Physics F111: Mechanics, Oscillations and Waves (MeOW)
Professor Debashis Bandyopadhyay
Physics Department, BITS Pilani
12-May23
PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 1
Simple Harmonic Oscillators
List of Problems from Ch. 3 A.P. French
Lect.Problems: 6.17, 12, 16 and 19; Tut Problems: 6.18, 13, 15 and 17
Suggest list of problems: 6.15, 6.16, 6.19  KK  , 1, 3, 4, 5, 6 and 14  French 
Not in course: 7, 8, 9, 10, 11 and 18
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 23
Simple Harmonic Oscillators
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 24
Simple Harmonic Oscillators
1 2
Moment of inertia of the rod about the pivot : I  Ml
3
You just pull and release. So, the oscillation will start. At any ins tan t of time,
let the position of the bar is   position from its vertical position.
l l
l
l
So, torque equation :   I  k .  k l.l  Mg     5kl  2Mg 
2 2
2
4
3  5kl  2Mg 
1 2
l
3
 Ml     5kl  2Mg    
 5kl  2Mg   0  0 
3
4
4Ml
4Ml
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 25
Simple Harmonic Oscillators
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 26
MR 2
2
Moment of inertia of the disk about the po int of suspension : I 
 Ml
2
1 2 2
R l
I
Time period of oscillation : T  2
 2 2
Mgl
gl
2
2
2 2
2
R

2l
2R

4

l
2
2
T 
.4 

2gl
gl
g
dT
dT
2R 2 2 4 2
 For T  Tmin :
 0  2T


0
2
dl
dl
gl
g
l  R
2
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 27
Simple Harmonic Oscillators: Floating object
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 28
 a  Let at any ins tan t of
time, the amount of portion of the cylinder immerged is z.


So,thebouncy force : F   r 2 z   w  g; Mass of the cylinder : m   r 2 L c ; r  d 2
2
2

r
g w
d2z
d
z
2
2
m 2   r   w  gz  2  0 z  0;0 

dt
dt
m
g w
.
L c
According to the question : mg   r 2  w gl   r 2 L c g
w L
g w
gL
g

 0 
.


c l
L c
L l
l
2
d2z
d
z g
2
 b  2  0 z  0  2  z  0  z  Acos 0t; At t  0, z   B,z  0
dt
dt
l
 z   B cos 0 t and z  B0 sin 0t  A0 sin 0t  A sign curve
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 29
Simple Harmonic Oscillators
Velocity will follow a sign curve with initial
velocity is in  direction.
At, t  0,   t   0;
T T 
At, t  ,    B0  A0
4 4
2T  T 
At, t 
,    0;
4
2
3T  3T 
At, t 
, 
   B00   A0
4  4 
At, t  T , T   0.
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 30
Solving harmonic oscillator differential equations using complex exponential
1.One may solve the differential equation that represents a simple
t
harmonic motion by assuming a trial solution: x e .  is unknown.
2.Put this trial solution in the differential equation and findout the
unknowns in terms of the parameters given in the differential euqation.
3. Now, replace the unknowns in the trial solution in terms of known
parameters. So, the trial solution is now the solution of the diff . eq n .
d2x
Let us take the differential equation as : 2  02 x  0 and
dt
corresponding trial solution as : x  t  e t
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 31
Solving harmonic oscillator differential equations using
complex exponential
So, we have :  2  02  x  0   2  02  0    i0
The most general solution is the linear combination of the
trial solutions containing two different roots.
 x  t   ae
i0 t
 be
 i0t
  a  b  cos 0t  i  a  b  sin 0t
Take real part of x  x   a  b  cos 0t  Acos 0t
Take imaginary part of x  x   a  b  sin 0t  B sin 0t
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA32a
Solving harmonic oscillator differential equations using
complex exponential
So, we have :  2  02  x  0   2  02  0    i0
The most general solution is the linear combination
of the trial solutions containing two different roots.
 x  t   aei t  bei t   a  b  cos 0t  i  a  b  sin 0t
0
 x  t   ae
i0 t
0
 be
 i0 t
also represents two rotating
vectors a and b in complex plane with angular speed
0 in anticlock wise and clock wise direction.
12-May-23PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 32b
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 33
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 34
a ) Differential eq. of motion along x : F  2kx
d2x
d 2x
2kx
d 2x
 m 2  2kx  2  
 2  02 x  0
dt
dt
m
dt
2k
m
 x   x  0 0 
;T0  2
m
2k
2
0
Since the oscillation is along x  axis,so, let us assign the 0 as  x .
Please note that  x is not function of x.
 Solun.: x  t   A0 cos  xt; At t  0, x 0   A0 and x  0,
So, x  t   A0 cos  xt
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 35
Differential eq. of motion along y :Transverse horizontal direction
 b  For small oscillation : y  l,let unstretched
length is
0
.
1/ 2
2




y
y
2
2
  0  y  0 1  2   0 1  2 
2 0
0 


y2
y2
   0 
 Spring force : F  k   k.
2 0
2 0
y
Here,  is very very small.So,Fy  F sin   F sin    k   .
2
0

Total Vertical force : 2Fy  2k 
 0
1
2
1
2

 l 
d2y
 y  m 2  2k   y
dt

 0
 l 
2k  l 
 y 
    x   ;l  l0
m  0
 0
1
2
1
2
y
2k  l   l 
c 
     ;l  l0
x
m  0  0
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 36
Differential eq. of motion along y : Transverse direction
y
l  l0
9
5
41
5 Tx

l0 


 e  If l  l0 , then, 
4
x
l
4
9l0
9 Ty
Note : We have taken sin   y l , not, y l 0 as in case l  l0
 d  x  Acos xt  1  and
y  B sin  y t  2  ; Initial conditions :
At,t  0,x  0 and x  A0 ;1  0 and A  A0 ; x  A0 cos xt;
At,t  0, y  0 and y  A0 2 

2
and B  A0 ; y  A0 cos  yt
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 37
Differential eq. of motion along y :Transverse direction
 d  x  A0 cos xt and
y  A0 cos  y t
Resultant motion (x-y vibrations) can be combine to get the above figure.
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 38
Diff. eq. of motion along y :Transverse direction: Non-linear oscillation
For small oscillation : y  l,let unstretched length is
1/ 2
2



y
y
 20  y 2  0  1  2   0  1  2
2 0
0 

y 2
; Spring force : F  k 
   0 
2 0
2
0
.



y2
,
 k.
2 0
k 3
y2 y
y
. ;Total Vertical force : 2Fy   2
So, Fy  F sin   k.
2l0 l
2 0
k 3
d2y
k 3
d2y
y  0  2  2 y  0 Non  linear oscillation


2
ml0
dt
ml0 l
dt
 Not in course 
http://www.glowscript.org/#/user/techforcurious/folder/TwoSpring/program/5.3-TwoSpring2D
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 39
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 47
dx
3.12.Let x is along x  axis and
 velocity  is along y axis.
dt
 a  For undamped oscillator the solution is x  A0 cos 0t
and hence x   A00 sin 0 t
1 2 1
So, total energy of the oscillator : E  mx  m02 x 2
2
2
x2
x2


 1  equation of an ellipse.
2
 2E m  2E m0


12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 48
3.12. b With damping ,the solution is : x  t   A0 e


 t
2
cos t and
2
2
2
 t 

x
x
 t
2
Energy, E  t    A0 e   E0 e ;

1
2
 2E m  2E m0



2

2
x
x


 1  equation of an ellipse with
 t
 t
2
2E0 e m
2E0 e m0

 

decreasing major and minor axes
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 49
Oscillation along the major and minor axes with
time. The elliptical path is constant w.r.t. time.
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 50
Oscillation along the major and minor axes with time. The elliptical
path is not constant wrt time. It is collapsing with time.
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 51
Extra Slides
Now,  0 
2
2

2
4
 0 
2
For large Q,   0 ; Q 

2
4Q
2
0
, For Q  5  small damping 
 

After ' n' oscillation, t  nT . So, An  A0 e
 An  A0 e

1 2
. .nT
2Q T
 An  A0 e


t
2
0
Q
 A  t   A0 e
 A0 e

2Q
t
2Q
n
Q

0 t
and En  A e
2
0

2 n
Q
 E0 e

2 n
Q
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 55
Over damped motion with initial conditions:
At, t=0, x=x0 with initial velocity=0
Case I . x  0   x0 and x 0   0; Solution: x  t   e

 t
2
C e
1
t
 C2 e
 t

At ,t  0, x  x0 and x  0  x0  C1  C2    1





x t 0  0    C1  C2     C1  C2   C1      C2      0    2  ;
2
2
2


x0
x0
 C1   2   
and C2   2   
4
4
x0
 x t  
e
4

 t
2
et  2     e t  2    
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 56
Over damped motion with initial conditions:
At, t=0, x=0 with initial velocity=v0
Case II . Initial conditions : At t  0, x( 0 )  0 ; x( 0 )  0
General solution: x  t  

 t
e 2

C1et  C2 e t

Using: At  0,x  0 and x( 0 )  0 ; x( 0 )  0 we have




 C1  C2  0 and C1      C2      0 ;
2
2


0
0
Solving  C1 
and C2  
: So, x( t ) 
2
2
2

0  2 t
e
et  e t 


12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 57
Example : Over damping : Show that the system x  4 x  3 x  0 is over damped and graph the solution
with initial conditions x  0   1,x  0   0. Which root controls how fast the solution returns to equilibrium?
4  16  4.3
Let us take the trial solution : x e .    4  3  0  1 , 2 
2
 1 , 2  1, 3; Roots are real and different. So the system is overdamped .
t
General solution : x  t  

 t
e 2

2

c1e1t  c2 e 2t 
4
 t
e 2

c1e 1t  c2 e 3t

Using initial consitions : x 0   1 x 0   0 we have c1  c2  1 and 3c1  5c2  0
3
1
1 5t 
 3 3t
Solving :c1  and c2    x   e   e  ; The term e 3t goes more
2
2
2
2

slowly compare to e 5t , so it controlls the rate at which x goes to zero.
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 58
Damped harmonic motion: Effect of large damping: Critical damping
When :
2
4
 02  Damping force  restoring force  Critical damping.
Case I : Initial condition,at t  0, x  x0 and x  0; Solun. x  t    A  Bt 

 t
e 2
Using 1st initial condition: x0  A; 2nd initial condition
x t 0  

2
 A  Bt  e


2
  
 x  t   x0  1  t  e
2 

t
 Be


2
t
 B
t 0


2

2
A B

2
x0  0; B 

2
x0
t
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 59
Critical Damping Different cases
Case II : Initial condition : x 0   0 and x  0 ; Solun. : x  t    A  Bt  e


t
2
Using 1 initial condition : A  0; 2nd initial condition
st
x t 0  

2
 A  Bt  e


t
2
 Be


t
2
 B  0  x  t   0 te


t
2
t 0
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 60
imit
t 
xcritical damping ( t )
xover damping ( t )
0
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 61
Example 1. Critical damping: Show that the system
x  4x  4x  0 is critically damped and graph the
solution with initial conditions x  0   1, x  0   0.
Let the trial solution : x
et .   2  4   4  0  1 ,  2  2, 2
General solutions : x  e 2t  c1  c2 t 
 Using initial consitions : c1  1 and c2  2  x  e 2t 1  2t 
 Critically damped motion
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 62
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 63
2
2
dE ke a
3.16.
 2  Power  P  t  ; x  A sin 2 t  x  A sin t
dt
c
2
a
Now,x

x

A
sin

t;
x


cos

t;
x

a


A

sin t
 
T
T
 Energy radiated in one cycle :  P  t  dt  
0
A  ke

c3
2
4
2 T


sin 2 t dt  E in 1 cycle
0
 E in 1 cycle 
 A  ke
2
c
3
0
ke2 a 2  t 
c
3
2 T
ke
dt  3
c
 a t  dt
2
0
A2 4 ke2 T A2 4 ke2 2

. 
3
c
2
c3
2
2
3
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 64
 1
2 2
m A

energy stored
2
b
Q

2


2


 
2 2 3
k

e
A
energy loss / cycle


3
c

3
3
mc
mc
Q 2

2
ke  2  2 ke 
 c  E  E0 e
 t
 E0 e
 n T

 mc 3
 2
 ke 


E 1  n T

 e
 ln 2  n T
E0 2
ln 2 ln 2 Q ln 2 Q ln 2
n



 T T . 0
0T
2
Q
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 65
 d  Take,  5  10
14
Hz for visible light.

31


3
9.1  10
3  10
mc
mc
7
Q 2  2


5.1

10
2
9

19
15
ke  ke 2 2 6  10 1.6  10
5  10
3
3


By defination of half  life : E  E0 e
 t
8


1  t
ln 2
 e t 
;
2

0
ln 2 Q ln 2 Q ln 2  5.1  10  ln 2
8
Now,  , t 



 1.13  10 sec
15
Q

0
2
2  5  10 
7
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 66
Example 1. Damped SHM : Show that the system : x  1x  3x  0,is underdamped find its
damped angular frequency and graph the solution with initial conditions : x 0   1,x 0   0.
e t .   2    3  0    
Let the trial solution : x
General solutions : x 
1
 t
e 2
Choose the real part : x 
 i
c1e


1
 t
e 2
11
t
2
 c2 e
i

 c1  c2  cos 

11 
t
2 
1
11
i
2
2


11 
 t
2 
1

 2t
e

1
 Using initial consitions : c1  1 and c2 
 x  1 

11
11


1

 11 
cos 
t
 2 


2
  t
Energy : E  t    A( t )  E  t   E0  e 2   E0 e  t




2
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 69
Oscillation of water column in V-tube
Let z is displacement at time t.
So,the pressure at the junction point on the left tube
is:P1  P0   g  l  z  ; P0  atmospheric pressure
Similarly : P2  P0   g  l  z cos  
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 70
Oscillation of water column in V-tube
d2z
 m 2    P2  P1   A    gz 1  cos   A
dt
1  cos  


d 2 z  gA
1 
 2 
1  cos   z  0;m   Al  1 
   Al
dt
m
cos 
 cos  
d 2 z  Ag  1  cos   cos 
d 2 z g cos 
 2 
z 0 2 
z 0
dt
 A  1  cos   l
dt
l
 0 
g
cos 
l
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 71
Simple Harmonic Oscillators
List of Problems from Ch. 3 A.P. French
Lect.Problems: 6.17, 12, 16 and 19; Tut Problems: 6.18, 13, 15 and 17
Suggest list of problems: 6.15, 6.16, 6.19  KK  , 1, 3, 4, 5, 6 and 14  French 
Not in course: 7, 8, 9, 10, 11 and 18
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 72
Simple Harmonic Oscillators
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 73
Simple Harmonic Oscillators
1 2
Moment of inertia of the rod about the pivot : I  Ml
3
l l
l
l
So, torque equation :   I   k  .  k l.l  Mg     5kl  2Mg  
2 2
2
4
3  5kl  2Mg 
1 2
l
3
 Ml     5kl  2Mg     
 5kl  2Mg    0   
3
4
4 Ml
4 Ml
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 74
Simple Harmonic Oscillators
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 75
MR 2
2
Moment of inertia of the disk about the po int of suspension : I 
 Ml
2
I
1 2 2
Time period of oscillation : T  2
 2  R  l  gl
Mgl
2

2
2
2 2
2
R

2l
2R

4

l
2
2
T 
.4 

2gl
gl
g
dT
dT
2R 2 2 4 2
 For T  Tmin :
 0  2T


0
2
dl
dl
gl
g
l  R
2
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 76
6.18.Torque about the po int of suspension :
 mgl

 mgl

 mgl

1
1
  
 Mgl  sin    
 Mgl    I total  I total  
 Mgl    0; I total  ml 2  Ml 2  MR 2
3
2
 2

 2

 2


mgl
 Mgl
2
and T  2
I total
I total
;
mgl
 Mgl
2
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 77
When the disk is attached via a frictionless bearing it cannot
rotate about its own axis and hence is not contributing any moment
1 2
'
2
of inertia. So, I total  ml  Ml
3
mgl
 Mgl
'
I total
2
 
and T  2
'
mgl
I total
 Mgl
2
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 78
MR 2
2
Moment of inertia of the disk about the po int of suspension : I 
 Ml
2
I
1 2 2
Time period of oscillation : T  2
 2  R  l  gl
Mgl
2

2
2
2 2
2
R

2l
2R

4

l
2
2
T 
.4 

2gl
gl
g
dT
dT
2R 2 2 4 2
 For T  Tmin :
 0  2T


0
2
dl
dl
gl
g
l  R
2
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 79
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 80
3.2.  1 Both the springs are identical and spring const is k.
2k
So,mx  kx  kx  2kx  x 
x  0  x  2x  0
m
2
m
1
m T0
T 
 2

.2

; So,the effective

2k
k
2
2
spring constant for parallel connection : keffective  k1  k2  ....
3.2. 2  General solun : Let the1st and the 2nd spring const are k1 and k2 .
Let the expansion of the 1st spring displacement is x1 . And the
displacement of the mass is x. So, expansion of the 2nd spring is x  x1
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 81
k1  k2
So, at the junction : k1 x1  k2  x  x1   x1  k1  k 2   k 2 x  x1 
x
k2
 k1  k2
 mx  k2  x  x1   k2  1 
k2


k1k2
x
x  
k1  k2

m  k1  k2 
k1k2
 mx 
x  0 ; T  2
; Here,k1  k2  k
k1  k2
k1k2
2m
m
T  2
 2T0 ;T0  2
k
k
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 82
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 83
3.3. Amplitude :5cm; Frequency : f 
10

Hz,  2 f  20 rad / sec
 a  The block will leave the platform when its acceleration becomes
equal to 'g'.Now,x  5 cos t  x  5 sin t  x  5 2 cos t   2 x
1000
2
2
Acceleration  g   x  x  g  
 2.5cm
2
20
2
b
At
x

2.5cm,x


5

sin

t


5

1

cos
t
 



  25 1  cos t  25   x  20 25  2.5
2
2
2
2
2
2
2
2

   2gh  7500  h  7500 2000  3.75cm
2
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 84
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 85
 a  Let at any ins tan t of
time, the amount of portion of the cylinder immerged is z.


So,thebouncy force : F   r 2 z   w  g; Mass of the cylinder : m   r 2 L c ; r  d 2
2
2

r
g w
d2z
d
z
2
2
m 2   r   w  gz  2  0 z  0;0 

dt
dt
m
g w
.
L c
According to the question : mg   r 2  w gl   r 2 L c g
w L
g w
gL
g

 0 
.


c l
L c
L l
l
2
d2z
d
z g
2
 b  2  0 z  0  2  z  0  z  Acos 0t; At t  0, z   B,z  0
dt
dt
l
 z   B cos 0 t and z  B0 sin 0t  A0 sin 0t  A sign curve
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 86
Simple Harmonic Oscillators
Velocity will follow a sign curve with initial
velocity is in  direction.
At, t  0,   t   0;
T T 
At, t  ,    B0  A0
4 4
2T  T 
At, t 
,    0;
4
2
3T  3T 
At, t 
, 
   B00   A0
4  4 
At, t  T , T   0.
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 87
1
MI of the rod about the CM point : I CM  ML2 ; MI about the position
12
of the nail : I n  I CM  Ml 2 ,where, l is the dis tan ce between the CM and
2
2
L L L
ML
ML
the po int of suspension    ; I n  I CM  Ml 2 

2 3 6
12
36
4ML2 ML2
L
In 

; Diff eq of motion :     I n   Mg.l.   Mg. .
36
9
6
MgL
9MgL
3g
2L
2
 
  0  
  0       0; 
 T  2
2
6 In
6  ML
2L
3g
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 88
MI of the hoop about the CM point : I CM  MR 2 ; MI about the position
d
of the nail : I n  I CM  MR  2MR ; R  radius 
2
Diff eq of motion :     I n   Mg.R.sin    Mg.R.
2
2
MgR
MgR
2
 
  0  


0




  0;
2
6 In
2MR
g
2R
d

 T  2
 2
2R
g
g
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 89
Oscillation due to torque
Total torque : I    r .k .r ; Here, I  mR
 mR 2   r 2 k   0  0 
2
r 2k
mR
2
1 2 1 2 1
1 2 2
2 2
Total energy : E  I   kx  mR   kr   const.
2
2
2
2
 mR 2   kr 2   0  mR 2   r 2 k   0  0 
r 2k
mR 2
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 90
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 91
dx
3.12.Let x is along x  axis and
 velocity  is along y axis.
dt
 a  For undamped oscillator the solution is x  A0 cos 0t
and x   A00 sin 0 t
1 2 1
Total energy of the oscillator : E  mx  m02 x 2
2
2
2
2
x
x


 1  equation of an ellipse.
2
 2E m  2E m0


12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 92
3.12. b  Introducing damping ,the solution will be modified as :
x  t   A0 e

 t
2
cos t and
2
2
x
x
E  t   E0 e  t ;

1
2
 2E m  2E m0


x2
x2


1
 t
 t
2
2E0 e m
2E0 e m0

 

 equation of an ellipse with decreasing major and minor axes
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 93
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 94
3.13.Given, x  Ae  t cos t
Now, eit  cos t  i sin t  Real part of eit  Imaginary part of eit
 cos t  Real part of eit  R.P.eit
Since we can never mix real part with imaginary part,
so, one can write : x  Ae
 t
cos t  R.P.Ae
 t
.e
it
 Ae
 t  it
dx
 t  it
  A    i  e
    i  x
dt
d2x
dx
2
and 2     i      i  x
dt
dt
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 95
d2x
dx
2
2
 2 
 0 x     i  x      i  x  02 x  0
dt
dt


2
2
2
2
2
2


     0    2i  i  x  0      0    i  2      0
This is possible when Real and Imaginary parts are separately  0
2    0   
   02 
2
4

and  2   2  02    0 
2
2
4
  2  02 
2
2
0
; So, x  Ae  t cos t is a possible solution of
the equation for  

2
and   02 
2
4
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 96
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 97
b
80
 a  mx  bx  kx  0  x  x  x  0  x  5bx  400x  0
0.2
0.2
 x   x  02 x  0;0  20
2
2
3

 b   2  02   02  02     0  20; b  m  0.2  20  4Ns / m
4
4
4

 t
0 0
20
2
2

 1; Now : A  t   A0 e ;t  10T  10
c Q  
 b m 4 0.2

 A  10T   A0 e
 A  10T   A0 e
 102
 .
2 
 102
 .
2 
3
; Now :  
0 ;b  4Ns / m;   0  20
2

 A0 e
0 102
2
.
3
0
2
 A0 e

102
3
 A0 e

20 
3
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 98
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 99
3.15. a  Now, E  t   E0 e
 t

E t 
e
 t

E0
0 2 f 2  256 512
Now, Q0 





ln 2
ln 2
b Q 
2 f

E  1
E0
1
  .1
  e    ln 2
2
2  512

 2Q0
ln 2
0.9
E 1
1
 t
4 
 3; E  E0 e 
 e  
 c  m  0.1kg ,k  0.9N / m;0 
0.1
E0 e
4
0
3
 b  m  0.1  0.25  0.025 kg sec  Q 

 12
 0.25
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 100
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 101
2
2
dE ke a
3.16.
 2  Power  P  t  ; x  A sin 2 t  x  A sin t
dt
c
2
a
Now,x

x

A
sin

t;
x


cos

t;
x


A

sin t
 
T
T
 Energy radiated in one cycle :  P  t  dt  
0
A  ke

c3
2
4
2 T


sin 2 t dt  E in 1 cycle
0
 E in 1 cycle 
 A  ke
2
c
3
0
ke2 a 2  t 
c
3
2 T
ke
dt  3
c
 a t  dt
2
0
A2 4 ke2 T A2 4 ke2 2

. 
3
c
2
c3
2
2
3
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 102
1
2 2
m A
3
3
energy stored
mc
mc
 2 2 2 2 3  2 
 b  Q  2
2
k e A 
energy loss / cycle
ke  2 ke 
3
c
1
 t
 n T
 n T
c
E

E
e

E
e


e
 ln 2  n T
 
0
0
2
ln 2 ln 2 Q ln 2 Q ln 2
n



 T T . 0
0T
2
Q
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 103
 d  Take,  5  10
14
Hz for visible light.

31


3
9.1  10
3  10
mc
mc
7
Q 2  2

 5.1  10
2
ke  ke 2 2 6  10 9 1.6  10 19 5  1015
3
3


By defination of half  life : E  E0 e
 t
8


1  t
ln 2
 e t 
;
2

0
ln 2 Q ln 2 Q ln 2  5.1  10  ln 2
8
Now,  , t 



 1.13  10 sec
15
Q

0
2
2  5  10 
7
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 104
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 105
3.17. Complete problem is under small oscillation condition.
Let y and y' be the shift of CM in the 1st and the 2nd column.
 r y    2r 
2
2
y
y'  y'  ; PE in 1st column :
4
2


h

y
h
y
2
2
2
PE 1st col .   r . h  y  g.
  r gh.   r g  hy  
2
2
2 

h  y'
h
2
2
PE 2nd col .    2r  g. h  y'  .
   2r  g.h.
2
2
2
2




y'
y
2
2
 PE 2nd col .    2r  g  hy' 
   r g  hy  
2 
8 


12-May-23PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA
106
PE Total
2
2




y
y
2
 PE 1st col .  PE 2nd col .   r g  hy     hy  
2  
8 

2
2

y
y 
5
5
2
2 2
2 2
  r    g   gr y  PE Total   gr y
8 
8
8
 2
Kinetic energy of the left and right arms at any time instant t :
2
2
1 2  dy 
1
1 2  dy 
2
 dy' 
K L   r  h   and K R    2r   h 
   r h  
2
2
 dt 
 dt  8
 dt 
2
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 107
Derivation of the Kinetic energy of the horizontal arm :
Radius of the horizontal arm at x  0 is r0 .
 Radius at x : r  x   r0  k.x ; Radius at the other end : x  l
r0
 r  l   r0  k.l  2r0  k 
l

 r  x   r0  k.x  r0  1 

x

l
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 108
3.17. Continued :
 dK H  Kinetic energy of a small section at x of
1 2
2
thickness dx  dK H   r  x    x  dx;
2
1 2
1 2 
2
 K H    r  x    x  dx    r0   1 
2
2

x 0
x 0
l
l
2
x 2
   x  dx;
l
How to calculate   x  ?The amount of liquid pass through a
particular section in left arm,must will pass through a
section at x in horizontal tube.
12-May-23PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA
109
dy
2
2 dy
2
 r
  r  x   x    r0
  r0  1 
dt
dt

2
0
  x    dy dt 

1 

1 2  dy 
 K H   r0   
2
 dt 
KETotal
2
2
x
   x
l
2
x
 ;  x  is the velocity at x  x.
l
2
1 2  dy  l l 2  dy 
x0  x 2  2  r0   dt  . 2  4  r0   dt 
1 
l

l
dx
1 2  5h  dy 
 K L  K R  K H   r0   l   
4
2  dt 

2
2
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 110
3.17. Continued :
2
 ETotal
1 2  5h  dy  5
 KET  PET   r0   l      g  r02 y 2  E  const.
4
2  dt  8

dE 1 2  5h  dy d y 5
5h
2 dy
  r0   l   . . 2   g  r0 y
 0; Take, l 
 given 
dt 2
2  dt dt
4
dt
2

2
2
1 2  5h 5h  d 2 y 5
d
y g
2
  r0     . 2   g  r0 y  0  h 2  y  0
2
2  dt
4
dt
2
 2
d2y g
d2y
g
2
 2 
y  0  2  0 y  0  0 
dt
2h
dt
2h
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 111
12-May-23PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA
112
12-May-23PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA
113
a ) Differential eq. of motion along x :
2
d2x
2k
d
x
2
F  2kx  2   x  0 x  2  02 x  0
dt
m
dt
2k
m
 x   x  0 0 
;T0  2
m
2k
2
0
Since the oscillation is along x  axis,so, let us assign the  as  x .
Please note that  x is not function of x. Solun. If at t  0,
x  0   A0 and x  0, then, x  t   A0 cos  xt
12-May-23
PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA
114
Differential eq. of motion along y :Transverse direction
For small oscillation : y  l,let unstretched length is
0
.
1/ 2
2




y
y
2
2
  0  y  0 1  2   0 1  2 
2 0
0 


y2
y2
   0 
 Spring force : F  k   k.
2 0
2 0
y
Here,  is very very small.So,Fy  F sin   F sin    k   .
2
0

Total Vertical force : 2Fy  2k 
 0
1
2

 l 
d2y
 y  m 2  2k   y
dt

 0
1
2
 l 
2k  l 
 y 
    x   ;l  l0
m  0
 0
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA
115
Differential eq. of motion along y : Transverse direction
y
l  l0
9
5
4
5 Tx

l0 


 c  If l  l0 , then, 
4
x
l
4
9l0
9 Ty
 d  x  Acos xt  1  and
y  B sin  y t  2  ; Initial conditions :
At,t  0,x  0 and x  A0 ;1  0 and A  A0 ; x  A0 cos xt;
At,t  0, y  0 and y  A0 2 

2
and B  A0 ; y  A0 cos  yt
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 116
Differential eq. of motion along y :Transverse direction
Resultant motion (x-y vibrations) is the combine of x and y oscillation in the above figure.
End of chapter 3
12-May-23 PHY F111: Mechanics, Oscillations and Waves, Professor Debashis Bandyopadhyay, Dept. of Physics, BITS Pilani, INDIA 117