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Oscillations and Waves
CHAPTER
2
Oscillation and Waves
2.1 INTRODUCTION
Oscillations are unbiquitous. It would be difficult to find something which never exhibits
oscillations. Atoms in solids, electromagnetic fields, multi-storeyed buildigs and share
prices all exhibit oscillations. In this course we shall restrict our attention to only the
simplest possible situations, but it should be borne in mind that this elementary analysis
provides insights into a diverse variety of apparently complex phenomena.
Periodic Motion :
The mation in which the particle repeats itself after a regular interval of time is known
as periodic motion.
Ex: Motion of a pendulum motion of a swing.
The periodic motion in which a particle moves to and fro motion about the mean position
over the same path is known as Oscillator.
Eg- Pendulum, balance wheel of a clock, Beating of heart etc.
Mechanism of ascillation:
Each oscillator has an equlibrium position, when it displaced from the mean position, it
experiences a force which is called restoring force which bring it back to the original
position However on reaching the position, it overshoots to the other side due to inertia
of motion. It again experiences restoring force and oscillation continues.
Types of Oscillator :
There are 4 types of oscillator
i. Simple Harmonic oscillator
ii. Damp harmonic oscillators
iii. Forced oscillator
iv. Copupled oscillators
2.2 SIMPLE HARMONIC MOTION
A particle is said to execute simple Harmonic Motion if the restoring force is directed
towards the mean position and its magnitude is directly proportional to the displacement
of particle from the mean position.
Let F = restoring force experienced by the particle when displaces a displacement (x)
from the mean position
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Engineering Physics
According to the definition of SHM,
F  – x i.e. [– sign because F is opposite to x]
F = – kx
[where k is a force constant]
 ma = – kx
 m



d2x
dt 2
d2x
dt 2
d2x
dt 2
d2 x
dt
2
  kx
  k mx
 k mx  0
 2 x  0
Here,  
k m
The above equation is known as differential equation for SHM.
The solution of above equation may be written as :-
x  A cos  t   
Energy of the body :
Potential Energy : By the defination of simple Harmonic Oscillator; potential Energy
of the particla is given by
x
x
x
x
0
0
0
0
E P   F  dx     kx  dx  cos 180    kx dx  k   x dx
1
 k x2
2
 k m  2 


1
 m2 x 2
2
 x  A cos  t   
1
E p  m2 A 2 cos 2  t   
2
Kinetic Energy :
1
E k  mv2 ,
2
V
dx
d
  A cos  t      A  sin  t   
dt
dt
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Oscillations and Waves
1
E p  m2 A 2 sin 2  t   
2
Total Energy of the particle :
E = Ek + Ep  1 m2 A 2 cos 2  t     1 m2 A 2 sin 2  t   
2
2
1
 m2 A 2 cos 2  t     sin 2  t    


2
1
E  m2 A 2
2
Total energy of the particle remain constant because, mass, angular velocity, and
amplitude are constant.
Q.
Show that the displacement verses Nelocity graph for a SHO is an ellipse.
Ans: x = A cos ,
x  A cos  t    ,
V   A  sin  t   
 we known that equation of ellipse :
x2
x
V
 cos 2  t   
 cos  t    ,
  sin  t    ,
2
A
A
A
V2
Q.
 sin 2  t    ,
x2

V2
1
A 2 2
A 2 A 2 2
this satisfies the equation of ellipse.
The differential equation of motion of a freely oscillation body is given by
2 2 x  182 x  0 .
Calculate the natural frequency of the body
d2x
dt
2
 9 2 x  0 ,
  3 ,
2  92 ,
2f  3
f  3 2  1.5 Hz
A body is executing SHM with time period 2 sec . At its mean position its speed is
0.05m/sec. What is its speed when it is at a distance of 0.04m from the mean position.
Ans: Given time period T  2 s
At mean position x = 0, v = 0.05m
At x = 0.04m
2 2
2
V 2  A 22 sin 2  t     A  1  cos  t    
Q.
 x2 
 A 22 1  2   2 A 2  x 2
 A 


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Engineering Physics
 A2  x 2 
V
At x = 0, V = 0.05
0.05 
2
A2  0
2
 0.05  A 2
At x = 0.04m
 A 2  0.05
 A  0.05
2
 0.0052   0.04 2
 V = 0.03m (Ans.)
2
Q.
Calculate the displacement at which the kinetic energy is equal to the potential energy.
Ans: Given K.E. = P.E.
V
1
1
m2 A 2 sin 2  t     m2 A 2 cos 2  t   
2
2
 sin 2  t     cos2  t     1  cos2  t     cos2  t   
 1  x 2 A2  x 2 A2
 2 x2 A2  1
 x
A
2
(Ans.)
2.3 DAMPED HARMONIC OSILAROR (DHO)
The ossilator whose amplitude in successive oscillation goes on decreasing due to presence
of damping force or dissipative force is known as Damped Harmonic Osilator.
If the velocity of the particle increases, then damping force also increases (F d). Hence
[  damping force & velocity are opposite to eachother]
Fd   V
 Fd   b V
where b = damping constant.
In damped harmonic osillator, when a particle displaces a displacement x, it experiences
two opposing forces. First one is restoring fore (F r ) and another is damping fore (F d).
Hence total force,
F = Fr + Fd
ma
= – kx – bv

m
d2x
dt

d2 x
dt
2
2
  kx  b
 2
dx
dt
dx
 2 x  0
dt

d2x
dt
2

b dx k
 x0
m dt m
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Oscillations and Waves
b
= damping coefficient.
2m
This above equation is known as differential.
Its solution may be written as
where  

x  et  A1e

2 2t
 B1e
2  2t



where A1 & B1 are two constants:
The solution present three distinct cases depending on the value of  and  .
Case-1 (  )
if  >  then the oscillator is known as under Damped Osillator..
if  <  then the oscillator is known as over Damped non-osillator..
if  =  then the oscillator is known as Critical Damped non-osillator..
 >  (Under Damped Osillator)
Case-1
In this case  >  , so 2  2  12  say 

2  2  i1
Now the solution of Damped Harmonic Osillator become
x   A cos  i t     e t .
The above solution contains cosine term which represent for the periodic motion of the
particle with time period, T1 
But we know
So
T
2
2

1
2  2
2

T < T1

1 1

f f1
 f1  f
Hence the frequency of damped ossillator is less than undamped osillator.


 t
The amplitude term Ae
which decreases exponentially with time.
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x(t)
Amplitute/displacement
Engineering Physics
time
Case-II    (Over Damped Non-Osillatation)
In this case    , so 2  2   ve   2  say 

 2  2  
Now the solution of Do over Damped Non-Osillator become

x  e t  A1e

2 2 t
 B1e 
2  2 t

t
t
 t 
  e  A1e  B1e


where A1 and B1 are constant.
  t
  t
x   A1 e    B1 e    .




    t
  t
& B1 e   which decreases
The above solution contain amplitude term A1 e
exponentially with time.
x(t)
t
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Oscillations and Waves
Case-III      (Critical Damped Non-Osillatation)
In this case    , so 2  2 = 0

 2  2  
Now the solution of Damped Harmonic Osillator is x  e t  A1  B1  .
The above solution contain amplitude term et  A1  B1   which decreases asymptotically
with time.
x(t)
t
DECREMENT :
The ratio between two successive maxima of damped harmonic ossillator is known as
decrement.
Decrement  D  
Aet
 t  T
Ae  1 
D  eT1
Logarithimic Decrement :
When the decrement is expressed in terms of logarithm, then it is called logarithimic decrement.
Logarithim Decrement   T1 
2
2  2
Q.1.
.
A natural angular frequency of simple Harmonic Osillator of mass 2kg is 0.8 rad/sec. It
under goes criticaly damped motion when taken to a visus medium. Find its damping
force on the oscillator when its speed is 0.2 m/sec.
Ans: Given that m - 2kg
angular frequency  = 0.8 rad/sec., V = 0.2 m/sec
Given that criticaly damped so   
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Engineering Physics
b
b
b



  0.8
2m
2m
4
 b = 0.8 × 4
 b = 3.2
= bv = 3.2 × 0.2 = 0.64
d
Q.2. The timp period for SHO is 2s. It subjected to damping force proporsal to speed with
samping coefficient 1 set–1. Find cogarithimic decrement and time period.
Ans: Time period of SHO = 2S., T = 2S.
we know
F
Fd  bv 
T 

1 1
V...., B  1 sec1 , f    0.5 ,   2f  2    0.5  3.14
2m
T 2
2
 3.14 2  12
 2.1109sec
logarithimic decrement = 1 × 2.1109 = 2.1109
Q.3.
1
of its initial value after 10
10
number of Oscillation. If time period is 2 sec. Then calculate damping coefficient &
logarithimic decrement.
The amplitude of an under damped Oscillator falls to
Ans: The equation of under damped Oscillator is x  Aet cos  1t   
The amplitude term is Aet
After 10 number of Oscillation the amplitude term is Ae t 10T 
Aet
1

 t 10T  10
Ae

et
1

 t  20  10
A 
  t 
t   20
10
 t  20 
 10et  e 
  10 t   t  20  .
2.4 F ORCED OSCILLATOR
The Oscillator which is subjected to external periodic force is known as forced Oscillator.
If the particle displace a displacement x from mean position, then it experiences three
forces. One is restoring force, F r = – kx, damping force Fd = –bV.
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Oscillations and Waves
External periodic force experienced, Fe  F0 cos 0 t , where F 0 = magnitude of force.
Now the total force experienced by the particle, F = F r + F d + F e
 ma   kx  bv  F0 cos 0 t


d2x
dt 2
d2x
dt


2
2
d2x
dt
2
  kx  b
dx
 F0 cos 0 t
dt
k
b dx F0
d2x
dx F
x
 cos 0 t  2  2 x  2  0 cos 0 t
m
m dt m
dt m
dt
 2 x  2
d2 x
m
 2
dx F0
 cos 0 t
dt m
F
dx
 2 x  0 cos 0 t
dt
m
dt
This is known as differential equation for Forced Osillator. Its solution may be written as
x = xc(t) + xp(t)

where xc(t) = Complementary solution  e t  A1e

And xp(t) = Particular solution

F0 m

2

2
 02
2 2t
 B1e 
2  2t



cos  0 t   

4 2 02
where  is phase-difference between Osillator and external periodic force.
The complementary solution contain et which dies away exponentially with time.
After long time the forced Osillator contain only particular solution. At that time the
solution is known as steady state solution of forced Oscillators.
Resonance :
The motion is which a body osillate with its natural frequency under the influence of
external periodic force having same frequency is called resonance. In this case the
frequency of forced Oscillator (  ) is nearly equal to the frequency of external periodic
force (  0). Hence the amplitude of forced Oscillator is maximum.
Mathmaticaly A max 
F0 m
20
Condition for resonance :
The amplitude of forced Oscillator is maximum when demomenator of expression for
amplitude is minimum.
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Engineering Physics
So

d  2
  02
d0 



2

 42 02   0

d  2
  02

d0 

   dd 0 4202  0
2

 2 2  02   20   42  20  0


  40 2  02  8 20  0


 2  02  2 2


 40 2  02  8 20
 02  2  2  2
 0  2  2 2
Sharpness of Resonance :
The vamplitude of forced Oscillator is maximum at the frequency and decreases rapidly
as the frequency increases or decreases from the resonance frequency. This is known
as sharpness of Resonance.
Zero damping curve
Weak damping curve
Moderate damping curve
Heavy damping curve
f = fn
f
2.5 COUPLED OSCILIATOR
When two or more body oscillates in such way that osicllation of one influences the
oscillation of other, then the resulting oscillation is known as coupled oscillation and the
bodies to constitute a coupled system.
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Oscillations and Waves


l
l
A
in 
mg s
m

y
B
90°
90°
gc

os
s
mg c o
x
mg sin 

mg
mg
Consider a coupled system consisting of two identical pendulum of mass connected by
a spring of spring constant k. Due to some external agent, let the pendulum ‘A’ displaces
a displacement x and the pendulum B displaces a displacement y. So change in length
of the spring is x–y.
The restoring force on A due to spring, F AS = – k(x – y)
The restoring force on B due to spring, F BS = k(x – y)
The restoring force on pendulum A due to give g, FAg   Mg sin    mg x 
The restoring force on pendilum B due to gravity ‘g’, FBg   mg sin    mg y 
The total force experienced by the pendulum A,
F = F As + F Ag ,
m
d2x
dt

d2x
2

ma   k  x  y   mg x 
  k  x  y   mg x 

d2x
dt
2

k
 x  y  g x 
m
k
 x  y  g x   0
m
(2.1)
dt
Similarly, Total force experienced by the pendulum B, F = F Bs + F Bg
2
 ma  k  x  y   mg y 

d 2 xy
dt
2

k
 x  y  g x 
m
m
d 2 xy
dt 2
 k  x  y   mg x 
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Engineering Physics

d 2 xy

k
 x  y  g y   0
m
(2.2)
dt
This above two equations [(2.1) and (2.2)] are coupled equation which can be recast os
convert in to a pair of decoupled equation in terms of new variables.
Now adding and substracting above two equation, we get.
d2
g
 x  y   x  y  0
(2.3)
2

dt
d2
dt
2
2
 x  y 
2k
 x  y  g   x  y  0
m
(2.4)
Now putting x  y  , x  y  
Hence Eq(2.3) becomes

d 2
dt
2
 12   0
And Eq(2.4) becomes


(2.5)
d 2
dt
2

2k
  g   0
m
d   2k


 g    0
2
m

dt
2
d2 
dt
2
 22   0
(2.6)
The above equation [(v) & (vi)] are decoupled equation in terms of new variable ‘  ’ and ‘  ’
where 1  g  ,   2k m  g 
Normal co-ordinate :
The new-co-ordinate by which a couple system describe is known as normal co-ordinate.
Normal code of vibration :
When the oscillation is expressed in terms of normal co-ordinates is called normal
mode of vibrations.
It is of two type.
(i) In-phase mode of vibration (ii) Out of phase of vibration.
In-Phase mode of vibrations :
If both pendulum oscillates with same phase, in same displacement, same distances is
called In-phase mode of vibration.
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Oscillations and Waves
Out-phase mode of vibration :
If both pendulum oscillates with same displacement but in opposite direction, then this
type of oscillation is called out phase mode of vibration.
x
x =y
y
y
x
(Out-phase mode of vibration)
(In-phase mode of vibration)
Q.
If two pendulum, each mass 0.6 kg and length 1m are connected by a spring of spring
const. 0.3Nm–1. Find the normal mode of frequency of coupled Oscillator.
Ans: Given m = 0.6 kg. l = 1m, k = 0.3 Nm–1.
1  g  
9.8
 3.13
1
2  0.3
2k
g  
 3.13 = 1 + 3.13 = 3.28.
m
0.6
2 
2f1  3.13
 f1 
3.13
3.13

 0.498Hz
2  3.14 6.28
2f 2  3.28
3.28
 0.52 Hz .
6.28
A forced harmonic oscillation have equal displacement at 1  300 Hz, 2  400 Hz .
Find the resourant frequency at which the displacement is maximum.
Given frequencies 1  300 Hz, 2  400 Hz
The displacement is maximum
Also given equal displacement
 f2 
Q.
Ans.
F0 m
 2  12 
2
 4212

F0 m
2  22  4222
Resonance frequencies r  0  2  2 2
... (i)
32
Engineering Physics
2  2  22
 2  2  22  2  2  22
Putting the value of 2 2 in eq(i)
1


2
2  12

 42 12
1
 2  22 
 2  22  4 2 22  2  12  4 2 22

2
 4222
  22  12  4 2 22  4 2 22  0

 12  22  42 22  22  0
2.6 WAVES – INTRODUCTION
Awave is a disturbance in a medium that caries energy without a net movement of particles.
A wave:
• transfers energy.
• usually involves a periodic, repetitive movement.
• does not result in a net movement of the medium or particles in the medium (mechanical
wave).
There are some basic descriptors of a wave. Wavelength    is distance between an
identical part of the wave.
one oscillation
(frequency is number of
oscillations per second)
Amplitude is maximum displacement from neutral position. This represents the energy
of the wave. Greater amplitude carries greater energy.
Displacement is the position of a particular point in the medium as it moves as the wave
passes. Maximum displacement is the amplitude of the wave.
Frequency (f) is the number of repetitions per second in Hz, s–1
Period (T) is the time for one wavelength to pass a point. T = f–1
Oscillations and Waves
33
The velocity    of the wave is the speed that a specific part of the wave passes a
point. The speed of a light wave is c.
We will deal with two types of waves:
• A transverse wave has the motion of the medium perpendicular to the movement
of the wave pulse.
• A longitudinal wave has the motion of the medium parallel to the movement of the
wave pulse.
For most waves, the particles of the medium move in a repetitive way that results in no
net displacement.
A transverse wave has the displacement of the particles in the medium moving
perpendicular to the direction of the wave’s movement
Physics – Waves
Examples of transverse waxes:
• Water waves (ripples of gravity waves, not sound through water)
• Light waves
• S-wave earthquake waves
• Stringed instruments
• Torsion wave
The high point of a transverse wave is a crest. The low part is a trough.
A longitudinal wave has the movement of the particles in the medium in the same
dimension as the direction of movement of the wave. Examples of longitudinal waves:
• Sound waved
• P-type earthquake waves
• Compression wave
34
Engineering Physics
Longitudinal waves create areas of compression where particles are pushed together
(higher density), and rarefaction where particles are pulled apart (lower density)
Sound waves are often represented by a transverse wave (sinusoidal wave).
Pressure
wavelength
wavelength
Distance
condensation
Both a transverse and longitudinal wave can be described with a displacement time
graph. Why’?
If a single point of the medium is examined over time, its motion will be periodic.
3
amplitude
2
1
0
-1
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
time
-2
-3
2.7 WAVES
The displacement position graph takes a picture of the medium at a specific time. The
displacement is for the medium. The position is for the progress of the wave. Why
does the same graph describe both types of waves?
displacement
of the medium

distance measured
along the wave
35
Oscillations and Waves
As a wave passes a point, the speed of the wave will be measured by the repeated,
motion. If the time is measured between two crests in a wave, the speed is the wavelength
divided by the period.
Ex:
A person is standing on a dock. The person starts a clock as one crest passes them. As
the fifth crest passes, the watch reads 3.5 s. A crest takes 4.7s to pass along the 3.2 m
of the dock. What can you describe quantitatively about the wave?
2.8 ONE DIMENSIONAL WAVES
We will examine one dimensional waves such as a transverse wave on a rope or spring,
and longitudinal waves on a spring (slinky).
As a mechanical wave reaches the end of its medium, it will reflect. The energy it
contains will not just disappear.
The reflection will vary for a hard (fixed) boundary, and for a soft (flexible or movable)
boundary.
The reflected wave will be upright for a soft boundary, and inverted for a fixed boundary.
A wave that reaches a change in its medium, will be have its speed changed as it
passes into the new medium (refraction), and it will also reflect at the new medium (a
type of boundary).
Traveling into a slower medium is like a hard boundary Traveling into a faster medium
is like a soft boundary. Watch the speed of the refracted wave, and the nature of the
reflected wave
2.9 WAVES – STANDING SOUND WAVES
A sound wave is a longitudinal wave that is often drawn as a transverse sine wave.
Standing sound waves are used in musical instruments. An instrument resonates.
Pressure
wavelength
wavelength
Distance
condensation
36
Engineering Physics
If an object can vibrate at a natural frequency, it can resonate if an oscillation causes it
to vibrate at its natural frequency.
The resonating object will amplify-the affect of the original vibration.
A musical instrument has a method to vary its resonant frequency.
Resonance is an example of a standing wave. This wave can be of various harmonics.
A harmonic describes how many standing waves are present.
We will examine resonant standing waves in two situations: strings and pipes with air.
A standing wave is a wave that is traveling back and forth and reflecting on a medium.
The boundary conditions are important because it affects the way that the wave reflects.
We will start with situations with both ends (reflection points) are hard reflectors (fixed):
sinusoidal, longitudinal
For a standing wave, this means the fixed ends will form a node. This is used in stringed
instruments.


f1 

The fundamental has a length of 1  2 ,
1 2
For second harmonic:  2  2 2 ,
f2 
 

2 
For the nth harmonic:  n  2 n ,
fn 
 n

 n 2
A pipe that is open at both ends, will form a standing wave with a soft reflection at both ends.
3/2 
4/2 
5/2 
The formulas to calculate the frequencies are the same as for a double closed pipe.
Why?
37
Oscillations and Waves
A pipe with one end closed is the most common for instruments. It had one hard reflection
and one soft reflection. The first harmonic has a node at the closed end and an antinode at the open end.
1  4


The fundamental has a length of 1  4 , f1    4
 3
f2 

 2 4
For first harmonic :  2  4 3 ,
1
  2n  1 

n
4
How does a pipe organ have different harmonics for different notes?
How does a flute or a trumpet have different harmonics for different notes?
What is the “sweet spot” on a bat or club?
For the nth harmonic :  n  4 2n , f n 
2.10 EQUATION OF
PROGRASSIVE WAVE:
Displacement
Consider a progressive wave travel from left to right from the origin.
The wave has reached B after travelling a distance AB = x from A.
A
O
B
X-Axis
Direction of wave
As the particle execute SHM, the distance ment of the particle at origin ‘O’ at any in
start ‘t’ can be written as
y  A sin t
where A = amplitude of wave.
Similarly a particle is vibrating at point ‘B’ at a distance ‘x’ from origin.
y  A sin  t    .
where  = phase difference
2
x

2 
2 

 2
 y  A sin  t 
x   A sin  t 
x
 
 

 T

38
Engineering Physics
 t x
 v x
2
 A sin 2   
 A sin 2  t     A sin
v t  x .

T 
  
This is the equation of propressive wave travelling along + ve x axis.
Correspondingly, a wave travelling in the – ve direction of x-axis may be represented as
2
y  A sin
v t  x .

In terms of kx,
y  A sin  t  kx  along + ve x-axis
(2.7)
y  A sin  t  kx  along –ve x-axis.
Differentiating above equation (2.7) twice w.r.t. t, we get
2 y
  A2 sin  t  kx 
2
dt
2 y

2
  2 y
dt
Again differentiating equn(2.7) twice w.r.t. x, we get
2 y
  Ak 2 sin  t  kx 
dx 2

2 y
2
  k2y
dx
From equns(2.8) and (2.9), it is clear that :
2 y
dt 2

2  2 y
k 2 dx 2

2 y
dt 2
 v2
(2.8)
(2.9)
2 y
dt 2
This is called differential equation of progressive wave.
2.11 SUPER POSITION OF WAVE
The phenomenon in which two or more waves moving simulataneously combine is
known as super position.
When more than one waves simuntareously propagate through a region, the resultant
wave function is linear combination of thewave function of indivisual waves :
Coherent super position :
The superposition between the wave is said to be coherent if the phase difference
between them is constant.
In this case the intensity of resultant wave is differ from sum of intensities of indivisual
wave.
39
Oscillations and Waves
In-Coherent Super position :
The super position between the wave is said to be in-coherent if the phase difference
between them is not constant or frequently changes.
In this case the intensity of resultant wave is equal to the sum of intensities of indivisual wave.
2.12 TWO BEAM SUPERPOSITION
Let how waves having same frequency but different amplitude and different phases
emitted fr om two sour ces. The indivisua l wa ve r epr esented a s
y1  A1 sin  t  kx  1  and y 2  A 2 sin  kx  t  2 
So resultant value, Y  Y1  Y2
 A1 sin  kx  t  1   A 2 sin  kx  t  2 
 A1 sin  kx  t   cos 1  sin 1  cos  kx  t 
 A 2 sin  kx  t   cos 2  sin 2  cos  kx  t 
 sin  kx  t   A1 cos 1  A 2 cos 2   cos  kx  t 
 A1 sin 1  A 2 sin 2 
= put A1 cos 1  A 2 cos 2  A cos 
(2.10)
and A1 sin 1  A 2 sin 2  A sin 
Putting the value we have
(2.11)
Y  sin  kx  t   A cos   cos  kx  t   A sin 
Y  Asin  kx  t   
Resultant Amplitude :
Squaring Eq(i) & (ii) we have
A12 cos 2 1  A 22 cos 2 2  2A1A 2 cos 1  cos 2  A 2 cos 2
A12 sin 2 1  A 22 sin 2 2  2A1A 2 sin 1  sin 2  A 2 sin 2
A12  A 22  2A1A 2 cos  2  1   A 2
 A  A12  A 22  2A1A 2 cos  2  1  .
Coherent superposition : In this case the phase difference is constant.
Case-I : Intensity shows maximum value if cos  2  1   1
 cos  2  1   cos 2n 
 2  1  2n  , where n = 0, 1, 2 ....
40
Engineering Physics
So maximum amplitude
A max  A12  A 22  2A1A 2  1 ,
A max 
 A1  A 2 2
2
A max
  A1  A 2  .
Case-II
Intensity shows minimum value if cos  2  1   1
 cos  2  1   cos  2n  1 
 2  1   2n  1 
So minimum amplitude,
A min  A12  A 22  2A1A 2 ,
A min 
 A1  A 2 2
A min  A1  A 2 .
Case-III
If A1 = A2 then
A max  A12  A12  2A12 = 2A1 = 2A2.
Intensity (I)  4A12  4A 22 .
If A1 = – A2, then
A min  A12 = 0.
Incoherent superposition :
In this case phase difference is not constant. The term cos  2  1  changes rand only
between +1 to –1. So the time average of phase.
A  A12  A 22 .
41
Oscillations and Waves
SOLVED EXAMPLES
Example-1.
A simple harmonic motion is represented by
x =10 sin (20 t + 0.5)
Write down its amplitude, angular frequency, frequency, time period and initial
phase, if displacement is measured in metres and time in seconds. [Himachal 09C]
Soln. Given x = 10 sin (20 t + 0.5)
Standard equation for displacement in SHM is x = A sin (t + 0)
Comparing the above two equations, we get
(i) Amplitude, A = 10 m.
[  A and x have same units]
(ii) Angular frequency, = 20 rad s–1.
(iii) Frequency, v =
 20 10


= 3.18 HZ.
2 2 
2 2 


= 0.314 s.
 20 10
(v) Initial phase, 0 = 0.5 rad.
Example-2.
A particle executes SHM with a time period of 2 sand anfpzitude 5 em Find (i)
displacement iii) velocity and (iii) acceleration, after 1/3 second; starting from
the mean position.
Soln. Here T = 2 s, A = 5 cm, t = 1/3 s
(i) For the particle starting from mean position,
(iv) Time period, T =
D·ISPIacernent, x = A sin t = A sin
 5sin
(ii) Velocity, v 
2
t
T
2 1

3
  5sin  5 
= 4.33 cm.
2 3
3
2
dx 2A
2
2  5


cos t 
cos  5  3.14  0.5  7.85cms 1
dt
T
T
2
3
(iii) Acceleration, a 
4  9.87  5

dv 42 A
2
sin
 2 sin t 
4
3
dt
T
T
= 9.87 × 5 ×
3
= 42.77 cm s–2.
2
42
Engineering Physics
Example-3.
A body oscillates with SHM according to the equaiion :
x(t) =5cos (2t + /4),
where t is in sec. and x in metres. Calculate
(a) Displacement at t = 0 (b) Time period (c) Initial velocity
Soln. Given x(t) = 5 cos(2t + /4)
We compare with standard equation, x(t) = Acos(t + 0)
(a) Displacement at t = 0, x(0) = 5cos

1
5
 5

m.
4
2
2
2
= 2
T
 Time period, T = 1 s.
(b) Clearly,  = 2 or
(c) Velocity, v =


dx
= –5 sin  2t   × 2
4
dt

Initial velocity at t = 0, v = –10sin
 10
=
m / s.
2
4
Example-4.
A body oscillates with SHM according to the equation,
x = (5.0m) cos [(2 rad s–1)t + /4].
At t = 1.5 s, calculate (a) displacement, (b) speed and (c) acceleration of the body.
Soln. Here  = 2 rad s–1, T = 2/ = 1 s, t = 1.5 s
(a) Displacement,
x = 5.0 cos(2 × 1.5 + /4)= 5.0 cos (3+ /4)
= – 5.0 cos /4 = – 5.0 × 0.707 = –3.535 m.
dx d

[5.0 cos (2t + /4)]
dt dt
= – 5.0 × 2 sin (2t + /4) = – 5.0 × 2 sin (2 × 1.5 + /4)
(b) Velocity, v =
= + 5.0 × 2 sin /4 = 5.0 × 2 ×
22
× 0.707 = 22.22 m s–1.
7
dv
d
=
[–10 sin (2t + /4)]
dt
dt
= –202 cos (2t + /4) ,
= – 42 [5.0 cos (2 × 1.5 + /4)]
= – 4 × 9.87 × (–3.535)
[Using (a)]
= 139.56 m s–1.
(c) Acceleration, a =
43
Oscillations and Waves
Example-5.
The equation of a simple harmonic motion is given by y = 6 sin 10 t + 8 cos I0
t, where y is in em and t in sec. Determine the amplitude, period and initial phase.
Soln. Given y = 6 sin 10 t + 8 cas 10 t
...(1)
The general equation of SHM is
y = a sin (t + ) = a sin rot cos  + a cos rot sin 
= (a cos ) sin t + (a sin ) cos t
...(2)
Comparing equations (1) and (2), we get
a cas = 6
...(3)
a sin = 8
...(4)
and
t = 10 t or  = 10
2 2

 0.2s.
 10
Squaring and adding (3) and (4), we get a 2 (cos2  + sin2 ) = 62 + 82
= 36 + 64 = 100 or a 2 = 100
 Amplitude; a = 10 cm
 Time period, T 
8
= 1.3333
6
 Initial phase,  = tan–1 (1.3333) = 53°8'.
Example-6.
A particle executes S.H.M. of amplitude 25 cm and time period 3 s. What is the
minimum time required for the particle to move between two points 12.5 em on
either side of the mean position ?
Soln. When the particle starts from mean position, its displacement at instant t is given by
y = A sin  t
Given A = 25 cm, T = 3 s, y = 12.5 cm
2 2


rad s–1
T
3
Dividing (4) by (3), we get,

12.5 = 25 sin
or
sin
tan =
2
t
3
2 12.5 1
t

3
25 2
2t 
1

or t  s
3
6
4
 Time taken by the particle to move between two points 12.5em on either side of
mean position is given by

2t  2 
1 1
 s  0.5s
4 2
44
Engineering Physics
Example-7.
The shortest distance travelled by a particle executing SHM from mean position
in 2 s is equal to ( 3 /2) times its amplitude. Determine its time period.
Soln. Here t = 2 s, y = ( 3 / 2 ) A, T = ?
As y = A sin t = A sin


2
t
T
3
2  2
A  A sin
2
T
or
sin
4
3


 sin
T
2
3
4 

T
3
Example-8.
The time-period of a simple pendulum is 2 s and it can go to and fro from equilibrium
position at a maximum distance of 5 em If at the start of the motion the pendulum
is in the position of maximum displacement towards the right of the equilibrium
position, then write the displacement equation of the pendulum.
Soln. The displacement in SHM is given by y = A sin (t + 0)
Given T = 2 s, A= 5 cm
2 2

  rad s–1
T
2
y = 5 sin(t + 0)

At time t = 0, displacement y = 5 ern. Therefore, 5 = 5 sin (× 0 + 0)
or sin 0 = 1 
0 = /2
Hence displacement equation for the pendulum is
=


y = 5 sin  t   = 5 cos t.
2

Example-9.
For a particle in SHM, the displacement x of the particle as ajunction of time t is
given as
x = A sin (2 t)
Here x is in em and t is in seconds.
Let the time taken by the particle to travel from x = 0 to x = A / 2 be t 1 and the time
taken to travel from x = A / 2 to x = A be t2. Find tl / t2
Soln. Here x = 0 at t = 0.
2
 2
 T = 1s
T
At t = t1, x = A/2. Therefore,
Also  
45
Oscillations and Waves
A
= A sin (2t1)
2

2t1 
6
1
= sin (2 t1)
2
1
t1  s
or

12
T 1
Time taken from x = 0 to x = A is I.  ss
4 4
T 1
1 1 1

s
t2    s
or
t1 + t2 =
or
4 4
4 12 6
t1 1 / 12 1
Hence


t2 1 / 6 2
or
Example-10.
A particle is moving with SHM in a straight line. When the distance of the particle
from the equilibrium position has values x l and x2 the corresponding values of
velocities are u1 and u2. Show that the time period of oscillation is given by
1/2
 x2  x2 
T  2  22 12 
 u1  u2 
Soln.
When
x  x1 , v  u1
When
x  x2 , v  u 2
As v   A2  x 2
u1   A2  x12
or
u12  2 ( A2  x12 )
...(1)
and u2   A2  x22
or
u22  2 ( A2  x22 )
...(2)

Substracting (2) from (1), we get
u12  u22  2 ( A2  x12 )  2 ( A2  x22 )  2 ( x22  x12 )
1/2
or
 u 2  u22 
   12
2
 x2  x1 
1/2
,
 x2  x2 
2
T
 2  22 12 

 u1  u2 
Example-11.
If the distance y of a point moving on a straight line measured from afixed origin
on it and velocity v are connected by the relation 4v2 = 25 – y2, then show that the
motion is simple harmonic and find its time period.
Soln. Given 4v2 = 25 – y2
or
v
1
26  y 2
2
46
Engineering Physics
Also velocity in SHM, v   A2  y 2
Comparing the above two equations, we find that the given equation represents SHM
 = 1/2 rad s–l.
o
f
a
m
p
l
i
t
u
d
e
A
=
5
Time-period, T 
a
n
d
2 2  2

 4s.

1
Example-l2.
A particle executing SHM along a straight line has a velocity of 4 ms –1 when at a
distance 3 mfrom the mean position and 3 ms–1 when at a distance of 4 mfrom it.
Find the time it takes to travel 2.5 mfrom the positive exiremity of its oscillation.
Soln. When y1 = 3 m, v1 = 4 ms–1
When y2 = 3 m, v2 = 4 ms–1
As
v   A2  y 2
or
16  2 ( A2  9)
 4   A2  32
...(1)
3   A2  42 or 9  2 ( A2  16)
Dividing (1) by (2), we get
and
or
...(2)
16 A2  9

or 16 A2  256  9 A2  18
9 A2  16
7 A2 = 256 –81 =175 or A2 = 25
A = 25 = 5m
From (1), 4 =  52  32   4
= l rad s–1
When the particle is 2.5 m from the positive extreme position, its displacement from the
mean position is y = 5 – 2.5 = 2.5 m
When the time is noted from the extreme position, we can write y = A cos t
2.5 = 5 cas (1 × t)

or
cos t 
2.5 1

  cos
5
2
3
Hence
t
 3.142

 1.047 s.
3
3
Example-13.
A particle executing linear SHM has a maximum velocity of 40 cm s –1 and a
maximum acceleration of 50 cm s–2. Find its amplitude and the period of oscillation.
Soln. Maximum velocity, vmax = A = 40 cm s-1
Maximum acceleration, a max = 2 A = 50 cm s–2

amax 2 A 50


vmax
 A 40
or  
5
rad s–1.
4
47
Oscillations and Waves
Amplitude, A 
vmax 40  4

32 crn.

4
Period of oscillation, T 
2 2  3.142  4

 5.03 s.

5
Example-l4.
The vertical motion of a huge piston in a machine is approximately simple harmonic
with a frequenoj of 0.50 5–1. A block of 10 kg is placed on the piston. What is the
maximum amplitude of the piston's SHM for the block and the piston to remain together?
Soln. Here y = 0.5 s–1, g = 9.8 In s–2
The maximum acceleration in SHM is given by a max = 2 A = (2 v)2 A = 42 v2 A
The block will remain in contact with the piston if a max < g or 42 v2 A < g
Hence the maximum amplitude of the piston will be
Amax 
g
9.8
 2
= 0.99m.
2 2
4 v
4 (0.5) 2
Example-15.
A block of mass one kg is fastened to a spring with a spring constant 50 Nm –1.
The block is pulled to a distance x = 10 cm from its equilibrium position at x = 0
on a frictionless surface from rest at t = 0. Write the expression for its x(t) and v(t).
Soln. Here m = 1 kg, k = 50 Nm–1, A = 10cm = 0.10 m
k
50

= 7.07 rad s–1.
m
1
As the motion starts from the mean position, so the displacement equation can be
written as x(t) = A sin t
or x(t) = 0.10 sin 7.07 t

and
dx
= 0.1 0 × 7.07 cos 7.07 t
dt
v(t) = 0.707 cos 7.07 t ms–1.
v(t) =
or
Example-16.
A block whose mass is 1kg is fastened to a spring. The spring has a spring constant
of 50 N m–1. The block is pulled to a distance x = 10 emfrom its equilibrium position
at x = 0 on a frictionless surface from rest at t = 0. Calculate the kinetic, potential
and total energies of the block when it is 5 cm away from the mean position.
Soln. Here m = l kg, k = 50 N m–1, A = 10 cm = 0.10 m, y = 5 cm = 0.05 m
1
1
Kinetic energy, Ek =
k(A2 – y2) = × 50 [(0.10)2 – (0.05)2] = 0.1875 J.
2
2
1 2 1
Potential energy, Ep = ky =
× 50 × (0.05)2 = 0.0625 J.
2
2
Total energy, E = Ek + Ep = 0.1875 + 0.0625 = 0.25 J.
48
Engineering Physics
Example-17.
A body executes SHM of time period 8s. If its mass be 0.1 kg, its velocity 1 second
after it passes through its mean position be 4 ms –l, find its (i) kinetic energy (ii)
potential energy and (iii) total energy.
Soln. Here m = 0.1 kg, T = 8 S
But
2 2 

 rad s–1
T
8 4
v = A cos t

4

When t = 1 s, v = 4 ms–1

1
  
 A cos   1   A 
4
2
4  4
or A 
16 2
m

2
2
 16 2 
1
1

Total energy, E  m2 A2   0.1     
 = 1.6 J.
2
2
 4    
1
1
Kinetic energy, Ek  mv 2   0.1 (4)2 = 0.8J.
2
2
Potential energy, Ep = E – Ek = 1:6 – 0.8 = 0.8 J.
Example-18.
A spring offorce constant 800 Nm–1 has an extension of 5 cm. What is the work
done in increasing the extension from 5 cm to 15 cm ?
Soln. Here k = 800 Nm–1, xl = 5 cm = 0.05 m, x2 = 15 cm = 0.15 m
W = Increase in P.E. of the spring
1
1
 k ( x22  x12 ) =
× 800 [(0.15)2 – (0.05)2] J = 8 J.
2
2
Example-19.
A particle of mass 10 g is describing SHM along a straight line with a period of
2 sand amplitude of 10 em What is the kinetic energy when it is (i) 2 cm (ii) 5 cm,
from its equilibrium position ? How do you account for the difference between its
two values ?
Soln. Velocity at displacement y is
v   A2  y 2
Given A = 10 cm, T = 2 s
Angular frequency,  
(i) When y = 2 cm,

K E.=
2 2

= rad s–2
T
2
v   100  4   96 cm s–l
1
1
mv2 =
× 10 × 2 × 96
2
2
= 480 2 erg.
49
Oscillations and Waves
(ii) When y = 5 cm,
v   100  25   75 cm s–l
1 2
1
mv = × 10 × 2 × 75 = 375 2 erg.
2
2
The KE. decreases when the particle moves from y = 2 em to y = 5 em. This' is due to
the increase in the potential energy of the particle.
Example-20.
At a time when the displacement is half the amplitude, what fraction of the total
energy is kinetic and what fraction is potential in S.H.M ?
1
1
Soln. Displacement =
amplitude or y =
A
2
2
1
2 2
Total energy of SHM, E  m A
2
1
2
2
2
Kinetic energy of SHM Ek  m ( A  y )
2

KE.=
2

1
 A 
 m2  A2    
2
 2  

 3 A2  3 1
1
3
2 2
 m2 
  . m A  E
2
4
 4  4 2
Potential energy of SHM,
Ep 
1
1
 A
m2 y 2  m2  
2
2
2
2
1 1
1
 . m2 A2  E.
4 2
4
Example-21.
A particle is executing SHM of amplitude A. At what displacement from the mean
position, is the energy half kinetic and half potential?
Soln. As EK = Ep

or
or
1
1
m2 ( A2  y 2 )  m2 y 2
2
2
A2  y 2  y 2
A2
y2 
2
or
or
2 y 2  A2
A
y
2
Thus the energy will be half kinetic and half potential at displacement
side of the mean position.
A
on either
2
50
Engineering Physics
Exc:mple-22.
A particle executes simple harmonic motion of amplitude A. (i) At what distance
from the mean position is its kinetic energy equal to its potential energy? (ii) At
what points is its speed half the maximum speed ?
Soln. The potential energy and kinetic energy of a particle at a displacement y are given by
1
E p  ky 2
2
1
Ek  k ( A2  y 2 )
and
...(1)
2
where A is the amplitude and k is the force constant.
(i) As Ek = Ep
A
1
1
  0.71 A
k ( A2  y 2 )  ky 2 or
2y2 = A2
or y  
2
2
2
= 0.71 times the amplitude on either side of mean position.

1
v
2 max
In general, kinetic energy
2
1
1 1
1 1 2
1

 mv 2  m  vmax   . mvmax
  Maximum kinetic energy
2
2 2
4 2
4

1
Ek   ( Ek ) max
or
...(2)
4
1
From equation (1), Ek  k ( A2  y 2 )
2
1 2
[Put y = 0]
 ( Ek ) max  kA
2
Putting these values in equation (2), we get
(ii) Here, v =
or
1
1 1
k ( A2  y 2 )   kA2
2
4 2
4y2 = 3A2
or
y= ±
3
A = ± 0.86 A
2
= 0.86 times the amplitude on either side of mean position.
Example-23.
A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched
string is 20.0 m If a transverse jerk is struck at one end of the string, how long
does the disturbance'1 pipe take to reach the other end ?
51
Oscillations and Waves
Soln.
2.50
kg m–1, T = 200 N
20.0
Speed of the transverse jerk is
Given m =
v
T
200  20.0

 1600  40 ms 1
m
2.50
Distance 20
 Time taken by the jerk to reaeh the other end Speed  40 = 0.5 s.
Example-24.
A steel wire has a length of 12.0m and a mass of 2.10 g. What should be the
tension in the wire so that the speed of a transverse wave on the wire equals the
speed of sound in dry air at 20°C which is 343 ms–1 ?
Soln. Speed of a transverse wave in the steel wire is given by
v
Given
T
T
2
or v 
or T = v2 m
m
m
v = 343ms–1, m = 2.10 kg m–1
T = (343)2 ×
2.10
= 2.06 × 104 N.
12.0
Example-25.
You have learnt that a travelling wave in one dimension is represented by afunction
y = f(x, t) where x and t must appear in the combination x – vt or x + v t, i.e.,
y = F (x ± vt). Is the converse true? Examine if the following functions for y can
possibly represent a travelling wave:
(i) (x – vt)
(ii) log[(x + v t)/x0]
(iii) exp [–(x + v t)/x0]
(iv) 1/(x + v t).
Soln. If y = f (x ± vt) represents a travelling wave, then the converse may not be true i.e.,
every function of x – vt or x + vt may not always represent a travelling wave. The
basic requirement for a function to represent a travelling wave is that it must be finite
for all values of x and t.
The functions (i), (ii) and (iv) are not finite for all values of x and t, hence they
cannot represent a travelling wave. Only function (iii) satisfies the condition to represent
a travelling wave.
Example-26.
A bat emits ultrasonic sound of frequency 100kHz in air. If this sound meets a
water surface, what is the wavelength of (i) the reflected sound, (ii) the transmitted
sound? Speed of sound in air = 340 ms–1 and in water = 1486ms–1.
Soln. Here v = 100 kHz = 105 Hz, a = 340 ms–1, w = 1486 ms–1
Frequency of both the reflected and transmitted sound remains unchanged. .
52
Engineering Physics
(i) Wavelength of reflected sound,  a 
a 340

= 3.4 × 10–3m.
v 105
(ii) Wavelength of transmitted sound,  w 
w 1486
 5 = 1.49 × l0–2 m.
v
10
Example-27.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the
wavelength of sound in a tissue in which the speed of sound is 1.7 kms –1 ? The
operating frequency of the scanner is 4.2 MHz.
Soln. Here  = 1.7kms–1 = 1.7 × 103ms–1, v = 4.2 MHz = 4.2 × 106 Hz
Wavelength,

 1.7  103

= 4.047 x10 m.
v 4.2  106
Example-28.
A transverse harmonic wave on a string is described by
y(x, t) = 3.0sin(36t + 0.018x + /4),
where x, y are in em and t in s. The positive direction of x is from left to right.
(i) Is this a travelling or a stationary wave ? If it is travelling, what are the speed
and direction of its propagation ?
(ii) What are its amplitude and frequency ?
(iii) What is the initial phase at the origin ?
(iv) What is the least distance between two successive crests in the wave?
Soln. Given:
y (x, t) = 3.0sin (36t + 0.018x + / 4)
...(1)
The standard equation for a harmonic wave is
 2 2

y ( x, t )  A sin  t 
x  0 

T

Comparing equations (1) and (2), we get
...(2)
2
2

 36,
 0.018, 0 
T

4
(i) The given equation represents travelling wave propagating from right to left
(as x term is + ve).
Speed of the wave,
A  3.0,
  / 2 1 / 0.018
36



= 2000cm s–1 = 20 ms–1.
T T / 2
1 / 36
0.018
(ii) Amplitude, A = 3.0cm
1 36 18

Frequency, v  
= 5.73 s–1.
T 2 3.14

53
Oscillations and Waves

rad.
4
(iv) Least distance between two successive crests is equal to wavelength.
2

= 349.0 cm = 3.49 m.
0.018
Example-29.
For the wave described in Exercise 15.8 displacement (y) versus (t) graphs for
x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does
the oscillatory motion in travelling wave differ from one point to another: amplitude,
frequency or phase ?
Soln. The transverse harmonic wave on a string is given by
y(x, t) = 3.0 sin (36t + 0.018x + /4)
Displacement-time graph for x = 0 :
For x = 0, we have y(0, t) = 3.0sin (36t + /4)
a = 3 cm, 0= /4 and  = 36 rad s-l
(iii) Initial phase at the origin,  
2 2 

 s
 36 18
The displacement at different instants of time will be as follows:
T

t
0
T
8
y
3
2
3
2T
8
3
2
3T
8
0
4T
8
3

2
5T
8
2
6T
8
3

2
7T
8
T
0
3
2
The y – t graph will be as shown in Fig.E.2.29
3
y(cm) 
3/ 2
0
T/8
2T/8 3T/8
4T/8
5T/8
6T/8
7T/8
T
3/ 2
–3
t(s)
Fig.E.2.29
Thus, the y – t graph is sinusoidal in nature.
Similarly, we can obtain y – t graphs for x = 2 cm and x = 4 cm. It is seen that all the
three graphs are sinusoidal. They have same amplitude and frequency, but they differ
in initial phases.
54
Engineering Physics
Example-30.
For a travelling harmonic wave
y = 2.0 cos (10 t – 0.0080x + 0.35)
where x and y are in cm and t in s. What is the phase difference between oscillatory
motion at two points separated by a distance of (i) 4 m, (ii) 0.5 m, (iii)
Soln. Given y = 2.0 cos (10t – 0.0080x + 0.35)
The standard equation of travelling harmonic wave is
 2t 2x

y  A cos 

 

 T

Comparing equations (1) and (2), we get

3
(iv)
?
2
4
...(1)
...(2)
2
2
2
2
m
m
= 0.0080 or  
cm 

0.0080
0.0080  100
0.80
Phase difference,
 
2
2
 path difference 
 x


(i) When x = 4 m

(ii) When x = 0.5 m

(iii) When x 
2
 0.80  4  3.2 rad.
2
2
 
 0.80  0.5  0.40 rad.
2
 

2

 
2 
   rad.
 2
3
4

 
2 3 3


rad.

4
2
(iv) When x 
Example-31.
The transverse displacement of a string (clamped at its two ends) is given by
2
y(x, t) = 0.06 sin
x cos (120 t)
3
where x, y are in m and t is in s. The length of the string is 1.5 m and its mass is
3.0 × 10–2kg. Answer the following:
(a) Does the function represent a travelling or a stationary wave?
(b) Interpret the wave as a superposition of two waves travelling in opposite
directions. What are the wavelength, frequency and speed of propagation of each wave ?
(c) Determine the tension in the string.
Soln. The equation for transverse displacement is
y (x, t) = 0.06 sin
2
x cos 120 t
3
...(i)
55
Oscillations and Waves
(a) The equation (i) represents a stationary wave as it involves the product of two
separate harmonic functions of x and t.
(b) Stationary waves are formed by the superposition of two waves of same frequency
travelling in opposite directions. Let the two waves be represented as
2
2
(t  x) and y2   A sin (t  x)


The resultant stationary wave is given by, y = y1 + y2
y1  A sin
2
 2

 A sin (t  x)  sin (t  x) 




2 x
2 t
cos


Comparing the equations (i) and (ii), we get
 2 A sin
and
2 2


3
or
= 3m
2
  120

or
= 60  = 60 × 3 = 180 ms–1
...(ii)
 180

= 60 Hz.

3
(c) Velocity of transverse wave in a string is given by
Frequency,
v

But
T
m
3.0  102
 2  102 kg m 1 ,   180ms 1
1.5
T = v2m = (180)2 × 2 × 10–2 = 648 N.
m
Example-32.
The transverse displacement of a string (clamped at its two ends) is given by
2
y ( x, t )  0.06 sin
x cos 120  t , were x, y are in m and t is in s.
3
(i) Do all the points on the string oscillate with the same
(a) frequency, (b) phase, (c) amplitude?
Explain your answers.
(ii) What is the amplitude of a point 0.375 m away from one end ?
Soln. (i) The transverse displacement is given by
y (x, t) = 0.06 sin
2
x cos 120 t
3
56
Engineering Physics
(a) The time dependent harmonic function cos 120 t of the stationary wave represents
its frequency. As this function does not depend on x, so frequency of oscillation of all
points on the string is same.
(b) The phase of all the points on the string is same for the reasons similar to (a).
(c) The amplitude of stationary wave is given by
2
A = 0.06 sin
x [time independent part]
3
As A depends on x, amplitude of all the points on the string is not same.
(ii) Now, amplitude at a point 0.375 m away from one end is given by
2
× 0.375 = 0.06 sin 0.7854
3
= 0.06 × 0.707 = 0.042 m.
A = 0.06 sin
Example-33.
Given below are some functions of x and t to represent the displacement (transverse
or longitudinal) of an elastic wave. State which of these represent (i) a travelling
wave, (ii) a stationary wave or (iii) none at all :
(a) y = 2 cos (3x) sin (10t)
(b) y = 2 x  t
(c) y = 3 sin (5x – 05t) + 4 cos (5x – 05t)
(d) y = cos x sin t + cos 2x sin 2t.
Soln. (a) As the function is the product of two separate harmonic functions of x and t, so it
represents a stationary wave.
(b) It cannot represent any type of wave.
(c) Here
y = 3 sin (5x – 0.5 t) + 4 cos (5x – 0.5t)
= 3 sin  + 4 cos 
[ = 5x – 0.5t]
If we put A cos  = 3 and A sin  = 4, then
y = a sin (+ )
It represents a simple harmonic travelling wave of amplitude,
1
A  32  4 2  5 and   tan (4 / 3)
(d) It represents the superpositi?n of two stationary waves, one represented by cos x
sin t and another by
Example-34.
A wire stretched between two rigid support vibrates in its fundamental mode with
a frequency of 45 Hz. The mass of the wire is 3.5 × 10–2 kg and its linear densityis
4.0 × 10–2 kg m–1. What is
(i) the speed of a transverse wave on the string, and (ii) the tension in the string?
Soln.Length of the wire,
Mass of the wire (M) 3.5  102

L=
= 0.875 m
Linear density (m)
4.0  102
57
Oscillations and Waves
Given : V = 45 Hz, L= 0.875 m, m = 4.0 × 10–2 kg m–1
1 T
2L m
(i) As
v

45 
1
T
2  0.875 4.0  102
1
T

2
(2  0.875) 4.0  102
or
T = (45)2 × (2 × 0.875)2 × 4.0 × 10–2 = 248 N.
(ii) Speed of the transverse wave,  = 2 v L = 2 × 45 × 0.875= 78.75 ms–1.
Example-35.
A steel rod 100 cmlong is clamped at its middle. The fundamental frequency of
longitudinal vibrations of the rod is given to be 2.53 kHz. What is the speed of
sound in steel ?
Soln. For the fundamental mode of vibration  = 2L = 2 × 100 = 200 crn = 2 m
Frequency, v = 2.53 kHz = 2530 Hz
 Speed of sound,  = v  = 2530 × 2 = 5060 ms–1 = 5.06 km s–1
Squaring, (45)2 =
EXERCISES
1.
2.
3.
Giving examples of each type, distinguish between periodic, harmonic and non-harmonic functions.
What is simple harmonic motion ? Write its any four properties.
Write down the differential equation for S.H.M. Give its solution. Hence obtain expression
for the time period of S.H.M.
4. Prove that the displacement equation x(t) =  cos t + b sin t
represents a simple harmonic motion. Determine its amplitude and phase constant.
5. What is meant by simple harmonic motion ? Give any two examples. Write its differential form.
6. Derive an expression for the instantaneous velocity and acceleration of a particle executing S.H.M.
7. Derive an expression for the instantaneous velocity of the particle executing S.H.M. Find
the position at which the particle velocity is (i) maximum and (ii) minimum.
8. What is S.H.M. ? Show that the acceleration of a particle in S.H.M. is proportional to its
displacement. Also write expression for the time-period in terms of acceleration.
9. Show that in simple harmonic motion (S.H.M.), the acceleration is directly proportional to
its displacement at the given instant.
10. The relation between the acceleration a and displacement x of a particle executing SHM is
 p
a     y; where p and q are constants.
q
What will be the time period T of the particle ?
58
Engineering Physics
11. Find an expression for the·total energy of a particle executing S.H.M.
12. Show that the total energy of a body executing S.H.M. is constant.
13. Showthat the total energy of a particle executing simple harmonic motion is directly
proportional to the square of amplitude and frequency.
14. A body is executing simple harmonic motion. At what distance from its mean position, its
energy is half kinetic and half potential ?
15. Show that the horizontal oscillations of a massless loaded spring are simple harmonic.
Deduce an expression for its time period.
16. Show that when a body is suspended from a spring and is pulled down a little and released,
it executes S.H.M. Also find an expression for its time period. Does it depend on acceleration
due to gravity?
17. What is an ideal simple pendulum? Derive an expression for its time period.
18. What is a simple pendulum ? Show that motion executed by the bob of the pendulum is
S.H.M. Derive an expression for its time period.
19. Show that for small oscillations the motion of a simple pendulum is simple harmonic.
Derive an expression for its time period. Does it depend on the mass of the bob?
20. Prove that if a liquid taken in a U-tube is disturbed from the state of equilibrium, it will
oscillate harmonically. Find expressions for the angular frequency and time period.
21. A ball of mass m fits smoothly in the cylindrical neck of an air chamber of volume V. The
neck area is A. Show that the oscillations of the ball in the neck of the air chamber are
simple harmonic. Calculate the time-period.
22. Show that the angular oscillations of the balancewheel of a watch are simple harmonic.
Hence deduce an expression for the time-period of its oscillations.
23. A cylindrical piece of cork of base area A and height h floats in a liquid of density l. The
cork is depressed slightly and then released. Show that the cork oscillates up and down
h
simple harmonically with a time period T = 2  g , where P is the density of the cork.
1
24.
25.
26..
27.
28.
What are free and damped oscillations? Show them graphically.
With the help of examples, differentiate between free oscillations and forced oscillations.
Briefly explain the principle underlying the tuning of a radio receiver.
What are coupled oscillations? Give examples.
Show that in a S.H.M. the phase difference between displacement and velocity is /2,
and between displacement and acceleration it is .
29. Draw the graphical representation of simple harmonic motion, showing the
(a)
displacement-time curve.
(b)
velocity-time curve and
(c)
acceleration-time curve.
30. Define the term wave motion. Give four important characteristics of wave motion.
31. What are mechanical, electromagnetic and matter waves? Give an example of each
type.
Oscillations and Waves
59
32. What are transverse and longitudinal wave motions? Give an example for each type.
33. Mention the important properties which a medium must possess for the propagation of
mechanical waves through it.
34. Through what type of media, can (i) transverse waves and (ii) longitudinal waves be
transmitted? Explain.
35. Given below are some examples of wave motion. State in each case if the motion is
transverse, longitudinal or a combination of both.
(a) Motion of a kink in a long coil spring produced by displacing one end of the spring
sideways.
(b) Wave produced in a cylinder containing water by moving its piston back and forth.
(c) Wave produced by a motorboat sailing in water.
(d) Ultrasonic waves in air produced by a vibrating quartz crystal.
36. Derive a relation between wave velocity, frequency 1and wavelength.
37. On the basis of dimensional considerations, write the formula for the speed of transverse
waves on a stretched string.
38. On the basis of dimensional considerations, write the formula for the speed of transverse
waves in a solid.
39. Discuss the effect of temperature and density on the velocity of sound in air.
40. Why does sound travel faster in moist air than in dry air?
41. What is a plane progressive harmonic wave ? Establish displacement relation for a harmonic
wave travelling along the positive direction of X-axis.
42. What is wave motion ? Derive the equation of a harmonic wave.
43. What do you mean by phase of a wave? Discuss the phase change with (i) time and
(ii) position. Hence define time period and wavelength of a wave.
44. State and illustrate the principle of superposition of waves.
45. What are stationary waves ? State the necessary condition for the formation of stationary
waves.
46. Write any three characteristics of stationary waves.
47. State the laws of vibrations of stretched strings.
48. Prove analytically that in the case of an open organ pipe of length L the frequencies of
vibrating air column are given by
v = n(/2L),
where n is an integer.
49. Distinguish between transverse and longitudinal waves.
50. What is the effect of pressure on the speed of sound? Justify your answer.
51. What is the nature of sound waves in air? How is the speed of sound waves in atmosphere
affected by the (i) humidity (ii) temperature ?
52. Give any three differences between progressive waves and stationary waves.
53. What is a progressive wave? Derive an expression which represents a progessive wave.
54. What is the difference between velocity of wave and velocity of a particle ? Obtain the
relation for the velocity of wave in wave motion.
60
Engineering Physics
55. From the equation y = r sin
56.
57.
58.
59.
60.
61.
62.
63.
64.
65.
66.
2
(vt – x), establish the relation between particle velocity,,

and wave velocity.
Show that simple harmonic motion may be regarded as the projection of uniform circular
motion along a diameter of the circle. Hence derive an expression for the displacement of
a particle in S.H.M.
Explain the relation in phase between displacement, velocity and acceleration in SHM,
graphically as well as theoretically.
Derive expressions for the kinetic and potential energies of a harmonic oscillator. Hence
show that total energy is conserved in S.H.M.
Find the total energy of the particle executing S.H.M. and show graphically the variation
of P.E. and K.E. with time in S.H.M. What is the frequency, of these energies with
respect to the frequency of the particle executing S.H.M ?
Show that for a particle in linear S.H.M., the average kinetic energy over a period of
oscillation is equal to the average potential energy over the same period.
At what distance from the mean position is the kinetic energy in simple harmonic oscillator
equal potential energy?
What is a spring factor? Derive the expression for resultant spring constant when two
springs having constants k1 and k2 are connected in (i) parallel, and (ii) in series.
On the basis of spring model, explain the propagation of a sound in (i) air and (ii) solids.
In reference to a wave motion, define the terms (i) amplitude, (ii) time period, (iii)
frequency, (iv) angular frequency, (v) wavelength, (vi) wave number, (vii) angular wave
number and (viii) wave velocity.
What are stationary waves ? Explain the formation of stationary waves graphically.
Obtain an expression for a stationary wave formed by two sinusoidal waves travelling
along the same path in opposite directions and obtain the positions of nodes and antinodes.
An incident wave and a reflected wave are represented by
2
2
(t  x) and y2  a sin (t  x)


Derive the equation of the stationary wave and calculate the positions of nodes and
antinodes.
67. Discuss the formation of standing waves in a string fixed at both ends and the different
modes of vibrations.
68. Discuss the formation of harmonics in a stretched string. Show that in case of a stretched
string the first four harmonics are in the ratio 1 : 2 : 3 : 4.
y1  a sin
61
Oscillations and Waves
69.
70.
71.
What is the difference between progressive wave & stationary wave?
What is the characteristic of wave?
Aprogressive wave is represented by y  4sin  0.2x  2t   6  . Here all the quantity is in
S.I. unit. Calculate amplitude, time period frequency, phase, speed, wave length of the wave.
Ans: Given progressive wave is y  4sin  0.2x  2t   6 
Amplitude = 4.
2
1
f 

In the equation   2f  2  2f
 f  0.318
2

1
1
T 
 3.14 . K = propagation vector = 0.2.
f 0.318
72.
2
2
 0.2
 
 31.4 .

0.2
Find the speed of the longitudinal wave propagating in the medium of desity 1.5 ×
103gm/m and bulk modulus 6× 107 dyn/am2.
Bulk Modulus
6  107

Densily
1.5  103
73.
A uniform spring of length 1m, mass 10gm is under tension of 160N. Find the speed of
the transvers wave in the string?
T
m 10
Ans:  
,  = mass per unit length    ,
  16 .


1
74.
Find the ratio of maximum intensity & minimum intensity of two waves amplitude 2 &
3 respectively to be coherent.
Ans: A1 = 2, A2 = 3. Maximum intensity = 4.
75.
Two waves of same frequency have amplitude two units & there units respectively,
when coherent superpositior takes place, the intensity of resultant wave is I 1. The
intensity becomes I2 when they superpose in coherent. Find ratio of I1 and I2.
Ans: Given amplitude A1 = 2 units, A2 = 3 units.
when coherent superposition takes place.
2
I1  A1  A 2 

I2
A12  A 22
Ans:
speed 


In yougs double slit experiment the distance between two slit is 0.5m. The wave length
of light used   5  105 cm and distance between slit & screen is 50 cm. Calculate
fring width.
Ans:

D 5  105  102 x

.
2d
2  0.5
62
Engineering Physics
OBJECTIVE TYPE QUESTIONS
1.
2.
3.
If 0 is the natural frequency of a body and k is the damping constant, then its quality
factor is
(a) 0/k
(c) 20/k
(b) 0/2k
The effect of the damping force on a harmonic oscillator is to
(a)
reduce angular frequency and amplitude of vibration,
(b)
increase amplitude and angular frequency of vibration,
(c)
reduce angular frequency and increase amplitude of vibration,
(d)
increase angular frequency and reduce amplitude of vibration.
Relaxation time is the time in which the amplitude A0 of a damped oscillator falls to
(a)
A0
3
(c)
A0

A0
(d) A0e.
e
If T be the ~elaxation time and , the damping coefficient, then
(b)
4.
(a)
5.
 
2

(c)   1

(d)  = constant.
2
The quality factor of a series L-C-R circuit is
(b)

(a)
1 1
R LC
(c)
RL
C
1 L
(d) RLC
R C
The resonant frequency of a forced oscillator of natural frequency 0 in a medium of
damping coefficient  is
(b)
6.
12
 2 2 
(c) r   0  
4

(a)
r = 0
(b)

2 
r   02  
2

12
(d) r   2  02
63
Oscillations and Waves
7.
8.
If F 0 be the amplitude of the applied force and k the restoring force per unit displacement,
then the amplitude of the forced oscillator for  = 0 is
(a)
k
F0
(c)
F0
k
(b)
kF 0
(d)
1
kF0
If  and 0 be the applied frequency and natural frequency of a driven oscillator, then the
velocity resonance occurs at
(a)
9.
 > 0
(c)  
0
2
(b)
 = 0
(d)   20
The phase difference between external force and the velocity of the forced oscillator is
(a)

2
(c) 
3
2
10. For higher sharpness of the resonance curve, the Q value must be
(a)
small
(c) medium
(b)
large
(d) none of these.
11. If  be the damping constant and T*, the periodic time of a damped oscillator, then
logarithmic decrement of motion is
(b)
0
(d)
(a)
 T*
2 2
(c)
T*


4T *
12. If the damping constant A is small, the resonance curve is
(a)
flat
(c) unchanged
(b)
sharp
(d) neither of these.
13. The amplitude of a damped oscillator with damping factor  varies as
(b)
T *
(d)
(a)
e t
(c) e
t

t
2
1
(b)
(d) e t
e2
14. In terms of Q, the logarithmic decrement  is given by
64
Engineering Physics
(a)


2Q
(c) 2Q

Q
(d) 2Q
2
15. The relaxation time of a dampen oscillator is given by
(b)

(a)

(b)
2
0Q
0
2Q
(c)
2Q
0
1
(d) 2 Q
0
16. The energy decay time  of a damped oscillator is
2

(c)  

2
(b)
(d) none of these.
  2
17. In terms of Q, the energy decay time is given by
(a)

(a)

T0Q
2
(c)
(b)

2
T0Q
(d)   2T0 Q

2T0
Q
18. The number of complete oscillations executed by a damped oscillator in time  (energy
decay time) is
(a)
n
0 
2
(c) n  20 

0
(d) 2
2
0
19. If  be the damping constant and 0 the natural frequency of the damped oscillator, then
for oscillatory motion,
(a)
(e)   20
  20
(b)
n
(b)
  20
20. The frequency * of a damped oscillator is
(d)  
0
2
65
Oscillations and Waves
1/2
(a)

2 
0  1  2 
 40 
2 

(c)  1   
 42 
0 

(b)

2 
0  1  2 
 40 
(d)
0
1/2

2 
1

 42 
0 

21. In case of a forced oscillator, transient beats arc observed in
(a)
Steady state
(b) middle stage
(c)
early state
(d) none of these.
22. Transverse waves can be produced in
(a)
gas
(c) solid
(b)
liquid
(d) semiliquid.
23. If u and v be respectively the particle velocity and wave velocity, then
(a)
u = 
dy
dx
(c) u  
1 dy
 dx
1 dy
1
dy
(d)  
 dx
u
dx
24. The velocity v of longitudinal wave in air is proportional to
(b)
u=
(a)

(c)
mT
(b)
(d) none of these.
E
25. Velocity of transverse wave in a vibrating string is
(a)
T /m
(c)
mT
(b)
m/T
(d)
m T
ANSWERS
1 (a)
2 (a)
3 (b)
4 (a)
5. (b)
6 (b)
7 (c)
8 (b)
9 (b)
10 (b)
11 (a)
12 (b)
13 (c)
14 (a)
15 (c)
16 (c)
17 (a)
18 (a)
19 (c)
20 (a)
21 (b)
22 (c)
23 (a)
24 (b)
25 (a)