Apparent change in volume
4. Convert θ20C water
Δ Q3> Δ Q1+ Δ Q2
Cu glass
Vessel
1. ApparentExpansion of liquid
00C water
00C ice
Δ Q2=m1Lf
J
J
cal
Lv=Lsteam=540 g =540 4.2 g =540 4200 kg
Heat supplied at constant rate
Graph & equation
1. Water has maximum density at 4 C (minimum volume)
2. On heating,
0 OC
4OC water contracts
O
Graphs
above water expands
3.
4C
-coefficent of linear
expansion
Δ θ -temperature change
A -Area of rod
3. Convert
J
J
cal
Lf=Lice=80 g =80 4.2 g =80 4200 kg
Sice= 1 cal
=2.1 J0 = 2100 J0
gC
kg C
2 g0C
ANOMALOUS EXPANSION OF WATER
Y-Young‛s Modulus
ERROR IN SCALE READING DUE TO
EXPANSION OR CONTRACTION
J =
J
=
2. Swater= 1 cal
g0C 4.2 g0C 4200 kg0C
S=f(T)
= (1- Δθ)
ll
Δ Q1=m1Sice θ 1
S
C
I
S
Y
H
P
H
A
L
L
A
W
J Kg-1K-1
if speecific heat is variable
O
Thermal Stress in a rigidly fixed rod
Joule
Kg Kelvin
00C ice
WATER EQUIVALENT
: β : = 1:2:3
=3
SIunit-
Boiling
Δθ:mLv
Lv-Latent heat
of vapourisation
Melting
Δθ :mLf
Lf-Latent heat
of fusion
[or use Kelvin instead of 0C]
SUPERFICIAL/AREA EXPANSION
Δθ
l1‛,
2
1. ΔQ=ms Δ T
S-specific heat capacity
1.Δ A=Aβ Δ θ
β -coefficant of area
expansion
2.Al =A(1+βΔ θ)
unit
/ 0c or / k,dimension-[K-1]
3.β = Δ A
A ΔT
4.β =2
Δl
change state of body
2. Convert -θ 0C ice
1
+
l2,
change temperature of body
+
1
1. m1 g ice [-θ,0C] mixed with m2g water [-θ2 0C]
+
l1,
Heat Supplied (ΔQ)
Area of hole increases
body expands on
heating.Expansion of area of body is
independent of shape and size of hole
/ 0c or / k, dimension-K-1
Whatever be the change
in temperature, the
difference in length
remains constant
l1 1=l2
|
Problem solving methodology
+
2. ll = l (1+ Δ θ)
3. = Δ l
unit
lΔ θ
l =l+ Δl
heat
Δθ
1 calorie=4.2J
Fe
EXPANSION OF CAVITY
l
ICE-WATER SYSTEM
Fe
THERMAL EXPANSION
A-LINEAR
CALORIMETRY
Cu
When equaling multiply with 4200 for Δ Q
ie, mgh= m L + 4200
>
)
(if Lf is in calorie
g
f
2. Kinetic energy to Heat energy
K.E= 1 mv2 =
> Δ Q=m Lf
2
calorie
if Lf is in
g
then
1 mv2
=
>m Lf+ 4200
2
HEAT TRANSFER
1. Conduction
Heat flows from hot end to cold end, medium is necessary,
slow process
ΔQ = Rate of flow of heat
Δθ
ΔQ
Δt
=K A
Δt
l
A = Area of cross section
watt
watt
Unit of ‘K‛=
or
Δθ
= Temperature gradient
metre0C
metre K
l
‘K‛ depends on the nature of material
K = coefficient of thermal
conductivity
THERMAL PROPERTIES OF MATTER
Electrical Conduction
1) current, I=
2)
dq
dt
2) H= θ1- θ2 = Δθ
R
R
3) Thermal resistance,
θ1
Rn
θ2
k1
l2
θn-1
kn
R1
ii) Parallel Combination
R 1=
R1
R2
l1
K1A
R2 =
.........
R2
l2
K2A
Find Req=R1+R2+........
θ1
Rn
From that find ‘Keq .
Here (V1-V2) is same
Relation between E & E λ=
>E =
A2
kn
l
=
>
θ1
TEMPERATURE OF INTERMEDIATE JUNCTION
θ1-θj θ2-θj θ3-θj
=0
+
+
R1
R2
R3
l1
θ1
θ j=
l3
k1,A
k ,A
3
θ3
+
θ2
+
θ3
R1 R2 R3
1 + 1+1
R1 R2 R3
θ1
E= σe (T -T0 )
4
4
Watt
m3
E= σ T4
σ
ΔQ = σ AT4
,
Δt
Stefan‛s constant
value of σ
Dimension
ΔQ
Δt
5.67+ 10-8 W/m2K4
[σ[ =MT-3K-4
Radiant power
θ0
Δt
θ2 + θ1
2
(- θ[
0
θ2
θ0
Δt=time
surrounding temperature
θ
0
ΔQ =σ eA (T4-T 4)
0
Δt
WIEN‛S LAW
Wien‛s displacement law
1
T
λ mT =b
λm
Eλ
b=Wien‛s constant
λ m T1=λ m T2
1
2
Area =A=
0
∫E
λ
d λ=E= σ T4
[ [
A
T
Hence 1 = 1
A2
T2
b
[dimension]=[b]=LK
Eλ
T3
T1<T2<T3
T2
λm1>λm2>λm3
T1
λm3 λm2 λm1
λ
NEWTON‛S LAW OF COOLING
-dT
dt
directly proportional to excess
of temperature of the body
over that of surrounding.
(T-T0)
T=Temperature of body
T0=Temperature of surounding
4
λ
value- 2.89+ 10-3 mK
“As the temperature of the body increases, the
wavelength at which the spectral intensity (E) is
maximum shift towards left.”
STEFAN‛S LAW
Emissive power of a black body
fourth power of absolute temperature
[(
θ1> θ2
In the presence of a surrounding
T0) (general body)
Rate of cooling
I 1+ I 2 + I 3 = 0
l2
θj
0
KIRCHHOFF‛S LAW
Ratio of emissive power to absorptive power is same
for all surfaces at the same temperature and is
equal to the emissive power of a perfectly balck
body at that temperature.
E1
E2
E =E
a1 = a2 =.........= A
Here,Temp Difference same
θ1
-[ θ2- θ1 [
=K
Δt
ΔQ =σ A (T4-T 4)
0
Δt
E= σ (T4-T04)
If e=1 , it indicates a perfect black body
Find 1 = 1 + 1 +....
R1
R2
Req
from that find Keq
k2,A
EQUATION FOR PROBLEM SOLVING
In the presence of a surrounding
(T0) (black body)
λm
Eλ d λ
If e=0 , means general body radiates no energy
l
R2 =
K2A2
θ2
∫
value of e =
>0<e<1
l
R1 =
K1A1
R3
Watt
m2
Energy radiated by a general body
e=
Energy radiated by a black body
replace to resistors
R1
R2
unit
EMISSIVITY (E)
An
θ2
Energy radiated
area+ time
Energy radiated
Spectral emissive power(Eλ)= area+ time+ wavelength unit
A1
k2
1 = 1 + 1 +......
Req R1 R2
Emissive power(E)=
Here ‘H‛ is same
θ2
k1
|
e=emissivity
EMISSIVE POWER/INTENSITY OF THERMAL RADIATION
‛
ii) Parallel Combination
ΔQ = eA σ T4
Δt
S
C
I
S
Y
H
P
H
A
L
L
WA
θn
replace with resistors
Here ‘I‛ is same
Wind blows from land to sea
during night
Absorptive, reflective and Transmitted power
Energy absorbed
Q
Q
Qr
Absorptive power(a)= a =
Q
Energy incident
Qa
Energy reflected
Q
Reflective power(r)= r =
Q
Energy incident
Qt
Energy transmitted
Q
Transmitted power(t)= t =
Q
Energy incident
a+r+t=1
ln
=
>
l1
Req=R1+R2+.......+Rn
θ3
k2
Land Breeze
RADIATION
i) Series Combination
i) Series Combination
.........
l
R = KA
5) Combination of conductors
5) Combination of resistors
R2
Sea Breeze
Wind blows from sea to land
during day time
4) H= θ1- θ2 = θ1- θ2 = KA (θ1- θ2(
R
l
(l/KA)
(V -V ) =σ A (V1-V2)
V -V
I= 1 2 = 1 2 A
l
R
l
NEWTON‛S LAW OF COOLING
For ordinary body E= e σ T4
Natural convection takes place due to the effect of gravity
dQ
dt
1) Heat current, H=
I= ΔV ( ΔV= Vhigh-Vlow)
R
R1
Requires a medium, actual movement of fluid,occus naturally or
forced.
Thermal Conduction
3) electrical resistance, R= l
A
4)
CONVECTION
THERMAL PROPERTIES
OF MATTER
OHM‛S LAW OF CONDUCTION
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DEGREES OF FREEDOM
f
(b) at high temperature,
f
=3
=
=
5
7
a)
CP −C V = R
(MSH)
b)
CP − C V = R
(GSH) M
R
R
d) C
f
=1 +=-1
e)
=
P
1 Vibrational mode f = 2
CP
Ideal gas is composed of polyatomic
molecule that
has 4 vibrational modes.
Total degree of freedom is
b) 14
c) 8
vrms =
d) 6
R
CP-CV=
28
b)
R
CP-CV=
14
2
d
=
3kBT
m
I v1 I + I v2 I + ....... + I vnI
n
8RT =
πM
v avg =
n 1+ n 2
CP mix
CV mix
R
c) CP-CV=
7
d) C
=
Q3
Consider a mixture of n moles of helium gas and
2n moles of oxygen gas (molecules taken to be rigid)
as an ideal gas. It‛s Cp/CV value will be:
a) 19/13
-CV= R
P
b) 67/45
8P
πd
vmp =
c) 40/27
d) 23/15
Q5
a)
5
RT
2
b)
3
9
RT c)
RT
2
2
BOYLE’S
LAW
d)
3RT
Q6
The rms speeds of the molecules of Hydrogen,
Oxygen & Carbondioxide at the same temparature
are VH, VO and Vc respectively then:
a) VH >VO>VC
b) VC >VO>VH
c) VH =VO>VC
d) VH =VO=VC
CHARLE’S
LAW
PV
V
b) r2
W = Pdv
U
= 1
QP
c) r
d) r
T
T
. PV = constant, if T =Cosntant
. P 1V1 = P2V2 ,When gas changes it‛ s
state under constant
temperature.
. v 1 = v 2 ,When gas change its state
T1 T2 under constant pressure.
. P α T;
. P1
T1
=
P
T
P2
T2
KINETIC THEORY OF GAS
PRESSURE
OF GAS
PV
1
2
mn Vrms
3
-
mn v 2
IDEAL
GAS
LAW
PV=nRT
R=8.314 JK-1mol-1
P
V
. V α T; v = constant; P = constant.
T
1
2
2 πd n
d = diameter of molecules.
P/T
P
W
= 1- 1
QP
1
d2
1
r2
T
P
n = no. of molecules per
unit volume
GAY
LUSSAC’S
LAW
V/T
V
a) r3
d) 11RT
QP= U + W
U= ncv T
MEAN FREE PATH
The mean free path of molecules of gas,
(radius r) is inversely proportional to
c) 9RT
FIRST LAW OF
THERMODYNAMICS
07
λ mean =
b) 15RT
05
Vmps:Vavg:Vrms = 1 : 1.14 : 1.228
2RT
2P
=
Mo
d
A gas mixture consists of 2 moles of 02 and 4 moles
of Ar at temparature T. Neglecting all vibrational
modes, the total internal energy of the system is
a) 4RT
VELOCITY OF GAS
Average distance travelled by
molecules between two
successive collision
Consider a gas of triatomic molecules. The molecules
are assumed to be triangular made up of massless rigid
rods whose vertices are occupied by atoms.The internal
energy of a mole of the gas at temperature T is:
Q4
06
Speed possessed by maximum
number of molecules of gas.
2
3P
n1cP1 + n2cP2
Most probable
speed:
Arithmatic mean of speed of
molecules of gas at given
temperature.
v avg =
n 1+ n 2
S
C
I
S
Y
PH
H
A
L
L
WA
If CP and Cv denoted the specific heats
of unit mass of nitrogen at constant pressure
and volume respectively, then
Average Speed:
v1 + v2 +............... + vn
n
3RT
=
M
f
1 KT
1 Molecule per f = 2
f
Total for a molecule = -K T
2
3
Monoatomic Molecule = -K T
2
f RT
Total for a mole= 2
3
Monoatomic=-R T
2
(mole)
5
Diatomic= -R T
2
(mole)
3
Translatory Kinetic energy=-R T
2
(mole f= 3)
n1cv1 + n2cv2
CPmix =
mix
2
= 1 +-
a)
Square root of mean of square
of speed of different molecules,
2
Cv
2
Q2
Root Mean
square speed:
vrms =
7
Dia = 5
4
Tri = 3
V
(a) Linear f= 5
(b) Non-linear f= 6
a) 12
5
3
Mono- = -
R
==2
-1
c) C
For triatomic gas,
Q1
CVmix =
f
LAW OF EQUIPARTITION OF ENERGY
MIXING OF GASES
SPECIFIC HEAT CAPACITY
. For monoatomic gas, f
. For diatomic gas,
(a) at room temperature,
03
02
01
= constant; V = constant.
, When gas change its state
under constant Volume.
Relation between pressure
and Kinetic Energy.
E =
3
PV
2
Specific heat of Solid = 3R
WATER = 9R
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Classification of waves
Waves
Necessity of medium
The medium should have elasticity,
inertia and uniform density
Characteristics of Wave
Energy propagation
=
> The particles of the medium are executing
simple harmonic motion.
Progressive
waves
One Dimensional
Transverse
waves
Non-mechanical
waves(EMW)
Stationary
waves
Two Dimensional
Longitudinal
waves
=
> The phase of vibration of the particle keep
on changing.
=
> Wave carries energy and momentum.
A
[
A Amplitude
Oscillating term
phase
wave length
ω Angular frequency
Initial phase
2π or 2π ν
ω =_
Direction
T
Propagation
_
1 μ Vω2A2
dK = _
dU= _
_
1 μ Vω2A2
dt dt
Rate of an power Transmission
Case 2
massive
string
x
l
V= ρπr2
K Wave Constant
2π
K= _
M-Mass of block
m-mass of string
ρ-density of material
of string
Case 3
Time taken to reach the pulse at top
Case 1
l
v
[
[
[
ρ
_
Mg 1V= _
μ
ρ-density of the liquid
-density of the material
μ-Linear
mass density of
the string
M
_
_ρ
C
C
C
amax=w2A
P
I=4_
π r2
If vmax of the particle = n x vof wave
R
R
T
T
Transverse-wave in a rod
2
Vmax=wA
C
Energy
2 2 2
_
I = time
+ Area = 2π ν a ρ ν
a=w y
V=w A2-y2
I a ν2 a2
Longitudinal wave in a rod
Medium should posses the property of
rigidity
Medium should posses the property of
elasticity
Transverse waves can be polarised
Longitudinal waves can not be polarised
wA=n
n= 2πA
x l
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l
2πA= n x l
l= 2πA
n
Sound waves travel through air,vibration
of air column in organ pipes vibration of
air column above the surface of water in
the tube of resonance apparatus.
A= nl
2π
Factors affecting velocity of sound
→ Pressure
mass of string m
velocity at bottom
Mg
V1 = μ
M
velocity at Top
→ Density
• Velocity of sound in air
is independent of pressure
V2
2
→ Temp
Vα 1
V=
V1
V1
V2
V1 _
_
M
= 1 = _
ρ2
ρ1
=
V2
m
=
T1
T2
M+m
g
V2= (M+m)
μ
Velocity of Longitudual Wave
massless
string
[
C
These waves can be transmitted through
solids,liquids and gases because for
these waves propagation,volume
elasticity is necessary.
Intensity of wave
Acceleration
of particle
velocity of particle
Temp Cofficient(α)
Increase in velocity of sound for 1oC
or 1K rise in temperature of gas
r2
_
r1
ρ
T=Mg 1- _
C
C R C R CR
Minimum
Transverse waves can be transmitted
through solids,they can be setup on the
surface of liquids.But they can not be
transmitted through liquids and gases.
Velocity at any Point
Velocity, v= T
μ
Time taken, t=
l
T
μ
E
V= _
ρ
(E=Elasticity of the medium; ρ=density of the medium)
(1) As solids are most elastic while gases least i.e. ES>EL>EG. So
the velocity of sound is maximum in solids and minimum in gases
+
V1 _
T
_
= T1
V
2
μ= ρA
Time period
T
l
T=2 _
g
M
_
T
Frequency
Time taken to reach the pulse at Top
l
T=Mg
m
μ= _
Maximum
pressure and
density
V= gx
Velocity of Transmission wave in a string.
T
V= _
μ
C
Case 4
2
4
1
dx
Velocity of particle=-Velocity wave+ Slope of the graph
It travels in the form of compression
(C) and rarefaction (R).
Movement of string of a sitar or violin,
movement of the membrane of a tabla or
Dholak,movement of kink on a rope,waves
set-up on the surface of water.
WAVES
Rate of Energy Transmission
T
k
S
C
I
S
Y
H
P
H
A
L
L
WA
Important Terms
ν
dt
Longitudinal wave in a fluid
It travels in the form of crests (C) and
troughs (T).
y(x,t)=a sin(ωt +
- kx +
- 0/0)
(
x
(v) y=A sin ω (t- _
ν
v
T
Angular frquency
Particles of the medium.
V
C
[
t -_
x
(iii) y=A sin 2π _
T
2π (ν t-x)
(iv) y=A sin _
Amplitude
2
2
d
1 d
_y = _
_y
2
2
dx ν dt2
Particles of the medium vibrates in the
direction of wave motion.
Transverse wave on a string
The general equation of a plane progressive wave with initial phase is
Displacement
Particles of the medium vibrates in a
direction perpendicular to the direction
of propagation of wave.
dy _
dy
_
=- ω + _
Equation of progressive wave
2π x)
(ii) y=A sin(ωt- _
Longitudinal waves
Three Dimensional
=
> The velocity of the particle is not equal to
velocity of wave.
(i) y=A sin(ωt-kx)
Transverse waves
Vibration of particles
Dimension
Mechanical
waves
can travel in
vacuum
All travelling wave satisfy a differential
equation called wave equation.
Classification of waves based on vibration of particles
According to
wave is a disturbance which propagates
energy and momentum from one place to
another without the transport of medium.
μ
= l
T
T= Tension in the string
μ= Linear mass density
Velocity Sound (air)
P _ Newton
V= _
ρ
P γ _ Laplace
V= _
ρ
Value of α =0.608
=0.61
m/s
C
o
Humidity
Humidity ↑
Speed of sound
↑
Sound travels faster in moist air
than in dry air
C
γ= P
CV
Mono atomic
γ= 5/3
diatomic
γ= 7/5
Relation between △X and
△Φ
△Φ= 2π △X
l
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Wave combination of string
Reflection of Mechanical waves
1) From rarer to denser medium
Echo
2) From rarer to denser medium
Source
Incident wave
Incident wave
Reflection from rigid end/denser medium→ Phase change by π
-x
Rigid boundary
Transverse
wave
Rarer
+x
Free boundary
-x
+x
Observer
d
Source at distance “d” from screen
Denser
t= dv + dv = 2d
v
Reflection from free end/rarer medium→ No phase change
Reflected wave
Transmitted wave
Persistance of hearing for
human ear is o.1 sec
S
C
I
S
Y
H
P
H
A
L
L
A
W
Conditions for echo:
if t > 0.1 ⇒ 2d > 0.1 ⇒ d >
Transmitted wave
v
v
20
Reflected wave
Principle of superposition
The displacement at any time due to number of waves meeting simulatoneously at a point in a medium is the
vector sum of individual displacements due each one of the waves at that point of same time
Superposition
Incident wave y1= a1 sin(ωt-k1x)
Incident wave yi= aisin(ωt-k1x)
Reflected wave yr= ar sin(ωt-k1(-x)+π)
Reflected wave yr= ar sin(ωt-k1(-x)+0)
= -ar sin(ωt+k1x)
= ar sin(ωt+k1x)
Transmitted wave yt= atsin(ωt-k2x)
Beat
• Constructive
• Destructive
Standing wave
Condition:•Two waves of same frequency, same wavelength, same velocity
•Resultant intensities will be different from the sum of intensities
of each wave seperately
y1= a1 sin ωt, y2= a2 sin(ωt + ϕ)
ϕ-Phase difference between two waves
y=y1+y2 ⇒ y= A sin(ωt + θ)
A= a12+a22+2a1a2cosϕ
tanθ=
Intensity
a2sinϕ
a1+a2cosϕ
A2
I = I1 + I2 + 2 I1
I1
I2
=
a1
a2
2
Imax
Imin
=
(a1 + a2)2
(a1 - a2)2
=
( I1 + I 2)2
( I1 - I 2)2
sound waves travelling in same medium with slightly different frequencies
superimpose on each other.
The intensity of resultant sound at particular position rises and falls regularly
with time.
The phenomenon of variation of intensity of sound with time at a particular
position is called beats.
Point to remember:-
i) For Constructive interference:-
ϕ = O, 2
, 4
, ---, 2
1) One beat:-
Maximum intensity
(at t=0)
n when n = 0, 1 , 2, ---
x=O, λ, 2λ, --- , nλ , when n= 0, 1, ---
Imax = I1 + I2 +2 I1I2
Becomes minimum
intensity
2
2
= ( I1 + I2) = (A1+A2)
ii) For Destructive interference:-
when ϕ=π, 3π, 5π, --- (2n-1)π ; where n=1,2,3,---
3
x= _ , _
,
2
2
Imin=I1+I2-2 I1I2
=
> Imin=( I1- I2)2
Again acheives
maximum
_
(2n-1) 2 , where,n=1,2,3,-----
(A1-A2)2
Beat period:-
One beat is formed
Time interval between two sucessive beats (ie.two sucessive maximum of sound)
is called beat period.
2
ϕ
I2 cos ϕ
A2 = a12 + a22 + 2a1a2 cos
r4π2
Beats:-
WAVES
Interference of sound wave
•This is due to the interference of waves
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• In a string
• In an open pipe
• In a closed pipe
Interference
Transmitted wave yt= atsin(ωt-k2x)
Beat frequency:No.of beats produced per second
Beat frequency:-
n= n1- n2
Beat period:- T=
1
= n 1n
Beat frequency
1 - 2
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End correction:1 l - 3l
e= _
1
2
2
(
Let n2 is the unknown frequency of tuning fork B, and this tuning fork B produce x beats per
second with another tuning fork of known frequency n1.
As number of beat/sec is equal to the difference in frequencies of two sources, therefore n2 = n1 ± x
By filing
If B is filed, its frequency increases
?) A Source of unknown frequency produces 4 beat/s when sounded with a
source of Known frequency 250 Hz. The second harmonic of the source
of unknown frequency gives 5 beat/s when Sounded with a source of
frequency 513 Hz. The unknown frequency is?
b) 246 Hz
c) 240 Hz
508
d) 260 Hz
250
246
492
5 be
at
21
t
bea
Standing Waves:
1:2:3
T
l=
n
x
(n+1) → node
l
2
= n
2l
l=
n
x
1
+ (n-1)
(n+1) → antinode
n → node
= n
2l
l=
n
x
l
4
S . NO
Note
Distance between two adjacent node & antinode is
4
WAVES
3
Relation between loudness and intensity
Resonance: The phenomenon of making a body vibrate with it‛s natural frequency under the
influence of another vibrating body with the same frequency is called resonance.
n=1,3,5....
Distance between two adjacent node & antinode is
7 λ
_
l4+e =
Comparative Study of Stretched Strings, Open Organ Pipe and Closed Organ Pipe
y=2a sin(kx) cos(ωt)
4
Unison: If the two frequencies are equal then vibrating bodies are said
to be in unison.
1:2:3
Closed pipe
5
_
l3+e =
(iii) Similarly if n2 = 2nn1, it means n2 is n-octave higher n1 is n octave lower.
n → antinode
l
2
4
1
(ii) If n2 = 23n1 it means n1 is 3-octave higher or n1 is 3-octave lower.
y=2a cos(kx) sin(ωt)
Open pipe
=
2
(i) If n2 = 2n1 it means n2 is an octave higher than n1 or n1 is an octave lower than n2.
y=2a sin(kx) cos(ωt)
2l
th
n
3
Octave: The tone whose frequency is double the fundamental frequency is defined as Octave.
•When two progressive waves (both longitudinal and transverse) having same amplitude, time
period,frequency moving along a straight line in opposite direction a superpose a new wave is
formed. It is called stationary Or standing wave.
= n
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513
Hence unknown frequency is 254 Hz
String
3
_
S
C
I
S
Y
H
P
H
A
L
L
A
W
n..................4
Solution:
254
Phase difference between 2 particle at both sides of node is 180
or
Strain and pressure is maximum at node and minimum at antinode
v
v
Fundamental frequency or
1st harmonic
2
Frequency of
or 2nd harmonic
n2 = 2n1
n2 = 2n1
1 overtone
1 overtone
3
Frequency of
or 3rd harmonic
2nd overtone
n1 =
2l
st
n3 = 3n1
Frequency ratio of
overtones
2:3:4.....
5
Frequency ratio of
harmonics
1:2:3:4.....
6
Nature of waves
Transverse
stationary
4
o
Stretched string Open organ pipe
1
2
4
Parameter
n1 =
2l
st
n3 = 3n1
2nd overtone
2:3:4.....
1:2:3:4.....
Longitudinal
stationary
Closed organ pipe
n1 =
L
I0=10-12W/m2
log10 Intensity
unit W/m2
unit(dB)
I0=Threshold intensity
I
I0
dB=10+ log10
Δ L = change in loudness
Δ I = change in intensity
v
4l
Missing
n3 = 3n1
1st overtone
3:5:7.....
1:3:5:7.....
Longitudinal
stationary
L1=10+ log10
I1
I0
L2=10+ log10
I2
I0
[
( II
L2-L1=10 log
Δ L =10 log
2
0
( II
2
1
( II
-log
(
a) 254 Hz
(
If B is loaded with wax so its frequency decreases
l2+e =
_
l1 +e=
4
l
1
0
I1
L1
I2
L2
[(
By loading
(
Resonance tube experiment
Frequency Increasing
Determination of Unknown Frequency
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Doppler Effect
Case 4
(source is stationary, listener is moving
away from the source)
Whenever there is a relative motion between a source
of sound and the listener, the apparent frequency of sound
heard by the listener is different from the actual
frequency of sound emitted by the source.
VS
S
C
I
S
Y
H
P
H
A
L
L
A
W
VS=0
General equation
(when both source & listener are moving)
(
Case 9
(source is moving away from stationary wall)
Case 5
(source is moving away from the listener,
listener is stationary)
(
Case 2
(The source is stationary & listener is approaching the source)
Case 10
Case 6
(source and listener moving in same direction)
VL=0
l= V+VL
l= _
V VS
x
)θ
VS COS θ
1
1=
V+VL
_
y
z
V+Vs
(
)
θ2
VS
VL
Case 3
(source & listener are approaching each other)
(
Note:
B
VL
l
A
(
V-V
= _L
V
l=
x
)θ
VS COS θ
1
y
)
_L
s
θ1
COS
( V+V
V-V COS
θ2
L
_
s
COS θ1=
VL
COS θ2=
θ1
x
_
x2+y2
y
_
x2+y2
(
A
( V-V
V+V
l
B
(
V+V
= _L
V
(
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l=
(
V+V
l= _L V-Vs
VS
θ2
VS
s
VL
z
VL
VS
θ2
L
_
Case 11
Case 7
(source and listener moving in opposite direction)
VL
COS
( V-V
V+V COS
(
(
S
V
l= _
V+ VS
(
VS
V
l= _
V- VS
(
( V+V
V-V
l= _S VL=0
(
(
Sound
VL=0
(
V-Vs
V-V
l= _L V
VS
Case 1
(listener is stationary & source is approaching the listener)
(
VL
(
+
V+
- VS
V- V
l= _L
V+V
l= _s (
(
VL
Case 8
(source approaching a stationary wall)
Beat(Δ) =lB-lA
=
[V+V -V+V ]
V
L
L
2VL
Δ= _
V
WAVES
4
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TIME PERIOD OF S.H.M
k
T=2π
F=-kx
F=-kx
F=-kx
m
k
T=2π
T=2π
COMBINATIONS OF SPRINGS
T1
k1
T=2π
k1 k2
T2
k2
TS=
k2
T2
k1+k2
m
m
T=2π
keq
T12+T22
T1
T2
k1
k2
m
1
k1
Tp
2
=
1
T1
2
+
k
1
T2
2
m
k2
keq
eg:
if
k
q=120o,q=60o
keq= k+2k=3k
4
2q
m
m
T=2π
m
keq
2m
T=2π
2
3k
CUTTING OF SPRINGS
ka
1
k1L1=k2L2
L
k
OSCILATION OF
LIQUID COLUMN
L
OSCILATION OF
FLOATING BODY
H
σ,m
x
Liquid (r)
x
x
T=2π
h
T=2π
m
rgA
σH
rg
σ=Density of Body
r=Density of liquid
k1=
kL
T=2π
T=2π
h
g
r=Density of liquid
A=Area of U-tube
L1
k2=
L1
k(L1+L2)
L1
L2
k2=
kL
L2
k(L1+L2)
L2
eg:
k
L
m
2rgA
k2
k1
L= L1+L2
k1=
d 2x
dt
Second‛s pendulam
T=2 second
2
d 2x
l=1 meter
= -kx
dt2
dt
-k
m
=
x
k
x =0
m
+
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k
keq= k1+k2
m
TS2=T12+T22
m1+m2
keq= k+2kcos2q
k1 T1
keq=
m1m2
d 2x
m
k
SPECIAL CASE
2) Parallel
1) Series
m
k+rHg
T=2π
m
k
mrel.
T=2π
mrel=
ma= -kx
l
F=-kx
Liquid
m
k
DIFFERENTIAL
EQUATION OF S.H.M
l
g
T=2π
k
x
H
Liquid
m
k
T=2π
6
↑a
m
F=-kx
5
OSCILLATIONS 01
4
L1:L2:L3=1:2:3
k1=
k2=
k3=
k(L1+L2+L3)
L1
k6
2
k6
3
= 3k
= 2k
=
k6
1
= 6k
Concept of geffective
T=2π
2
d 2x
l
dt
geff
2
w=
2
Pendulum in lift
Case 1-Moving up with constant acceleration ‘a‛
geff=g+a
↑a
w x =0
2
w=
k
m
Eg:
d 2x
g-a
dt2
Case 3-Moving with constant velocity
+ 80 x =0
w =80
2
a=0
w= 80
2p = 80
l
T=2π
m = 4. Kg
K = 320 N/m
2
4 d x + 320 x =0
dt2
l
T=2π
0
geff=g-a
a
k
m
Φ =Initial phase difference
g+a
Case 2-Moving down with constant acceleration ‘a‛
a
+
A=Amplitude of SHM
l
T=2π
↑
3
x
↑
2
1
SIMPLE PENDULUM
g
T
T= 2p
Case 4-Free fall
80
a=g
geff=g-a=g-g=0
T
Pendulum in a truck moving with
constant acceleration
a
geff=
a
g
T=2π
Pendulam in Water
σ=Density of Bob
r=Density of liquid
r
(
geff=g 1T=2π
σ
(
l
g 1-
(
r
σ
(
g2+a2
l
g2+a2
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PROJECTION OF CIRCULAR MOTION
w
x
1
1l
w
x=Asin(wt+Φ0)
-A
2
x=-A
wt
x=0
-A
2
Φ
t=0
>
>
0
x=0
w
From 0 to A
A
2) X= A
0
2) X= A
p
A
=AcosΦ0
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A
2
>
p
x=Asin(wt+ 4 )
x=A
2
t=0
1st Quadrant
3) X=
3A
3A
2
=AsinΦ0
2
Φ =60O= p3
2
Φ =45O= p4
0
p
x=Acos(wt+ 4 )
p
x=Asin(wt+ 3 )
1l 2l x=A
x=0
OSCILLATIONS 02
A
A
2
3A
2
2
x=A
Velocity of SHM
x=Asin(wt+Φ0)
Graphical Representation of Velocity
Start from mean position
x=Asin(wt+Φ0) Φ0 = 0
Graphical Representation of Displacement
3A
Two particles executing SHM meet at X=
2
2
Start from extreme position
x=Acos(wt+Φ0),Φ0 = 0
w
3A
2
=AsinΦA
B
Φ
Φ
A
2
A
2
3A
2
=AcosΦB
Φ =30 =
B
O
p
6
A→
2
v=
Aw
2
v
v=0
B
x=0 X=
3A
2
x=0
q1
x=A
T
6
T
12
w
q2
Phase difference
x=0
Φ=30o+30o=60o or 300o
w
=
)
∆q
q1-q2
2p
)T
x=0
T
8
x=0
x=A
A
T
6
∆t=
x=A
A
4
2
T
12
3A
2
x=A
A=
V12x22-V22x12
V12-V22
v
3A w
v=
v=-Awsin(wt+Φ0)
v=Awcos(wt+Φ0)
vmax=Aw
a) x=0
d) x= A
→
t=0
2
Calculation of time
for particle B
2
A -x
A
p
A
3A
v= w
c) x=
A
B
Φ =60O= 3
t=0
x
2
b) x=
for particle A
3A
Start from mean position
x=Asin(wt+Φ0), Φ0 = 0
Start from extreme position
x=Acos(wt+Φ0) Φ0 = 0
x
v=Awcos(wt+Φ0)
X=
0
0
x=0
X=
Φ =60O= p3
t=0
x=Acos(wt+ 3 )
Φ =45O= p4
t=0
Φ
A
=AcosΦ0
2
1) X= A
2
p
=AsinΦ0
2
2
Start from extreme position
0
0
x=Asin(wt+ 6 )
2 Quadrant
w
x=Acos(wt+Φ0)
Φ
0
t=0
1
1l
Φ =30O= p6
t=0
A
Eg: Particle lies at x=
[t=0] and move towards A
2
Projection/shadow of uniform circular motion in y axis is SHM
2
x=+A
nd
4thQuadrant
2l
+A
2
w
A
=AsinΦ0
2
1) X= A
2
>
2
w
2
l
Start from mean position
x=Asin(wt+Φ0)
1st Quadrant
>
x
INITIAL PHASE FROM POSITION & DIRECTION
3rdQuadrant
2nd Quadrant
1st Quadrant
v1
v2
x1
x2
w=
V12-V22
x22-x12
PHYSICS
WALLAH
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Acceleration of SHM
v=Awcos(wt+Φ0)
Calculation of Time period
and amplitude
v
<
x=Asin(wt+Φ0)
Graphical Representation of Acceleration
p
2
p
vmax=Aw
2
a=-Aw2sin(wt+Φ0)
amax=-Aw2
a<
amax
w= v
max
2p amax
=
T vmax
<
2
Phase difference between x and a=
p
OSCILLATIONS
03
p
Start from extreme position
v=-Awsin(wt+Φ0), Φ0 = 0
v
vmax2
A= a
max
v
PHYSICS
WALLAH
x
p
Phase difference between x and v=
Phase difference between v and a=
amax=Aw2
Start from mean position
v=Awcos(wt+Φ0), Φ0 = 0
a=-Aw2cos(wt+Φ0)
a=-Aw sin(wt+Φ0)
2
a
vmax
T =2p a
max
2
Energy of SHM
T.E
P.E
K.E
1
1
K.E= mv2 = mw2(A2-x2)
2
2
KEmax=
KE=0
x=-A
1 mw2A2
2
x=0
KE=0
2
1
PEmax= mw2A2
2
x=-A
2
2) x=
E
2
PEmax= 1 mw2A2
2
x=0
x=A
1 mw2A2
2
-A
A
2
K.E= 3E
4
P.E= E
4
K.E= E
A
2
A
2
A
Total mechanical energy(E) =
1 mw2A2
2
Note: In SHM, if particle oscilate with
3) x=0
PE=0
Total mechanical energy =
A
K.E=P.E=
x=A
1
P.E= mw2x2
1) x=
P.E=0
frequency
with 2
w
w, then the K.E & P.E oscilate
4) x=A
K.E= 0
P.E=E
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