Let us start with
Complete Kinematics of SHM
m1
M
m2
Periodic Motion
● It is the motion that repeat itself after fixed
interval of time.
Ex:Rotation of earth, revolution of earth
Periodic Motion
Time Period:
● The time interval in which a particle repeats motion.
Oscillatory Motion
● To & fro motion about a mean position. Examples:
(a) Motion of pendulum
(b) To & fro motion of
ball
Oscillatory Motion
● All oscillatory motions are
periodic, but not vice-versa.
Periodic Motion
Oscillatory
Motion
The circular motion of a particle with constant
speed is
A
periodic but not simple harmonic
B
simple harmonic but not periodic
C
periodic and simple harmonic
D
neither periodic nor simple
harmonic
The circular motion of a particle with constant
speed is
A
periodic but not simple harmonic
B
simple harmonic but not periodic
C
periodic and simple harmonic
D
neither periodic nor simple
harmonic
Solution:
In circular motion of a particle with constant speed,
particle repeats its motion after a regular interval of
time but does not oscillate about a fixed point. So,
motion of particle is periodic but not simple
harmonic.
SHM
● It is a special case of oscillatory motion in which
acceleration is always directed towards mean
position & is proportional to displacement from M.P.
SHM
Motion parameter of SHM
● F = –kx
E.P.
M.P.
E.P.
Motion parameter of SHM
● Time Period(T):
Time taken to complete one oscillation
● Frequency(f):
It is the number of oscillations per second
● Angular frequency (𝜔):
Position
●
x = A sin (𝜔t)
Velocity
●
v=
= A𝜔 cos 𝜔t =
Acceleration
Differential equation of SHM
Energy of SHM
Energy of SHM
TE
Remember!!!
Kinetic Energy oscillates with double
the frequency of oscillation.
𝜔KE= 2 𝜔SHM
TKE = TSHM/2
TE
Phase
● The argument of sin function is called or phase.
● It gives an idea about location of particle.
t=0
x = A sin
Phase
t=0
t=0
t=0
Motion parameter of SHM
t=0
t=0
t=0
t=0
Phase
(i)
x = A sin 𝜔t
(ii)
x = A sin
(iii)
x = A sin (𝜔t + 𝜋)
(iv)
x = A sin
(v)
If particle has shifted
mean
position, then equation is
x = x0 + A sin (𝜔t + ⲫ0)
t=0
(i)
(ii)
t=0
(iii)
(iv)
t=0
x0
t=0
x
t=0
Question Time
The equation of SHM is x = 10 sin
(i) amplitude
(iii) Angular freq.
(v) frequency.
Find
(ii) Initial phase
(iv) Time period
(vi) spring constant(m=2 kg)
The equation of SHM is x = 10 sin
Find
(vii) In how much time particle reaches +ve EP & MP
(viii) Find KE, U & TE at x = 5 m.
Solution:
x = 10 sin
(i) amplitude = 10
(ii) Put t = 0, Initial phase = 𝜋/3
(iii) 𝜔 = 𝜋/2
(iv) T =
(v)
(vi) k = m𝜔 2
Solution:
(vii) For +ve extreme
x = + 10
⇒ 10 = 10
(viii) For +ve extreme
The positions of particle at t=0 is shown. Find
equation of its SHM if Time Period T is 10 sec.
A
3m
-E.P.
8m
M.P.
+E.P.
B
t=0
C
10.5m
13m
D
3m
-E.P.
8m
M.P.
+E.P.
t=0
10.5m
13m
The positions of particle at t=0 is shown. Find
equation of its SHM if Time Period T is 10 sec.
A
3m
-E.P.
8m
M.P.
+E.P.
B
t=0
C
10.5m
13m
D
Solution:
x = x0 + A sin(cot + ⲫ)
X0 = 8 meter
A = 13-8 = 5 meter
At t = 0
x = x0 + A sin ⲫ
10.5 = 8 + 5 sin ⲫ
But in 1st quadrant & moving towards
right, ⲫ = 𝜋/6
A particle of mass 2.5 kg is released from rest
at x = 3m under the force F = -10x + 20.
Find equation of motion of particle.
Find angular velocity, Time period & mean
positions of two SHMs represented by
differential equations:
(i) d2x/dt2 + 9𝜋2 x = 0
Find angular velocity, Time period & mean
positions of two SHMs represented by
differential equations:
(ii) d2x/dt2 + 16𝜋2 x = 9
Find phase difference between two particles A
& B which have same time period but are
located at the shown positions at t=0.
M.
P
A/2
t=0
A
t=0
B
Two particles oscillate with period 3 sec & 8
sec. One is present at +ve extreme &
simultaneously other is at –ve extreme. In how
much time they will be in same phase?
M.P
t=0
t=0
A
1.2 secs
B
2.4 secs
C
3 secs
D
24 secs
M.P
t=0
t=0
Two particles oscillate with period 3 sec & 8
sec. One is present at +ve extreme &
simultaneously other is at –ve extreme. In how
much time they will be in same phase?
M.P
t=0
t=0
A
1.2 secs
B
2.4 secs
C
3 secs
D
24 secs
Solution:
For a particle executing SHM, If at x = 3m,
v = 4m/sec & at x = 4m , v = 3m/sec. Find 𝜔 & A.
A block of mass M is placed on a platform which
is tied to a spring (K) as shown. The complete
block + platform assembly oscillates simple
harmonically in vertical direction. Find max
amplitude such that block doesn't leave platform.
A
𝜔2/g
B
g/𝜔2
C
𝜔/g
D
g/𝜔
M
M
A block of mass M is placed on a platform which
is tied to a spring (K) as shown. The complete
block + platform assembly oscillates simple
harmonically in vertical direction. Find max
amplitude such that block doesn't leave platform.
A
𝜔2/g
B
g/𝜔2
C
𝜔/g
D
g/𝜔
M
Solution:
At extreme position weight provide a necessary
restoring force
That is mg ≥ 𝜔2A.
∴A < g/𝜔2
The displacement of a particle along the x-axis
is given by x = asin2ωt. The motion of the
particle corresponds to
A
simple harmonic motion of
frequency ω/π
B
simple harmonic motion of
frequency 3ω/2π
C
not simple harmonic motion
D
simple harmonic motion of
frequency ω/2π
The displacement of a particle along the x-axis
is given by x = asin2ωt. The motion of the
particle corresponds to
A
simple harmonic motion of
frequency ω/π
B
simple harmonic motion of
frequency 3ω/2π
C
not simple harmonic motion
D
simple harmonic motion of
frequency ω/2π
Calculate the time taken to travel from
(i) x = 0 to x = A/2
(ii) x = A/2 to x = A
A
T/12, T/6
B
T/8, T/8
C
T/4, T/4
D
T/6, T/12
Calculate the time taken to travel from
(i) x = 0 to x = A/2
(ii) x = A/2 to x = A
A
T/12, T/6
B
T/8, T/8
C
T/4, T/4
D
T/6, T/12
Shortcut
M.P
T/6
T/12
T/12
T/4
x = –A
T/4
T/6
T/4
x=0
T/4
x = +A
Solution:
Similarly,
Calculate the time taken to travel from
(i) 0 to A /
(ii) A /
to A
Calculate the time taken to travel from
(i)
(ii)
(HW)
Particle rotates in clockwise sense in a circle
of radius 4m at 𝜋/2 rad/s. Its initial position is
shown in the diagram. Find equation of
projection of particle an x-axis.
y
𝜋/6
x
Let us start with
SHM of Bodies in Equilibrium
MASTER
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CLASS
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SHM of bodies Equilibrium
● There should exist a point where net force on particle
is
This ismay
called
mean
● zero.
A particle
or as
may
notposition.
perform SHM about this
mean position.
SHM of bodies Equilibrium
● To check, displace the particle slightly from its mean
position by distance x.
● If a net extra restoring force come into picture to bring
the particle back to its mean position, then it will
definitely perform SHM
Fnet = - kx
A block is attached with spring as shown. Find
the time period for small oscillation.
A
B
M
C
D
A block is attached with spring as shown. Find
the time period for small oscillation.
A
B
M
C
D
A block is attached with spring as shown. Find
the time period for small oscillation.
A
B
C
M
D
A block is attached with spring as shown. Find
the time period for small oscillation.
A
B
C
M
D
A block is attached with spring as shown. Find
the time period for small oscillation.
A
B
C
D
A block is attached with spring as shown. Find
the time period for small oscillation.
A
B
C
D
Solution:Hence A constant force only shifts M.P. It does not impact
time period of SHM.
A block suspended in a lift with the help of a
spring as shown. Now the lift start
accelerating, find time period in two cases.
A
a
a
B
C
D
A block suspended in a lift with the help of a
spring as shown. Now the lift start
accelerating, find time period in two cases.
A
a
a
B
C
D
Two springs and a block are connected as
shown. Find time period for small oscillations.
A
B
k
k
M
C
D
Two springs and a block are connected as
shown. Find time period for small oscillations.
A
B
k
k
M
C
D
Solution:Series combination , Keq = k/2
Two springs and a block are connected as
shown. Find time period for small oscillations.
A
k
k
B
C
M
D
Two springs and a block are connected as
shown. Find time period for small oscillations.
A
k
k
B
C
M
D
Solution:Parallel combination Keq = k + k = 2k
Two springs and a block are connected as
shown. Find time period for small oscillations.
A
B
k
k
M
C
D
Two springs and a block are connected as
shown. Find time period for small oscillations.
A
B
k
k
M
C
D
Solution:It is a parallel combination
The springs and a block are connected as
shown. Find time period for small oscillations.
k
2k
4k
3k
M
6k
The springs and a block are connected as
shown. Find time period for small oscillations.
A
B
k1
k2
C
D
M
The springs and a block are connected as
shown. Find time period for small oscillations.
A
B
k1
k2
C
D
M
A block hangs from a spring with the help of
massless pulley as shown. Find period for
small oscillations.
A
B
C
M
D
A block hangs from a spring with the help of
massless pulley as shown. Find period for
small oscillations.
A
B
C
M
D
Solution:Lets displace the particles by x
When ‘m’ is pulled down by x spring will also
elongate by ‘x’
x
k
m
x↑
m
Solution:FR = T
T = kx [third law]
T
∴ FR = k x
m
∴ FR = [constant] x
k
kx
T
m
x
A block hangs from a spring with the help of
massless pulley as shown in two cases. Find
period for small oscillations.
(b)
(a)
M
M
(a)
M
(b)
M
Solution:(a)
k
T/2
T
m
Solution:(b)
By the principle of virtual work;
−Tx + 2TxP = 0
2TxP = Tx
k
xP
2T
T
m
x
m
Solution:-
Spring is extended by xp
Let's analyze the forces
kxp
2T
2T
T
T
T
Shortcut
Validity:
1. Pulley and spring should be massless.
2. The coefficient of tension on the block should be
1.
ni → coefficient of tension in the spring of spring constant
ki
M
M
A block hangs from a spring with the help of
massless pulley as shown in two cases. Find
period for small oscillations.
A
B
C
M
D
A block hangs from a spring with the help of
massless pulley as shown in two cases. Find
period for small oscillations.
A
B
C
M
D
Solution:
k
2T
^
T^
^T
k
m
A block hangs from springs with the help of
massless pulleys as shown. Find period for
small oscillations.
k
k
k
M
A
k/2
B
k/4
C
k/6
D
k/8
A block hangs from springs with the help of
massless pulleys as shown. Find period for
small oscillations.
k
k
k
M
A
k/2
B
k/4
C
k/6
D
k/8
A block hangs from springs with the help of
massless pulleys as shown. Find period for
small oscillations.
k
k
M
A
9k/5
B
k/5
C
9k/4
D
k/4
A block hangs from springs with the help of
massless pulleys as shown. Find period for
small oscillations.
k
k
M
A
9k/5
B
k/5
C
9k/4
D
k/4
A block of mass M is tied with the help of three
spring all at 1200 as shown. Find Time period
for small vertical oscillation of block.
A
k
k
M
k
B
C
D
A block of mass M is tied with the help of three
spring all at 1200 as shown. Find Time period
for small vertical oscillation of block.
A
k
k
M
k
B
C
D
Two blocks of masses m1 and m2 tied with the
help of spring as shown. Find the time period
of oscillations of both blocks in terms of
reduced mass.
m1
m2
A disc pulley (mass M, Radius R) hangs with
help of a spring of constant k and a string.
FInd time period for small oscillations of the
pulley assuming no slipping
A
B
C
D
A disc pulley (mass M, Radius R) hangs with
help of a spring of constant k and a string.
FInd time period for small oscillations of the
pulley assuming no slipping
A
B
C
D
A block of mass m floats with h part of its
height submerged under water. Find period for
small oscillations of the block.
A
B
𝜌
h
C
𝜌l
D
A block of mass m floats with h part of its
height submerged under water. Find period for
small oscillations of the block.
A
B
𝜌
h
C
𝜌l
D
Solution:
x0
EQUILIBRIUM
1
x0+x
x
FB
(Buoyant force)
mg
Since its in equilibrium
mg = FB
mg = ρl Ax0 g
2
FB
(Buoyant force)
mg
F = mg – FB
F = mg – ρl A(x0 + x)g
Solution:
A
F = mg – ρl Agx0 – ρlAgx
F = – ρl Ag x
L
Compare the equation
in step 2 with
Mass = density × volume
F = –[constant]x
m = ρAL
A block hangs from a spring with the help of
massless pulley as shown in two cases. Find
period for small oscillations.
k
h
A non-viscous fluid is filled in a frictionless Utube. The density of fluid is ρ. Find the time
period for small oscillations of the fluid if total
length of liquid column in tube is l.
A
Area of cross
section is A
B
L
R
C
D
L
R
A non-viscous fluid is filled in a frictionless Utube. The density of fluid is ρ. Find the time
period for small oscillations of the fluid if total
length of liquid column in tube is l.
A
Area of cross
section is A
B
L
R
C
D
Solution:
x
x
Lets displace the water by x (downward) in the left limb
But (A1x1 = A2x2)
This part of liquid will apply a restoring force
F = PA
F = –2ρgAx
‘–’ sign shows that force is opposite to displacement
Solution:
F = –[constant]x
Force = Pressure × Area
P=ρgh
2x
P = ρ g (2x)
F = 2ρgAx
where m → mass of the liquid
m = ρA(2L + πR)
x
2x
x
Consider a spherical planet of mass ‘M’ and
radius ‘R’ with a tunnel dig across as shown in
the figure. The distance of the tunnel from the
center is ‘a’. An object of mass ‘m’ is kept at
the center of tunnel. Then find the time period
for small oscillations.
A
B
a
C
C
D
a
C
Consider a spherical planet of mass ‘M’ and
radius ‘R’ with a tunnel dig across as shown in
the figure. The distance of the tunnel from the
center is ‘a’. An object of mass ‘m’ is kept at
the center of tunnel. Then find the time period
for small oscillations.
A
B
a
C
C
D
A ideal gas (P, V, T) is enclosed in a chamber
using a piston of mass M and Radius R as
shown. Find time period of piston for small
oscillation if the gas is (Home Work)
(i) Isothermal
(ii) Adiabatic
M,R
Find time period of mo if it is slightly displaced
along the axis of ring of Radius R and mass M.
m0
Let us start with
Angular SHM
Angular SHM
● There should exist a point where net torque on
particle is
zero. This is called as mean position.
● A particle may or may not perform SHM about
this mean position.
Angular SHM
● To check, displace the particle slightly from its
mean position by angle 𝜃 .
● If a net extra restoring torque come into picture
to bring the particle back to its mean position,
then it will definitely perform SHM
𝜏net ∝ 𝜃
𝜏net = - k𝜃
where I = MOI about axis of Rotation
Similarities
Linear SHM
Angular SHM
Similarities
Linear SHM
Angular SHM
● Restoring force is present
● Restoring torque is present
● F = −[constant] x
● a = −ω2x
●
τ = −[constant] θ
● α = −ω2 θ
Where I is MOI about the point
of suspension
Steps for finding the time period
Force
method
EQUIVALENT
Torque
method
Steps for finding Time period
Step 1 : Displace object by θ from MP
Step 2: Find restoring torque, and check if its proportional to θ
Step 3: Compare with τ = −[constant]θ
Step 4 :
I → moment of inertia about the point of suspension
Simple Pendulum
Simple Pendulum
● It is Independent of mass of
bob.
Physical Pendulum
Pivot
CM
Physical Pendulum
Pivot
𝜃
ℓ = distance of COM from
hinge point.
Let us know some
Important Points
Remember!!!
Time period of simple Pendulum
is Independent of mass of bob.
Second’s Pendulum
● Its length is 1m.
● Its Time period is 2 seconds.
Simple Pendulum in a Accelerated Frames
L
●
● a = acceleration of the point of
suspension w.r.t. ground
L
m
r
1. Time Period on Earth
geff= gearth =g
2. Time period on Moon
3. Time period in Space
gspace = 0
T=∞
It will not oscillate
4. Upward Acceleration
● geffective = g + a
a
● Time period decreases, (it starts
oscillating faster)
5. Downward acceleration
● geffective= g – a
● Time period increases, (it starts
oscillating slower)
a
6. Horizontal acceleration
L
→a
geff
Question
Time….
Container is completely filled with a liquid of
density ρl. A bob is suspended as shown. Find
time period of small oscillations. Volume of
the bob is V, and the length of the string is L
ρs
A
T = 2π√[ρsL/(ρs-ρl)g]
B
T = π√[ρsL/(ρs-ρl)g]
C
T = 2π√[ρlL/(ρs-ρl)g]
D
T = π√[ρlL/(ρs-ρl)g]
ρl
ρs > ρl
Container is completely filled with a liquid of
density ρl. A bob is suspended as shown. Find
time period of small oscillations. Volume of
the bob is V, and the length of the string is L
ρs
A
T = 2π√[ρsL/(ρs-ρl)g]
B
T = π√[ρsL/(ρs-ρl)g]
C
T = 2π√[ρlL/(ρs-ρl)g]
D
T = π√[ρlL/(ρs-ρl)g]
ρl
ρs > ρl
Solution:
A rod of mass m length L is hinged at its
topmost point as shown.It is allowed to
oscillate freely in the vertical plane Find time
period for small oscillations.
A
M,L
B
C
D
A rod of mass m length L is hinged at its
topmost point as shown.It is allowed to
oscillate freely in the vertical plane Find time
period for small oscillations.
A
M,L
B
C
D
Solution:
A rod of mass m length L is hinged at a
distance L/3 from the top as shown. It is
allowed to oscillate freely in the vertical plane.
Find time period for small oscillations.
A
L/3
B
M,L
C
D
A rod of mass m length L is hinged at a
distance L/3 from the top as shown. It is
allowed to oscillate freely in the vertical plane.
Find time period for small oscillations.
A
L/3
B
M,L
C
D
Solution:
A ring of mass M, radius R is hinged at its
topmost point as shown. Find time period for
small oscillations (Home Work)
(i) If ring oscillates in its plane
(ii) If ring oscillates perpendicular to its plane
A
,
Hinge
B
,
C
,
D
,
A ring of mass M, radius R is hinged at its
topmost point as shown. Find time period for
small oscillations
(i) If ring oscillates in its plane
(ii) If ring oscillates perpendicular to its plane
A
,
Hinge
B
,
C
,
D
,
Solution:
(i)
(ii)
I1
A child swinging on a swing in sitting position,
stands up, then the time period of the swing
will
A
Increase
B
Decrease
C
Remains same
D
Increases of the child is long and
decreases if the child is short
A child swinging on a swing in sitting position,
stands up, then the time period of the swing
will
A
Increase
B
Decrease
C
Remains same
D
Increases of the child is long and
decreases if the child is short
The bob of a simple pendulum is a spherical
hollow ball filled with water. A plugged hole
near the bottom of the oscillating bob gets
suddenly unplugged. During observation, till
water is coming out, the time period of
oscillation would
A
B
C
D
First decrease and then
increase to the original value
First increase and then
decrease to the original
value
Increase towards a saturation
value
Remain unchanged
The bob of a simple pendulum is a spherical
hollow ball filled with water. A plugged hole
near the bottom of the oscillating bob gets
suddenly unplugged. During observation, till
water is coming out, the time period of
oscillation would
A
B
C
D
First decrease and then
increase to the original value
First increase and then
decrease to the original
value
Increase towards a saturation
value
Remain unchanged
Solution:
Centre of mass of combination of liquid and hollow
portion (at position ℓ), first goes down (to ℓ + Δℓ) and
when total water is drained out, centre of mass regain its
original position (to ℓ),
∴ ‘T’ first increases and then decreases to original value.
∴ ‘T’ first increases and then decreases to original value.
Let us now differentiate between
Undamped & Damped Oscillation
Undamped Oscillation
● If no energy loss takes place then system oscillates with
its
𝜔0 =
natural frequency
● Energy = constant
● Amplitude A =
constant
● Time period = 2𝜋/𝜔
● x = A sin (𝜔t + ⲫ)
Damped Oscillation
When Energy is gradually lost in SHM and the
amplitude decreases with time then oscillation is called
Damped oscillation.
Special case of Damped Oscillation
● Dissipative force is proportional to velocity.
F = –bv
where b = damping coefficient
Special case of Damped Oscillation
● F = -kx - bv
This is a 2nd order differential equation.
● Its general solution is in the form
x = A0 e–bt/2m sin(ω′t +
δ)
Amplitude
x = A0 e–bt/2m sin(ω′t + δ)
Amplitude A = A0 e–
bt/2m
● Amplitude decreases exponentially with time
The object still oscillates sinusoidally
Displacement
T
x = A0 e–bt/2m sin(ω′t +
δ)
But the amplitude decreases within the
envelope of exponential curve
Amplitude
Let us understand
various types of damping
1. Undamped
Displacement
System oscillates at natural frequency with constant
Amplitude.
Time
2. Underdamped (light damping)
Frequency is reduced compared to the undamped case
with the amplitude gradually decreasing to zero.
Displacement
Light damping
Time
3. Overdamped (heavy damping)
The system slowly returns (exponential decay) to equilibrium
without oscillating. Example – Door that uses spring to close the
door once open
displacement
Heavy damping
Time
4. Critically damped
Equilibrium returns as quickly as possible w/o oscillation.
displacement
Critical damping
Time
Let us start with
forced Oscillation
Forced Oscillation
● Example: On a swing, father keeps on pushing his child
periodically to compensate for frictional energy losses
Forced Oscillation
m
F0sin 𝜔t
Forced Oscillation
Forced Oscillation
3 forces will act on the system:
1. Damping force
−bv
F0sin 𝜔t
m
2. Restoring force:
3. Periodic force:
− kx
F0sin(ωt)
Forced Oscillation
The motion will be complicated, but after a while, the
object oscillates with frequency ω of the applied force.
Displacement is given by
●
x = Asin(ωt+φ)
Forced Oscillation
●
(natural angular frequency)
● Less damping: If b is very less
Characteristics of Forced Oscillation
● In forced oscillation amplitude is independent of time
● It is a technique to convert damped oscillation to
undamped oscillation
● The energy lost due to the damping force is
compensated by the work done by the applied force.
Forced Oscillation
amplitude
● Variation of amplitude with angular frequency 𝜔
𝜔0
Frequency (𝜔)
Resonance
Resonance
The condition on angular frequency when the amplitude
becomes maximum is called resonance condition.
● Resonance occurs when
applied frequency is equal to
the natural frequency.
● We see that A is max when
∴ ω = ω0
OR f = f0
Resonance
● Army soldiers marching in Resonance with each other.
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