WAVES 2.

WAVES
Section A – Euqation of W ave, Particle 2. Non Mechanical waves : These are electro
Velocity and Acceleration magnetic waves. The electromagnetic waves do not
require a medium for propagation. Its speed in
1. WAVES : vacuum is a universal constant. The motion of the
Waves is distributed energy or distributed electromagnetic waves in a medium depends on the
"disturbance (force)" electromagnetic properties of the medium.
• Following points regarding waves :
(i) The disturbance (force) is transmitted from one point (i) Transverse waves
to another. If the disturbance travels in the x direction but the
(ii) The energy is transmitted from one point to another. particles move in a direction, perpendicular to the x
(iii) The energy or distrubance passes in the form of axis as the wave passes it is called a transverse waves.
wave without any net displacement of medium.
y v
(iv) The oscillatory motion of preceding particle is T
imparted to the adjacent particle following it. T
(v) We need to keep creating disturbance in order to 2Tsin
propagate wave (energy or disturbance) continuously.

x
(a) Waves classification y
The waves are classified under two high level
V
T T
headings : O x
1. Mechanical waves : The motion of the particle
constituting the medium follows mechanical laws i.e.
Newton's laws of motion. Mechanical waves figure - I
originate from a distrubance in the medium (such as
a stone dropping in a pond) and the disturbance Consider a sinusoidal harmonic wave travelling
propagates through the medium. The force between through a string and the motion of a particle as shown
the atoms in the medium are responsible for the in the figure Ist (only one unit of wave shown for
propagation of mechanical waves. Each atom exerts illustration purpose). Since the particle is displaced
a force on the atoms near it, and through this force from its natural (mean) position, the tension in the
the motion of the atom is transmitted to the others. string arising from the deformation tends to restore
The atoms in the medium do not experience any net
the position of the particle. On the other hand,
displacement.
velocity of the particle (kinetic energy) move the
Mechanical waves is further classified in two particle farther is zero. Therefore, the particle is
categories such that
pulled down due to tension towards mean position.
1. Transverse waves (waves on a string) In the process, it acquires kinetic energy (greater
2. Longitudnal waves (sound waves) speed) and overshoots the mean position in the
downward direction. The cycle of restoration of
 position continues as vibration (oscillation) of particle The vibration and wave motion are at right angle to
takes place. each other.
Three position along x-axis named "1", "2" and "3"
(ii) Longitudinal waves
are marked with three vertical dotted lines. At either
Longitudinal waves are characterized by the direction of two instants as shown, the positions of string
of vibration (disturbance) and wave motion. They particles have different displacements from the
are along the same direction. It is clear that vibration undisturbed position on horizontal x-axis. We can
in the same direction needs to be associated with a conclude from this observation that displacement in
"restoring" mechanism in the longitudinal direction. y-direction is a function of positions of particle in
x-direction. As such, the displacement of a particle
(b) Mathematical description of waves
constituting the string is a function of "x".
We shall attempt here to evolve a mathematical
Let us now observe the positions of a given particle,
model of a travelling transverse wave. For this, we
say "1". It has certain positive displacement at time
choose a specific set up of string and associated
t = t, At the next snapshot at t = t + t, the
transverse wave travelling through it. The string is
displacement has reduced to zero. The particle at
tied to a fixed end, while disturbance is imparted at
"2" has maximum displacement at t = t, but the same
the free end by up and down motion. For our
has reduced at t = t + t. The third particle at "3'
purpose, we consider that pulse is small in dimension;
has certain positive displacement at t = t, At t = t + t,
the string is light, elastic and homogeneous. The
it acquires additional positive displacement and
assumptions are required as we visualize a small
reaches the position of maximum displacement.
travelling pulse which remains undiminished when
From these observation, we conclude that
it moves through the strings. We also assume that
displacement of a particle at any position along the
the string is long enough so that our observation is
string is a function of "t".
not subjected to pulse reflected at the fixed end.
Combining two observations, we conclude that
For understanding purpose, we first consider a single
displacment of a particle is a function of both position
pulse as shown in the figure (irrespective of whether
of the particle along the string and time.
we can realize such pulse in practice or not). Our
objective here is to determine the nature of a y = f (x, t)
mathematical description which will enable us to We can further specify the nature of the mathematical
determine displacement (disturbance) of string as function by association the speed of the wave in our
pulse passes through it. We visualize two snap shots consideration. Let "v" be the constant speed with
of the travelling pulse at two close time instants "t" which wave travels from the left end to the right
and "t + t". The single pulse is moving towards end. We notice that wave function at a given position
right in the positive x-direction. of the string is a function of time only as we are
considering displacement at a particular value of "x".
Let us consider left hand end of the string as the
Y 12 3 origin of reference (x = 0 and t = 0). The displacement
in y-direction (disturbance) at x = 0 is a function of
t=t
x time, "t" only :
O
y = f(t) = A sin t
The disturbance travels to the right at constant speed
t  t  t "v'. Let it reaches a point specified as x = x after
O x
time "t". If we visualize to describe the origin of this
disturbance at x = 0, then time elapsed for the
distrubance to move from the origin (x = 0) to the
 point (x = x) is "x/v". Therefore, if we want to use
the function of displacement at x = 0 as given above, One wavelength

then we need to subtract the time elapsed and set A

Displacement
the equation is :
O
 x  x Distance x
y  f  t –   A sin   t – 
 v   v B

This can also be expressed as

Crest Trough Crest Trough
 vt – x   x – vt 
 f   –f   The wavelength  of a wave is generally taken as
 v   v 
the distance between two successive crests or two
y(x, t) = g(x – vt)
successive trough. To be more specific, it is the
using any fixed value of t (i.e. at any instant), this distance between two consecutive points on the wave
shows shape of the string. which have same phase.
If the wave is travelling in –x direction, the wave A displacement-time graph may also be drawn for a
equation is written as wave motion, showing how the displacement of one
x particle at a particular distance from the source varies
y (x, t) = f (t  )
v with time. If this is simple harmonic variation then
The quantity x – vt is called phase of the wave the graph is a sine curve.
function. As phase of the pulse has fixed value
x – vt = const. • Wave Length, Frequency, Speed
dx If the source of a wave makes f vibrations per second,
Taking the derivative w.r.t. time v
dt so they will the particles of the transmitting medium.
where v is the phase velocity although often called That is, the frequency of the waves equals frequency
wave velocity. It is the velocity at which a particular of the source.
phase of the distrubance travels through space. When the source makes one complete vibration, one
In order for the function to represent a wave wave is generated and the disturbance spreads out a
travelling at speed v, the quantities x, v and t must distance  from the source. If the source continues
appear in the combination (x + vt) or (x – vt). Thus to vibrate with constant frequency f, then f waves
(x – vt)2 is acceptable but x2 – v2 t2 is not. will be produced per second and the wave advances
a distance f  in one second. If v is the wave speed
then
(c) Describing Waves :
v=f 
Two kinds of graph may be drawn displacement -
distance and displacement-time. This relationship holds for all wave motions.

A displacement-distance graph for a transverse Frequency depends on source (not on medium), v

mechanical waves shows the displacement y of the depends on medium (not on source frequency), but
vibrating particles of the transmitting medium at wavelength depend on both medium and source.
different distance x from the source at a certain
instant i.e. it is like a photograph showing shape of
(d) Initial Phase :
the wave at that particular instant.
At x = 0 and t = 0, the sine function evaluates to
The maximum displacement of each particle from
zero and as such y-displacement is zero. However,
its undisturbed position is the amplutude of the wave.
a wave form can be such that y-displacement is
In the figure 1, it OA or OB.
not zero at x =0 and t = 0. In such case, we need to
 account for the displacement by introducting an
A 1
angle like :  Asin   sin =
2 2
y(x,t) = Asin (kx – t + )
where "" is initial phase. At x = 0 and t = 0.  3
= ,
y(0, 0) = A sin () 4 4
The measurement of angle determines following two To choose the correct phase angle  we displaced
aspects of wave form at x = 0, t = 0 : (i) whether to wave. Slightly in +ve x direction such that
the displacement is positive or negative and (ii)
whether wave form has positive or negative slope.
a
For a harmonic wave represented by sine function,
b
there are two values of initial phase angle for which
displacement at reference origin (x = 0, t = 0) is
positive and has equal magnitude. We know that
the sine values of angles in first and second quadrants
are positive. A pair of initial phase angles, say  = /3 In above figure Paticle at a is move downward
and 2/3, correspond to equal positive sine values
A
are : towards point b i.e. particle at x = 0 & y =
2
sin = sin ( – )
have negative velocity which gives
    2  1
sin  sin   –   sin    y
3  3  3  2  A cos( – kx  ) at
t
To choose the initial phase in between the two values
t = 0, x = 0
2
/3 & . We can look at a wave motion in yet is cos = – ve (from figure) ...(2)
3
from above discussion 3/4 gives sin + ve and cos
another way. A wave form at an instant is displaced
negative i.e.
by a distance x in very small time interval t then
then speed to the particle at t = 0 & x = 0 is in 3

upward +ve direction in further time t 4

v
Note

(0,0)

v
 x
at time t y  A sin  t  
 v

EXAMPLE 1
Find out the expression of wave equation which is moving y  A sint
x
A
is +ve x direction and at x = 0, t = 0 y =
2 y = A sin ( t + kx + )
Sol. Let y = A sin (t – kx + )
at t = 0 and x = 0
EXAMPLE 2 Which of the two forms is correct ? In fact, both
are correct so long we are in a position to accurately
If (t) & (kx) terms have same sign then the wave
interpret the equation. Starting with the first equation
move toward –ve x direction and vice versa and
and using trigonometric identity :
with diffierent initial phase.
We have,
y = A sin (wt – kx) Wave move toward  A sin (kx – t) = A sin ( – kx + t)
y = A sin (–kx + wt) +ve x direction
= A sin (t – kx + )
y = A sin (–kx – wt) Thus we see that two forms represent waves along
= A sin (kx + wt + p) Wave move toward
–ve x direction  
y = A sin (kx + wt) at the same speed  v   . They differ, however,,
 k
2. PAR TICL E VE LOCITY AND in phase. There is phase difference of "". This has
ACCELERATION : implication on the waveform and the manner particle
Particle velocity at a given position x = x is obtained oscillates at any given time instant and position. Let
by differentiating wave function with respect to time us consider two waveforms at x = 0, t = 0. The
"t". We need to differentiate equation by treating slopes of the waveforms are :
"x" as constant. The partial differentiation yields

particle velocity as : y(x, t)  kA cos(kx – t) = kA
x
  = a positive number and
vp=  t y(x,t)   t Asin(kx – t) = –A cos (kx – t)

y(x, t)  –kA cos( t – kx) = –kA
We can use the property of cosine function to find x
the maximum velocity. We obtain maximum speed
= a negative number
when cosine function evaluates to "–1" :
 vpmax = A
Forms of wave functions
The acceleration of the particle is obtained by
differentiating expression of velocity partially with y
v
respect to time :
y  A sin[kx – t]
 
 ap = v p  {– A cos(kx – t)}
t t x
O
vp
= –2 A sin (kx – t) = –2y
y  A sin[t – kx]
Again the maximum value of the acceleration can vp
be obtained using property of sine function : x
O
 apmax = 2A

3. DIFFER ENT FOR MS OF WAVE Exchange of terms in the argument of sine function
FUNCTION : results in a phase difference of .
Different forms give rise to bit of confusion about In the first case, the slope is positive and hence particle
the form of wave function. The forms used for velocity is negative. It means particle is moving from
describing wave are : reference origin or mean position to negative
y (x, t) = A sin (kx – t) extreme position. In the second case, the slope is
negative and hence particle velocity is positive. It
y(x, t) = A sin (t – kx + )
 means particle is moving from positive extreme The transverse velocity and transverse acceleration
position to reference origin or mean position. Thus of any point on the string do not reach their
two forms represent waves which differ in direction maximum value simultaneously. Infact, the
in which particle is moving at a given position. transverse velocity reaches its maximum value
Once we select the appropriate wave form, we can (A) when the displacement y = 0, whereas the
write wave equation in other forms as given here : transverse acceleration reaches its maximum
magnitudes (2A) when y = ± A
 t  further
y(x, t) = A sin (kx – t) = A sin k  x –  = A sin
 k 
 dy 
2  dx 
(x – vt) t cons tan t

Further, substituting for "k" and "" in wave equation, y
 = – kA cos (wt – kx + ) ...(3)
we have : x

 2 2  x t   2y
y (x, t) = A sin  x – t   A sin 2   –  = = – k2A sin (t – kx + )
  T   T x 2

If we want to represent waveform moving in ...(4)

negative "x" direction, then we need to replace "t" From (1) and (3)
by "–t".
y  y
–
t k x
4. THE LINEAR WAVE EQUATION :
 vp = – vw × slope
By using wave function y = A sin (t – kx + ), we
i.e. if the slope at any point is negative, particle
can describe the motion of any point on the string.
velocity and vice-versa, for a wave moving along
Any point on the string moves only vertically, and
positive
so its x coordinate remains constant. The transverse
x axis i.e. vw is positive.
velocity v y of the point and its transverse
y
acceleration ay are therefore.
A
B
 dy 
vy   
 dt  x cons tan t x1 x2
x

y
  t = A cos (t – kx + ) ...(1)
For example, consider two points A and B on the
 dv  y-curve for a wave, as shown. The wave is moving
ay   y  along positive x-axis.
 dt  x cons tan t
Slope at A is positive therefore at the given moment,
 vy 2 its velocity is negative. That means it is coming
 y
  = –2A sin (t – kx + ) ..(2) downward. Reverse is the situation for particle at
t  t2
point B.
and hence Now using equation (2) and (4)
vy. max = A
 2y k 2  2y  2y 1 2 y
ay.max = 2A 2
 2 2  
x  t x 2 v 2 t 2
 This is known as the linear wave equation or Comparing with linear wave equation, we see that
diffential equation representation of the travelling the wave function is a solution to the linear wave
wave model. We have developed the linear wave equation if the speed at which the pulse moves is 3
equation from a sinusoidal mechanical wave cm/s. It is apparent from wave function therefore
travelling through a medium. But it is much more it is a solution to the linear wave equation.
general. The linear wave equation successfully
describes waves on strings, sound waves and also
EXAMPLE 4
electromagnetic waves.
A wave pulse is travelling on a string at 2 m/s.
Thus, the above equation can be written as,
displacement y of the particle at x = 0 at any time t
2 y 2 y is given by
2
 v2 2 ...(i)
t x
2
y
The general solution of this equation is of the form t2  1
y(x, t) = f (ax ± bt) ...(ii) Find
Thus, any function of x and t which satisfies Eq. (i) (i) Expression of the function y = (x, t) i.e.,
or which can be written as Eq. (ii) represents a displacement of a particle position x and time t.
wave. The only condition is that it should be finite
(ii) Shape of the pulse at t = 0 and t = 1s.
everywhere and at all times. Further, if these
conditions are satisfied, then speed of wave (v) is  x
given by, Sol. (i) By replacing t by  t –  , we can get the desired
 v
coefficient of t b wave function i.e.,
v 
coefficient of x a
Thus plus (+) sign between ax and bt implies that 2
y 2
the wave is travelling along negative x-direction and  x
 t –  1
minus (–) sign shows that it is travelling along  2
positive x-direction.
(ii) We can use wave function at a particular instant,
say t = 0, to find shape of the wave pulse using
EXAMPLE 3
different values of x.
Verify that wave function
2 2
y at t=0 y 2
(x – 3t)2  1 x
1
is a solution to the linear wave equation x and y are 4
in cm.
at x=0 y=2
Sol. By taking partial derivatives of this function w.r.t x
x=2 y=1
and to t.

 2 y 12(x – 3t)2 – 4 Y
 , and 2
x 2 [(x – 3t) 2  1]3

 2 y 108(x – 3t) 2 – 36 1

 t2 [(x – 3t)2  1]3

–2 0 –2 –4
–4
 2 y 1  2x
or 
 x2 9  t2
 x=–2 y=1 EXAMPLE 5
x=4 y = 0.4 A sinusoidal wave travelling in the positive x
x = –4 y = 0.4 direction has an amplitude of 15 cm, wavelength 40
Using these value, shape is drawn. cm and frequency 8 Hz. The vertical displacement of
the medium at t =0 and x = 0 is also 15 cm, as shown
Similarly for t = 1s, shape can drawn. What do you
conclude about direction of motion of the wave from y(cm)
the graphs? Also check how much the pulse has 40
move in 1s time interval. This is equal to wave
15
speed. Here is the procedure.
x(cm)

2
y 2
 x at t = 1s
1 –   1
 2 (a) Find the angular wave number, period angular
frquency and speed of the wave.
at x=2 y = 2(maximum value)
(b) Determine the phase constant , and write a
at x=0 y=1
general expression for the wave function.
at x=4 y=1
2 2 rad 
Y Sol. (a) k    rad / cm
 40cm 20
2

1 1
t=1 T  s  = 2 f = 16 s–1
t=0
f 8
1
v = f  = 320 cm/s
(b) It is given that A = 15 cm
x
0
–2 2 4 6 and also y = 15 cm at x = 0 and t = 0
then using y = A sin (t – kx + )
The pulse has moved to the right by 2 units in 1 s 15 = 15 sin   sin  = 1
interval. Therefore, the wave function is

x 
Also as t – = constt. y = A sin (t – kx + )
2 2
Differentiating w.r.t time
   rad  
 (15cm)sin (16s – )t –   .x  
1 dx dx   20 cm  2
1–  0  2
2 dt dt
Section B – Speed of transverse wave on segment of length l moves to the left with speed
string, energy in waves v. The net force on the segment is in the radial
direction because the horizontal components of the
1. SPEED OF A TRANSVERSE WAVE ON tension force cancel.
A STRING
Consider a pulse travelling along a string with a
 F  2T sin   2T
r

speed v to the right. If the amplitude of the pulse is Where we have used the approximation sin   
small compared to the length of the string, the tension for small .
T will be approximately constant along the string. If  is the mass per unit length of the string, the
In the reference frame moving with speed v to the mass of the segment of length l is
right, the pulse in stationary and the string moves m =  l = 2R (as  l = 2R)
with a speed v to the left. Figure shows a small mv 2
segment of the string of length l. This segment From Newton's second law  Fr = ma =
R
forms part of a circular arc of radius R.
Instantaneously the segment is moving with speed v  v2 
or 2T = (2R)  R 
in a circular path, so it has centripetal acceleration  
v2/R. The forces acting on the segment are the
tension T at each end. The horizontal component T
 v
of these forces are equal and opposite and thus 
cancel. The vertical component of these forces point
radially inward towards the centre of the circular.
arc. These radial forces provide centripetal EXAMPLE 6

acceleration. Let the angle substended by the Find speed of the wave generated in the string as in
the situation shown. Assume that the tension in not
segment at centre be 2. The net radial force acting
affected by the mass of the cord.
on the segment is

v
l v2
ar  l 500 gm/m
R v
 

R T T
20 kg
R  
O Sol. T = 20 × 10 = 200 N
v O
(b)
(a) 200
v  20m / s
0.5

Fig. (a) To obtain the speed v of a wave on a

stretched string. It is convenient to describe the EXAMPLE 7
motion of a small segment of the string in a moving A taut string having tension 100 N and linear mass
frame of reference. density 0.25 kg/m is used inside a cart to generate a
Fig. (b) In the moving frame of reference, the small wave pulse starting at the left end, as shown. What
 should be the velocity of the cart so that pulse EXAMPLE 9
remains stationary w.r.t ground.
A uniform rope of mass 0.1 kg and length 2.45 m
hangs from a ceiling
(a) Find the speed of transverse wave in the rope at
a point 0.5 m distant from the lower end.
(b) Calculate the time taken by a transverse wave to
travel the full length of the rope.
Sol. (a) As the string has mass and it is suspended
T vertically, tension in it will be different at different
Sol. Velocity of pulse =  20m / s

points. For a point at a distance x from the free end,
   tension will be due to the weight of the string below
Now v PG  v PC  v CG
it. So, if m is the mass of string of length l, the mass

0 = 20 î  v CG
m
 of length x of the string will be,   x.
VCG  –20 iˆ m / s  l 

m m 
 T    xg  xg   
EXAMPLE 8  l   l 
One end of 12.0 m long rubber tube with a total
mass of 0.9 kg is fastened to a fixed support. A cord T
attached to the other and passes over a pulley and   xg

supports an object with a mass of 5.0 kg. The tube
is struck a transverse blow at one end. Find the time
T
required for the pulse to reach the other end (g = 9.8 or v  xg ...(i)

m/s2)

Sol. Tension in the rubber tube AB, T = mg At x = 0.5 m, x
T = (5.0) (9.8) = 49 N
v  0.5  9.8 = 2.21 m/s
or
Mass per unit length of rubber tube, (b) From Eq. (i) we see that velocity of the wave is
different at different points. So, if at point x the
0.9 wave travels a distance dx in time dt, then
= = 0.075 kg/m B
12
m
dx dx
dt  
A v gx
 Speed of wave on the tube,

T 49 t l
v   25.56 m / s dx
 0.075   dt  
0 0 gx
 The required time is,

AB 12 l 2.45
t   0.47s or t2 2 = 1.0 s Ans.
v 25.56 g 9.8
2. ENERGY CALCULATION IN WAVES :
dx
i.e. . Hence,
dt
(a) Kinetic energy per unit length
The velocity of string element in transverse direction dK 1
 v2 A2 cos2 (kx – t)
is greatest at mean position and zero at the extreme dt 2
positions of waveform. We can find expression of Here kinetic energy is a periodic function. We can
transverse velocity by differentiating displacement obtain average rate of transmission of kinetic energy
with respect to time. Now, the y-displacement is by integrating the expression for integral wavelengths.
given by : Since only cos2(kx – t) is the varying entity, we
y = A sin (kx – t) need to find average of this quantity only. Its
Differentiating partially with respect to time, the integration over intergal wavelengths give a value
expression of particle velocity is : 1
of " " . Hence, average rate of transmission of
y 2
vp  = – A cos (kx – t)
t kinetic energy is :

In order to calculate kinetic energy, we consider a dK 1 1 1

|avg   v2 A2  v2 A 2
small string element of length "dx" having mass per dt 2 2 4
unit length "". The kinetic energy of the element is
given by :
(b) Elastic potential energy
1 1 The elastic potential energy of the string element
dK  dmv 2p  dx2 A 2 cos2 (kx – t)
2 2 results as string element is stretched during its
oscillation. The extension or stretching is maximum
This is the kinetic energy associated with the element at mean position. We can see in the figure that the
in motion. Since it involves squared of cosine length of string element of equal x-length "dx" is
function, its value is greatest for a phase of zero greater at mean position than at the extreme. As a
matter of fact, the elongation depends on the slope
 of the curve. Greater the slope, greater is the
(mean position) and zero for a phase of elongation. The string has the least length when slope
2
is zero. For illustration purpose, the curve is
(maximum displacement). purposely drawn in such a manner that the
elongation of string element at mean position is
Now, we get kinetic energy per unit length, "KL", highlighted.
by dividing this expression with the length of small
y
string considered : V

dK 1 2 2 t=t
KL    A cos2 (kx – t)
dx 2
x
O
• Rate of transmission of kinetic
energy
The rate, at which kinetic energy is transmitted, is
obtained by dividing expression of kinetic energy by t  t  t
vp
small time element, "dt" : x
O
dK 1 dx 2 2
   A cos 2 (kx – t)
dt 2 dt
fig : The string element stretched most at
But, wave or phase speed, v, is time rate of position
equilibrium position
 Greater extension of string element corresponds to • Rate of transmission of elastic potential energy
greater elastic energy. As such, it is greatest at mean The rate, at which elastic potential energy is
position and zero at extreme position. This deduction transmitted, is obtained by dividing expression of
in contrary to the case of SHM in which potential kinetic energy by small time element, "dt". This
energy is greatest at extreme position and zero at expression is same as that for kinetic enegy.
mean position.
dU 1
• Potential energy per unit length  v2 A2 cos2 (kx – t)
dt 2
When the string segment is stretched from the length
dx to the length ds an amount of work = T (ds – dx) and average rate of transmission of elastic potential
is done. This is equal to the potential energy stored energy is :
in the stretched string segment. So the potential
energy in this case is : dU 1 1 1
|avg   v2 A2  v2 A 2
dt 2 2 4

ds
dy
(c) Mechanical energy per unit length
Since the expression elastic potential energy is same
as that of kinetic energy, we get mechanical energy
expression by multiplying expression of kinetic
energy by "2". The mechanical energy associated
x x + dx
with small string element, "dx", is :
U = T (ds – dx)
1 2
Now ds  (dx 2  dy 2 ) dE = 2xdK = 2x dmv p = dx2A2cos2 (kx – t)
2

  dy  2  Similarly, the mechanical energy per unit length is :

 dx 1    
  dx   dE 1
EL   2x 2 A 2 cos2 (kx – t)
from the binomial expansion dx 2

2 = 2 A2 cos2 (kx – t)

1  dy 
so ds  dx +   dx
2  dx 
(d) Average power transmitted
2
1 y The average power transmitted by wave is equal
U = T (ds – dx)  T   dx
2  x  to time rate of transmission of mechanical energy
or the potential energy density over integral wavelengths. It is equal to :

2 dE 1 1
dU 1   y  Pavg  |avg  2  v2 A 2  v2 A 2
 T  ...(i) dt 4 2
dx 2   x 
If mass of the string is given in terms of mass per
dy unit volume, "", then we make appropriate change
= kAcos (kx – t)
dx in the derivation. We exchange "" by "s" where
and T = v2  "s" is the cross section of the string :
Put above value in equation (i) then we get
1
Pavg  sv2 A2
dU 1 2 2 2
  A cos2 (kx – t)
dx 2
(e) Energy density Ph ase differen ce betwe en t wo
Since there is no loss of energy involved, it is particles in the same wave :
expected that energy per unit length is uniform The general expression for a sinusoidal wave
throughout the string. As much energy enters that travelling in the positive x direction is
much energy goes out for a given length of string. y(x, t) = A sin (t – kx)
n
This average value along unit length of the string Eq of Particle at x 1 is given by y1 = A sin
length is equal to the average rate at which energy (t – kx1)
is being transferred. Eqn of particle which is at x2 from the origin

The average mechanical energy per unit length is y2 = Asin (t – kx2)
equal to integration of expression over integral Phase difference between particles is k(x2 – x1) =
wavelength 


1 1 2 2 Kx =  x 
EL|avg = 2x v2 A2 = v A k
4 2

We have derived this expression for harmonic wave Note

along a string. The concept, however, can be
extended to two or three dimensional transverse
waves. In the case of three dimensional transverse
waves, we consider small volumetric element. We, Section C – Super position principle interfer-
then, use density, , in place of mass per unit length, ence of waves
. The corresponding average energy per unit
volume is referred as energy density (u) : 1. PRINCIPLE OF SUPERPOSITION :

1 This principle defines the displacement of a medium

u vw 2 A 2
2 particle when it is oscillating under the influence of
two or more than two waves. The principle of
(f) Intensity
superposition is stated as :
Intensity of wave (I) is defined as power transmitted
"When two or more waves superpose on a medium
per unit cross section area of the medium :
particle than the resultant displacement of that
A2 1 medium particle is given by the vector sum of the
I  sv2  vw 2 A 2 individual displacements produced by the component
2s 2
waves at that medium particle independently."
Intensity of wave (I) is a very useful concept for
  
three dimensional waves radiating in all direction Let y 1 , y 2 ,....... y N are the displacements
from the source. This quantity is usually referred in produced by N independent waves at a medium
the context of light waves, which is transverse particle in absence of others then the displacemnt
harmonic wave in three dimensions. Intensity is of that medium, when all the waves are superposed
defined as the power transmitted per unit cross at that point, is given as
sectional area. Since light spreads uniformly all     
around, intensity is equal to power transmitted, y  y 1  y 2  y 3  .......  y N
divided by spherical surface drawn at that point with If all the waves are producing oscillations at that
source at its center. point are collinear then the displacement of the
medium particle where superposition is taking place
can be simply given by the algebric sum of the v
individual displacement. Thus we have
y = y1 + y2 + ..............+yN
The above equation is valid only if all individual v
displacements y1, y2 ........... yN are along same
straight line.
(a) Applic atio ns of Principle of
A simple example of superposition can be
Superposition of Waves
understood by figure shown. Suppose two wave
There are several different phenomenon which
pulses are travelling simultaneously in opposite
takes place during superposition of two or more
directions as shown. When they overlap each other
wave depending on the wave characteristics which
the displacement of particle on string is the algebric
are being superposed. We'll discuss some standard
sum of the two displacement as the displacements
phenomenons, and these are :
of the two pulses are in same direction. Figure
(1) Interference of Wave
shown (b) also shows the similar situation when
the wave pulses are in opposite side. (2) Stationary Waves
y (3) Beats
v
y2
v
(4) Lissajou's Figures (Not discussed here in detail.)
y1 y2
y1 Lets discuss these in detail.
x
(b) Interference of Waves
y
v Suppose two sinusoidal waves of same wavelength
and amplitude travel in same direction along the same
y2 straight line (may be on a stretched string) then
y1 x
superposition principle can be used to define the
v resultant displacement of every medium particle.
y The resultant wave in the medium depends on the
v
y2 extent to which the waves are in phase with respect
v
y1 y2 to each other, that is, how much one wave form is
y1
x shifted from the other waveform. If the two waves
y are exactly in same phase, that is the shape of one
v
wave exactly fits on to the other wave then they
y2 combine to double the displacement of every
y1 x medium particle as shown in figure (a). This
v phenomenon we call as constructive interference.
If the superposing waves are exactly out of phase
y1 + y2 or in opposite phase then they combine to cancel all
v
the displacements at every medium particle and
v
medium remains in the form of a straight line as
shown in figure (b)
 y y

Wave I
x x

–A
(b)

This phenomenon we call destructive interference.

y
Thus we can state that when waves meet, they
A
interfere constructively if they meet in same phase
and destructively if they meet in opposite phase. In
x
either case the wave patterns do not shift relative
–A to each other as they propagates. Such superposing
waves which have same form and wavelength and
have a fixed phase relation to each other, are called
y coherent waves. Sources of coherent waves are
A called coherent source. Two indepedent sources can
never be coherent in nature due to practical
Wave II limitations of manufacturing process. Generally all
x
coherent sources are made either by spliting of the
wave forms of a single source or the different
sources are fed by a single main energy source.
In simple words interference is the
y
phenomenon of superposition of two coherent waves
+A
travelling in same direction.
We've discussed that the resultant
x
displacement of a medium particle when two
–A
coherent waves interfere at that point, as sum or
difference of the individual displacements by the
two waves if they are in same phase (phase
difference = 0, 2, .....) or opposite phase (phase
y difference = , 3,.....) respectively. But the two
2A waves can also meet at a medium particle with
phase difference other then 0 or 2, say if phase
Resultant
x difference  is such that 0 <  < 2, then how is
Wave
the displacement of the point of superposition given
–2A (a)
? Now we discuss the interference of waves in
details analytically.
(c) Analytical Treatment of Interference
2
of Waves x  2n  x = n


For constructive interference

S1
x1 Inet = ( I1  I 2 )2
A1 sin(t  kx) y1  A1 sin(t  kx1)
When I1 = I2 = I
y 2  A 2 sin(t  kx2 )
x2 Inet = 4 I
S2
A 2 sin(t  kx) Anet = A1 + A2
When superposing waves are in opposite phase, the
resultant amplitude is the difference of two
Interference implies super position of waves. amplitudes & Inet is minimum; this is known as
Whenever two or more than two waves destructive interference.
superimpose each other they give sum of their
For Inet to be minimum,
individual diplacement.
cos  = – 1
Let the two waves coming from sources S1 & S2
 = (2n + 1) 
be
where n = {0,1,2,3,4,5...........}
y1 = A1 sin ( t + kx1 )
y2 = A2 sin (t + kx2) respectively. 2
x = (2n + 1) 
Due to superposition 
ynet = y1 + y2

ynet = A1 sin ( t + kx1) + A2 sin (t + kx2)  x = (2n  1)
2
Phase difference between y1 & y2 = k(x2 – x1)
For destructive interfence
i.e.,  = k(x2 – x1)
Inet = ( I1 – I 2 )2
2
As  = x (where x = path
 If I1 = I2
difference &  = phase difference) Inet = 0
Anet = A1 – A2
Anet = A12  A 22  2A1A2 cos 
( I1  I 2 )2
 A net 2  A12  A 22  2A1A 2 cos  Ratio of Imax & Imin =
( I1 – I 2 )2

 Inet = I1 + I2 + 2 I1I 2 cos (as I  A2) Generally,

When the two displacements are in phase, then the
Inet = I1 + I2 + 2 I1I 2 cos
resultant amplitude will be sum of the two amplitude
& Inet will be maximum, this is known of constructive If I1 = I2 = I
interference. Inet = 2I + 2Icos
For Inet to be maximum

cos = 1   = 2n Inet = 2I(1 + cos ) = 4Icos2
2
where n = {0,1,2,3,4,5...........}
EXAMPLE 10 0.5cm 1cm
Wave from two source, each of same frequency 0.5cm 1cm 1cm
1cm 1cm 1cm
+ =
and travelling in same direction, but with intensity 1cm 1cm 1cm 2cm

in the ratio 4 : 1 interfere. Find ratio of maximum to At t = ½ s

minimum intensity. (g) (h) (i)

2
 I1 
2  1 2 1cm
I max  I1  I2  I2    2 1 1cm

1cm
    2cm 1cm
+ =
Sol. I min  I1 – I2   I1   2 –1  = 9 : 1 1cm 2cm 1cm 2cm
 – 1
I2 At t = 2s
  (j) (k) (i)

1
In every s , each pulse (one real moving towards
EXAMPLE 11 2
A triangular pulse moving at 2 cm/s on a rope right and one imaginary moving towards left travels
approaches an end at which it is free to slide on a a distance of 1 cm, as the wave speed is 2 cm/s.)
vertical pole. (b) Particle speed, vp = |– v (slope)|
2 cm/s
1
Here, v = wave speed = 2 cm/s and slope =
2
1 cm
 Particle speed = 1 cm/s Ans.
2 cm 1cm 1cm
EXAMPLE 12
Figure shows a rectanglar pulse and triangular pulse
approaching each other. The pulse speed is 0.5 cm/s.
1 Sketch the resultant pulse at t = 2 s
(a) Draw the pulse at s interval until it is completely
2
reflected.
2cm
(b) What is the particle speed on the trailing edge at
the instant depicted ? –1 0 2 3
–2 1
Sol. (a) Reflection of a pulse from a free boundary is x(cm)
really the superposition of two identical waves Sol. In 2 s each pulse will travel a distance of 1 cm.
travelling in opposite direction. This can be shown The two pulses overlap between 0 and 1 cm as shown
as under. in figure. So, A1 and A2 can be added as shown in
figure (c).
1cm 1cm 1cm
+ =
2cm 1cm 1cm 2cm 2cm 1cm
(a) A1 2cm
At t = ½ S
–1 0 1
(a) (b) (c) + (c) 2cm
A1
(b)
2cm 2cm
A2 A2
2cm 2
1cm 1cm 1cm 0 1 –1 0 1 2
+ =
2cm 1cm 1cm 2cm 2cm 1cm Resultant pluse
at t = 2s

At t = 1 s
Note
(d) (e) (f)
Section D – Reflection and transmission be- Reflection of wave pulse (a) at a fixed end of a string
tween 2 string and (b) at a free end. Time increases from top to
bottom in each figure.
1. REFLECTION AND TRANSMISSION
IN WAVES : When a wave arrives at this free end, the ring slides
the rod. The ring reaches a maximum displacement.
(i) When a pulse travelling along a string reaches the
At this position the ring and string come momentarily
end, it is reflected. If the end is fixed as shown in
to rest as in the fourth drawing from the top in figure
figure (a), the pulse returns inverted. This is bacause
(b). But the string is stretched in this position, giving
as the leading edge reaches the wall, the string pulls increased tension, so the free end of the string is
up the wall. According to Newton's third law, the pulled back down, and again a reflected pulse is
wall will exert an equal and opposite force on the produced, but now the direction of the displacement
string as all instants. This force is therefore, directed is the same as for the initial pulse.
first down and then up. It produces a pulse that is
inverted but otherwise identical to the original. (ii) The formation of the reflected pulse is similar to the
The motion of free end can be studied by letting a overlap of two pulses travelling in opposite directions.
ring at the end of string sliding smoothly on the rod. The net displacement at any point is given by the
The ring and rod maintain the tension but exert no principle of superposition.
transverse force.

(a) (b)

Fig (a) : shows two pulses with the same shape, one inverted
with respect to the other, travelling in opposite
directions. Because these two pulses have the same
shape the net displacement of the point where the
string is attached to the wall is zero at all times.
Fig (b) : shows two pulses with the same shape, travelling in
oppoiste directions but not inverted relative to each
other. Note that at one instant, the displacement of
(a) (b) the free end is double the pulse height.
(iii) REFLECTION AND TRANSMISSION Now to find the relation between Ai, Ar, At we
BETWEEN TWO STRING : consider the figure (b)
Here we are dealing with the case where the end point Incident Power = Reflected Power + Transmitted
is neither completely fixed nor completely free to move Power
As we consider an example where a light string is Pi = Pr + Pt
attached to a heavy string as shown is figure a.
2 2 f 2 A i 21v 1  2 2f 2 A r 21v1  2 2f 2 A t 2 2 v 2 ...(i)
If a wave pulse is produced on a light string moving
towards the friction a part of the wave is reflected and
T T
a part is transmitted on the heavier string the reflected Put 1 = v 2 and 2 = v 2
1 2
wave is inverted with respect to the original one.
in equation (i) their
T
v1 
1
T Ai 2 A r 2 A t 2
v2   
2 v1 v1 v2
(v 2 ,  2 ) v1> v2
( v1, 1 )
v1 2
Ai 2  A r 2  At .......(ii)
y  A i sin(t – k 1x) v2

v2 Maximum displacement of joint particle P (as shown

y  At sin(t – k 2 x)
in figure) due to left string
At
= Ai + Ar
Ar Maximum displacement of joint particle due to right
v1 y  Ar sin(t  k 1x   ) string = At

figure (a) At the boundary (at point P) the wave must be

continuous, that is there are no kinks in it. Then we
must have Ai + Ar = At ...(iii)
On the other hand if the wave is produced on the
from equation (ii) & (iii)
heavier string which moves toward the junction a
part will the reflected and a part transmitted, no v1
inversion in waves shape will take place. Ai – Ar = v A t ...(iv)
2
The wave velocity is smaller for the heavier string
from eq. (iii) & (iv)
lighter string

v1  2v 2 
y  Ai sin( t – k 1x)
At =  v  v  A i
 1 2
v2
P
1 2  v 2 – v1 
Ar =  v  v  A i
 1 2

v2
v1
Ar At
P Note

y  Ar sin(t  k1x) y  At sin(t – k 2 x)

figure : (b)
Section E – Equation of standing wave (Sta- particles. Thus on superposition of two coherent
tionary waves) waves travelling in opposite direction the resulting
interference pattern, we call stationary waves, the
1. STANDING WAVES : oscillation amplitude of the medium particle at
In previous section we've discussed that when two different positions is different.
coherent waves superpose on a medium particle, At some point of medium the resultant amplitude is
phenomenon of interference takes place. Similarly maximum which are given as
when two coherent waves travelling in opposite R is maximum when cos kx = ± 1
direction superpose then simultaneous interference
if all the medium particles takes place. These waves 2
or x  N [N  I]
interfere to produce a pattern of all the medium 
particles what we call, a stationary wave. If the two
N
interfering waves which travel in opposite direction or x=
2
carry equal energies then no net flow of energy takes
place in the region of superposition. Within this  3
region redistribution of energy takes place between or x = 0, , , .....
2 2
medium particles. There are some medium particles
where constructive interference takes place and and the maximum value of R is given as
hence energy increases and on the other hand there Rmax= ± 2 A ...(6)
are some medium particles where destructive
 3
interference takes place and energy decreases. Now Thus in the medium at position x = 0, , , ,
2 2
we'll discuss the stationary waves analytically.
........... the waves interfere constructively and the
Let two waves of equal amplitude are travelling in
amplitude of oscillations becomes 2A. Similarly at
opposite direction along x-axis. The wave equation
some points of the medium, the waves interfere
of the two waves can be given as
destructively, the oscillation amplitude become
y1 = A sin (t – kx) minimum i.e. zero in this case. These are the points
[Wave travelling in +x direction] ...(1) where R is minimum, when
and y2 = A sin (t + kx) [Wave travelling in –x cos kx = 0
direction] ...(2)
2 x 
When the two waves superpose on medium particles, or  (2N  1)
 2
the resultant displacement of the medium particles
can be given as 
y = y1 + y2 or x = (2N + 1) [N  I]
4
or y = A sin (t – kx) + A sin (t + kx)
 3 5
or y = A [sint cos kx – cos t sin kx + sin or x , , ...........
4 4 4
t cos kx + cos t sin kx]
or y = 2A cos kx sin  t ...(3) and the minimum value of R is given as
Equation (3) can be rewritten as Rmin = 0 ...... (7)
y = R sin t ...(4)  3 5
Thus in the medium at position x = , ,
Where R = 2 A cos kx ...(5) 4 4 4
Here equation (4) is an equation of SHM. It implies ......... the waves interfere destructively and the
that after superposition of the two waves the medium amplitude of oscillation becomes zero. These points
particles executes SHM with same frequency  and always remain at rest. Figure (a) shows the oscillation
amplitude R which is given by equation (5) Here we amplitude of different medium particles in a
can see that the oscillation amplitude of medium stationary waves.
particles depends on x i.e. the position of medium
 (7) (8)

(9)
figure (b)
Figure (a)
Based on the above analysis of one complete
In figure (a) we can see that the medium particles oscillations of the medium particles, we can make
at which constructive interference takes place are some interference for a stationary waves. These are:
called antinodes of stationary wave and the points (i) In oscillations of stationary wave in a region, some
of destructive interference are called nodes of points are always at rest (nodes) and some oscillates
stationary waves which always remain at rest. with maximum amplitudes (antinodes). All other
Figure (b) explain the movement of medium particles medium particles oscillate with amplitudes less then
with time in the region where stationary waves are those of antinodes.
formed. Let us assume that at an instant t = 0 all the (ii) All medium particles between two successive
medium particles are at their extreme positions as nodes oscillate in same phase and all medium
shown in figure - (b - 1). Here points ABCD are the particles on one side of a node oscillate in opposite
nodes of stationary waves where medium particles phase with those on the other side of the same node.
remains at rest. All other starts moving towards their (iii) In the region of a stationary wave during one
complete oscillation all the medium particles come
T in the form of a straight line twice.
mean positions and t = all particles cross their
4
(iv) If the component wave amplitudes are equal,
mean position as shown in figure (b – 3), you can then in the region where stationary wave is formed,
see in the figure that the particles at nodes are not no net flow of energy takes place, only redistribution
moving. Now the medium crosses their mean position of energy takes place in the medium.
and starts moving on other side of mean position (a) Different Equation for a Stationary
toward the other extreme position. At time t = T/2, Wave
all the particles reach their other extreme position as Consider two equal amplitude waves travelling in
shown in figure (b - 5) and at time t = 3T/4 again all opposite direction as
these particles cross their mean position in opposite y1 = A sin (t – kx) ...(11)
direction as shown in figure (b - 7). and y2 = A sin (t + kx) ...(12)
The result of superposition of these two waves is
y = 2A cos kx sin t ...(13)
Which is the equation of stationary wave where 2A
(1) (2) cos kx represents the amplitude of medium particle
situated at position x and sin t is the time sinusoidal
factor. This equation (13) can be written in several
ways depending on initial phase differences in the
(3) (4)
component waves given by equation (11)) can (12).
If the superposing waves are having an initial phase
difference , then the component waves can be
expressed as
(5) (6)
 y1 = A sin (t – kx) ...(14) Sol. General Equation of standing wave
y2 = – A sin (t – kx) ...(15) y = A cos t
Superposition of the above two waves will result where
y = 2A sin kx cos t ...(16) A = A sin (kx + )
Equation (16) is also an equation of stationary wave here =L
but here amplitude of different medium particles in
2
the region of interference is given by  k=
L
R = 2A sin kx ...(17)
Similarly the possible equations of a stationary wave  2 
can be written as A = A sin (kx + ) = A sin  x   
 L 
y = A0 sin kx cos ( t + ) ....(18)
at x = 0 node
y = A0 cos kx sin (t + ) ...(19)
 A = 0 at x = 0
y = A0 sin kx sin (t + ) ...(20)
 =0
y = A0 cos kx cos (t + ) ...(21)
Here A0 is the amplitude of antinodes. In a pure 2
eq. of standing wave = A sin x cos t
stationary wave it is given as L
A0 = 2A
EXAMPLE 14
Where A is the amplitude of component waves. If Figure shows the standing waves pattern in a string
we care fully look at equation (18) to (21), we can at t = 0. Find out the equation of the standing wave
see that in equation (18) and (20), the particle where the amplitude of antinode is 2A.
amplitude is given by
R = A0 sin kx ...(22) y
Here at x = 0, there is nodes as R = 0 and in equation 2A
(19) and (21) the particle amplitude is given as
A
R = A0 cos kx ...(23)
Here at x = 0, there is an antinode as R = A0. Thus x
we can state that in a given system of co-ordinates
–2A
when origin of system is at a node we use either
equation (18) or (20) for analytical representation
of a stationary wave and we use equation (19) or
Sol. Let we assume the equation of standing waves
(21) for the same when an antinode is located at the
origin of system. is = A sin (t + )
where A = 2A sin (kx + )
 x = 0 is node  A = 0, at x = 0
EXAMPLE 13
2A sin  = 0  =0
Find out the equation of the standing waves for the
following standing wave pattern. at t = 0 Particle at is at y = A and going towards
mean position.

x=L   5
x=0  =  
2 3 6
so eq. of standing waves is
2 x
(A) A sin x cos t (B) A sin L cos t  5 
L
y = 2Asin kx sin  t  
 6 
x x
(C) A cos 2 L cos t (D) A cos L cos t
EXAMPLE 15 dm =  dx
A string 120 cm in length sustains standing wave Velocity of particle at mean position
with the points of the string at which the displacement = 2A sin kx 
amplitude is equal to 3.5 mm being separated by
15.0 cm. The maximum displacement amplitude is 1
then d (KE) = dx . 4A2 2 sin2kx  d (KE)
X. 95 mm then find out the value of X. 2
Sol. In this problem two cases are possible : = 2A22 . sin2kx dx
/2
2 2 2
 d(K.E)  2A    sin 0
kxdx
A x=0 B B x=0
15cm / 2
15cm 2 2
Total K.E = A    (1 – cos2kx)dx
0
Case-I is that A and B have the same displacement
amplitude and Case-2 is that C and D have the same /2
amplitude viz 3.5 mm. In case 1, if x = 0 is taken at  sin 2kx  1
 A 22  x – = A 2 2
antinode then  2k  0 2
A = a cos kx
In case -2, if x = 0 is taken at node, then Note

A = a sin kx
But since nothing is given in the question.
Hence from both the cases, result should be same.
This is possible only when Section F – Stationary waves in strings,
a cos kx = a sin kx vibration in string wave, sono
meter wire
 A 3.5
or kx = or a    4.95mm
1. STATIONARY WAVES IN STRINGS :
4 cos kx cos  / 4

(a) When both end of string is fixed :

(b) Energy of standing wave in one loop
A string of length L is stretched between two points.
When all the particles of one loop are at extreme When the string is set into vibrations, a transverse
position then total energy in the loop is in the form
progressive wave begins to travel along the string. It
of potential energy only when the particles reaches
is reflected at the other fixed end. The incident and
its mean position then total potential energy converts
the reflected waves interfere to produce a stationary
into kinetic energy of the particles so we can say
total energy of the loop remains constant. transverse wave in which the ends are always nodes,
if both ends of string are fixed.
Total kinetic energy at mean position is equal to
total energy of the loop because potential energy at Fundamental Mode
mean position is zero. (a) In the simplest form, the string vibrates in one loop
Small kinetic energy of the particle which is in in which the ends are the nodes and the centre is the
element dx is antinode. This mode of vibration is known as the
fundamental mode and frequency of vibration is known
as the fundamental frequency or first harmonic.
x dx

/2

1
d (KE) = dmv 2
2
 The frequency f3 is known as third harmonic or

Since the distance between consecutive nodes is second overtone.
2
Thus a stretched string vibrates with frequencies,
 which are integral multiples of the fundamental
 L 1  1 = 2L
2 frequencies. These frequencies are known as
harmonics.
If f1 is the fundamental frequency of vibration, then
the velocity of transverse waves is given as, The velocity of transverse wave in stretched string

v T
v  1f1 or f1  ...(i) is given as v  . Where T = tension in the string.
2L 

 = linear density or mass per unit length of string.

First Overtone If the string fixed at two ends, vibrates in its
(b) The same string under the same conditions may fundamental mode, then
also vibrate in two loops, such that the centre is
1 T
also the node f ....(17)
2L 
2 2
 L  2 = L
2 n T
In general f = nth harmonic
2 

(n – 1)th overtone
  In general, any integral multiple of the fundamental
2 2 frequency is an allowed frequency. These higher
If f2 is frequency of vibrations frequenceis are called overtones. Thus, v1 = 2v0 is
the first overtone, v2 = 3v0 is the second overtone
v v etc. An integral multiple of a frequency is called its
 f2  
2 L harmonic. Thus, for a string fixed at both the ends,
all the overtones are harmonics of the fundamental
v frequency and all the harmonics of the fundamental
 f2  ...(ii)
L frequency are overtones.
The frequency f2 is known as second harmonic or
first overtone. (b) When one end of the string is fixed
Second Overtone and other is free :
(c) The same string under the same conditions free end acts as antinode
may also vibrate in three segments.

3 3 1. /4
 L
2

2
 3  L
3
If f3 is the frequency in this mode of vibration, then, 1 T
f fundamental or Ist harmonic
4 
3v
f3  ...(iii)
2L
 3 T
f IIIrd harmonic or Ist overtone
4 

2.   3 / 4 (2n  1) T 
In general : f  th
 ((2n + 1) harmonic,
4  

nth overtone)

S. No. Travelling waves Stationary waves

These waves advance in a medium with These waves remain stationary between
1
a definite velocity two boundaries in the medium.

In these waves, all particles except nodes

In these waves, all particles of the
oscillate with same frequency but
2 medium oscillate with same frequency
different amplitudes. Amplitude is zero at
and amplitude.
nodes and maximum at antinodes.

At any instant the phase of all particles

At any instant phase of vibration varies between two successive nodes is the
continuosly from one particle to the same, but phase of particles on one side
3 other i.e.,
 phase difference between two of a node is opposite to the phase of
particles can have any value between particles on the other side of the node,
0 and 2 i.e, phase difference between any two
particles can be either 0 or

In these wave, at no instant all the In these waves all particles of the
4 particles of the medium pass through medium pass through their mean position
their mean positions simultaneously. simultaneously twice in each time period.

These waves transmit energy in the These waves do not transmit energy in
5
medium. the medium.

Note
Exercise - 1 Objective Problems | JEE Main

Section A – Equation of W ave, Particle 5. The equation of a wave travelling along the positive
Velocity and Acceleration x-axis, as shown in figure at t=0 is given by
1. A transverse wave is described by the equation   y
Y = Y0 sin 2 (ft – x/). The maximum particle (A) sin  kx – t  
 6 1
velocity is equal to four times the wave velocity if 0
x
(A)  =  Y0/4 (B)  =  Y0/2   –0.5
(C)  =  Y0 (D)  = 2 Y0 (B) sin  kx – t – 
 6 –0.1

2. If the speed of the wave shown in the figure is 330    

(C) sin  t – kx   (D) sin  t – kx – 
m/s in the given medium, then the equation of the  6  6
wave propagating in the positive x-direction will be
- (all quantities are in MKS units) 6. The displacement produced by a simple harmonic
wave is :
10  x
y sin  2000  t   cm. The time period
  17 
and maximum velocity of the particle will be
respectively -
(A) 10–3 second and 200 m/s
(A) y = 0.05 sin 2  (4000 t – 12.5 x)
(B) 10–2 second and 2000 m/s
(B) y = 0.05 sin 2  (4000 t – 122.5 x)
(C) 10–3 second and 330 m/s
(C) y = 0.05 sin 2  (3300 t – 10 x)
(D) 10–4 second and 20 m/s
(D) y = 0.05 sin 2  (3300 x – 10 t)
Section B – Speed of transverse wave on
3. A transverse wave of amplitude 0.50 m, wavelength
string, energy in waves
1 m and frequency 2 hertz is propagating in a string
in the negative x-direction. The expression form of 7. Both the strings, show in figure, are made of same
the wave is material and have same cross section. The pulleys
(A) y(x,t) = 0.5 sin (2x – 4t) are light. The wave speed of a transverse wave in
(B) y(x,t) = 0.5 cos (2x + 4t) the string AB is v1 and in
(C) y(x,t) = 0.5 sin (x – 2t) CD it is v2. The v1/v2 is A
(D) y(x,t) = 0.5 cos (2x – 2t) (A) 1
(B) 2
B C
4. A wave pulse is generated in a string that lies along (C) 2
x-axis. At the points A and B, as shown in figure, if
(D) 1/ 2 D
RA and RB are ratio of wave speed to the particle
speed respectively then :
y 8. A block of mass 1 kg is hanging vertically from a
string of length 1 m and Mass/length =0.001 kg/m.
B V A small pulse is generated at its lower end. The
Pulse reaches the top end in approximately.
A
x (A) 0.2 sec
(B) 0.1 sec
(A) RA > RB (B) RB > RA 1m
(C) 0.02 sec
(C) RA = RB (D) 0.01 sec
(D) Information is not sufficient to decide.
9. A uniform rope of length 10 m and mass 15 kg hangs 12. The relation between frequency  wavelength 
vertically from a rigid support. A block of mass 5 and velocity of propagation vof a wave is-
kg is attached to the free end of the rope. A
transverse pulse of wavelength 0.08 m is produced 
(A) v   (B) =1
v
at the lower end of the 3 rope. The wavelength of
the pulse when it reaches the top of the rope will
v  
be- (C) =1 (D) + =1
 v v
(A) 0.08 m
(B) 0.04 m
(C) 0.16 m
(D) 0 m
Section C – Super position principle
interference of waves

10. A uniform rope having some mass hanges vertically 13. Two waves of equal amplitude A, and equal
from a rigid support. A transverse wave pulse is frequency travels in the same direction in a medium.
produced at the lower end. The speed (v) of the The amplitude of the resultant wave is
wave pulse varies with height (h) from the lower (A) 0 (B) A
end as: (C) 2A (D) between 0 and 2A

v v
14. When two waves of the same amplitude and
(A) (B) frequency but having a phase difference of ,
travelling with the same speed in the same direction
h h
(positive x), interfere, then
(A) their resultant amplitude will be twice that of a
v
v single wave but the frequency will be same
(B) their resultant amplitude and frequency will both
(C) (D) be twice that of a single wave
h
h (C) their resultant amplitude will depend on the phase
angle while the frequency will be the same
(D) the frequency and amplitude of the resultant
11. A wire of 102 kg m 1 passes over a frictionless light wave will depend upon the phase angle.
pulley fixed on the top of a frictionless inclined plane,
which makes an angle of 30° with the horizontal. 15. Two waves are represented by
Masses m and M are tied at two ends of wire such
y1 = a1 cos (t – kx) and
that m rests on the plane and M hangs freely
vertically downwards. The entire system is in y2 = a2 sin(t – kx + /3)
equilibrium and a transverse wave propagates along Then the phase difference between them is-
1
the wire with a velocity of 100 ms .
 
(A) (B)
3 2
m 1
(A) M=5 kg (B) 
M 4
5 
(C) (D)
m 6 6
(C) m=20 kg (D) 4
M
16. Standing waves are produced by superposition of 20. A Wave pulse on a string has the dimension shown
two waves in figure. The waves speed is v=1 cm/s. If point O
y1 = 0.05 sin (3t – 2x) and is a free end. The shape of wave at time t=3 s is:
y2 = 0.05 sin (3t + 2x)
Where x and y are measured in meter and t in v=1cm/s
second. Find the amplitude of particle at x = 0.5m 1 cm
[cos 57.3 = 0.54] O
(A) 0.54 m (B) 5.4 m 1cm 1cm 2cm
(C) 54 m (D) 0.054 m
O
17. If two waves are represented by :
1cm
y1=2 sin (4x – 300t) & (A) O (B)
y2 = sin (4x–300t – 0.2)
then their superposed wave will have angular
frequency -
1cm
(A) 150/ (B) 150 
2cm
(C) 300 (D) 600  (C) 1cm (D)

18. x1 = A sin (t – 0.1x) and O 1cm

 
x2 = A sin  t  0.1x  2 
 
21. A wave pulse, travelling on a two piece string, gets
Resultant amplitude of combined wave is– partially reflected and partially transmitted at the

junction. The reflected wave is inverted in shape
(A) 2A cos (B) A 2 cos  / 2 as compared to the incident one. If the incident wave
4
has wavelength  and the transmitted wave .
   (A)  >  (B)  = 
(C) 2A cos (D) A 21  cos 4 
2   (C)  < 
(D) nothing can be said about the relation of  and .
Section D – Reflection and transmission
between 2 string
22. Two sound waves are respectively
19. A pulse shown here is reflected from the rigid wall
A and then from free end B. The shape of the string y1 = a sin (t–kx) and y2 = b cos (t–kx).
after these 2 Reflection will be. The phase difference between the two waves is:
(A) /2 (B) /3
(C)  (D) 3/4
B A

23. Two waves are represented by the following

equations :
(A) A
(B) A y1 = 5 sin 2 (10t – 0.1 x) and
B B
y2 = 10 sin 2(20t – 0.2x)
Ratio of intensities I2/I1 will be -
(C) A
(D) A
(A) 1 (B) 2
B B
(C) 4 (D) 16
24. Figure shows a rectangular pulse and a triangular 28. The equation for the vibration of a string fixed at
pulse approaching each other along x-axis. The both ends vibrating in its third harmonic is given by
pulse speed is 0.5 cm/s. What is the resultant y=2 cm sin [(0.6 cm–1)x]cos [(500 s–1)t]
displacement of medium particles due to The length of the string is –
superposition of waves at x = 0.5 cm and t = 2 sec. (A) 24.6 cm (B) 12.5 cm
(C) 20.6 cm (D) 15.7 cm
y (cm)
0.5 cm/s 0.5 cm/s
2
29. The vibrations of a string of length 60 cm fixed at
1 both ends are represented by the equation
x (cm)
–2 –1 0 1 2 3 y = 4 sin (x/15) cos(96t),
where x and y are in cm and t in seconds. The
(A) 3.5 cm (B) 2.5 cm maximum displacement at x = 5 cm is–
(C) 4 cm (D) 3 cm (A) 2 3 cm (B) 3 2 cm

(C) 2 cm (D) 3 cm
Section E – Equation of standing wave
(Stationary waves)

25. A wave is represented by the equation y = 1 30. If a wave is represented by the following equation

0sin 2 (100t  0.02 X )  10sin 2 (100t  0.02 X ). The 2x 2vt

y = A cos sin then it is a :
 
maximum amplitude and loop length are respectively
(A) Progressive wave
(A) 20 units and 30 units (B) 20 units and 25 units
(B) Stationary wave
(C) 30 units and 20 units (D) 25 units and 20 units
(C) Longitudinal progressive wave
(D) Transverse progressive wave
26. The resultant amplitude due to superposition of two
waves
Section F – Stationary waves in strings,
Y1  5sin ( wt  kx) and y2  5cos ( wt  kx  1500 ) vibration in string wave, sono
meter wire
(A) 5 (B) 5 3
31. Two wave pulses travel in opposite directions on a
(C) 5 2 – 3 (D) 5 2  3 string and approach each other. The shape of the
one pulse in inverted with respect to the other.
(A) the pulses will collide with each other and vanish
27. The equation of stationary wave along a stretched
after collision.
string is given by
(B) the pulses will reflect from each other i.e., the
x pulse going towards right will finally move towards
y = 5 sin cos 40  t
3 left and vice versa.
Where x and y are in cm and t in second. The (C) the pulses will pass through each other but their
separation between two adjacent nodes is - shapes will be modified
(A) 1.5 cm (B) 3 cm (D) the pulses will pass through each other without
(C) 6 cm (D) 4 cm any change in their shape.
32. A wire of linear mass density 9x10 –3 kg/m is 34. In a stationary wave represented by y = a sin t
stretched between two rigid supports under a cos kx, amplitude of the component progressive
tension of 360 N. The wire resonates at frequency wave is :
210 Hz. The next higher frequency at which the
a
same wire resonates is 280 Hz. The number of loops (A) (B) a
2
produced in first case will be-
(A) 1 (B) 2 (C) 2a (D) None
(C) 3 (D) 4
35. The rate of transfer of energy in a wave depends
33. A stretched sonometer wire resonates at a (A) directly on the square of the wave amplitude
frequency of 350 Hz and at the next higher and square of the wave frequency
frequency of 420 Hz. The fundamental frequency (B) directly on the square of the wave amplitude
of this wire is : and square root of the wave frequency
(A) 350 Hz (B) 5 Hz (C) directly on the wave frequency and square of
(C) 70 Hz (D) 170 Hz the wave amplitude
(D) directly on the wave amplitude and square of
the wave frequency.
Exercise - 2 (Leve-I) Objective Problems | JEE Main

Section A – Equation of W ave, Particle 5. A sinusoidal progressive wave is generated in a

Velocity and Acceleration
string. It’s equation is given by y = (2 mm) sin (2x
1. The displacement of a particle of a string carrying – 100 t + /3). The time when particle at x = 4 m
a travelling wave is given by first passes through mean position, will be
y = (3 cm) sin 6.28 (0.50 x – 50 t)
1 1
where x is in centimeter and t is in second. The (A) sec (B) sec
150 12
velocity of the wave is-
(A) 100 m/s (B) 50 cm/s 1 1
(C) sec (D) sec
(C) 100 cm/s (D) 10 m/s 300 100

2. The equation of a progressive wave is 6. A transverse wave is described by the equation

y = A sin [2(ft – x/)]. The maximum particle
 4  velocity is equal the wave velocity if :
y = 0.4 sin 120t  5 x 
 
(A)  = A/4 (B) =A/2
Where distance is in meters and time is in seconds. (C)  = A (D)  = 2A
Calculate frequency and wavelength.
(A) 60 Hz, 2.5 m (B) 30 Hz, 3 m Section B – Speed of transverse wave on
(C) 90 Hz, 2.5 m (D) 60 Hz, 5 m string, energy in waves

7. A wire is 4 m long and has a mass 0.2 kg. The wire

 2x  is kept horizontally. A transverse pulse is generated
3. An equation y = acos2  2nt   represents a
   by plucking one end of the taut (tight) wire. The
wave with- pulse makes four trips back and forth along the cord
(A) amplitude a, frequency n and wavelength  in 0.8 sec. The tension is the cord will be -
(B) amplitude a, frequency 2n and wavelength 2 (A) 80 N (B) 160 N
a (C) 240 N (D) 320 N
(C) amplitude , frequency 2n and wavelength 
2
8. A string of 7 m length has a mass of 0.035 kg. If
a 
(D) amplitude , frequency 2n and wavelength tension in the string is 60.5 N, then speed of a wave
2 2
on the string is :
(A) 77 m/s (B) 102 m/s
4. When a plane wave train transverses a medium,
(C) 110 m/s (D) 165 m/s
individual particles execute a periodic motion given
 x 
by the equation y = 5 sin 4  4 t  16  where the Section C – Super position principle
 
interference of waves
lengths are expressed in centimeters and time in
seconds. The phase difference for two positions of 9. Three waves of equal frequencies having
the same particles which are occupied at a time amplitudes 10 m, 4m and m arrive at a given
interval 0.8 s apart is- point with successive phase difference of . The
(A) 72° (B) 144° amplitude of the resulting wave in  m is given by -
(C) 102° (D) 36° (A) 7 (B) 6
(C) 5 (D) 4
10. In the figure the intensity of waves arriving at D 14. Two pulses in a stretched string whose centers are
from two coherent sources S1 and S2 is I0. The initially 8 cm apart are moving towards each other
wavelength of the wave is  = 4 m. Resultant as shown in Figure. The speed of each pulse is 2 cm
intensity at D will be - S 4m s-1. After 2 second the total energy of the pulses will
1 D
(A) 4I0 be
(B) I0 3m (A) zero
(C) 2I0 (B) purely kinetic
(D) zero S2 (C) purely potential 8 cm

(D) partly kinetic and partly potential.

11. Two pulses in a stretched string, whose centres are
initially 8 cm apart, are moving towards each other
as shown in the figure. The speed of each pulse is 15. Two waves represented by y1 = a sin t and y2 = a
2 cm/s. After 2 s the total energy of the pulses will 
be: sin (t + ) with  = are superposed at any point
2
(A) zero at a particular instant. The resultant amplitude is
(B) purely kinetic 8 cm (A) a (B) 4a
(C) purely potential
(D) parely kinetic and parely potential (C) 2a (D) zero

12. Equations of two progressive waves at a 16. A harmonic wave is travelling on string 1. At a
certain point in a medium are given by junction with string 2 it is partly reflected and partly
y1 = a sin (t + 1) and y2 = a sin (t + 2). If transmitted. The linear mass density of the second
amplitude and time period of resultant wave formed string is four times that of the first string, and that
by the superposition of these two waves is same as the boundary between the two strings is at x =0. If
that of both the waves, then 1 – 2 is the expression for the incident wave is, yi = Ai cos
 2 (k1x – 1t)
(A) (B)
3 3 Then findout the expression for the transmitted
wave.
 
(C) (D)
6 4 1 3
(A) A i cos(2k1x – 1 t) (B) Ai cos(2k1x – 1t)
3 2
13. There are three strings RP, PQ, and QS as shown.
2
Their mass and lengths are RP = (0.1 kg, 2m), (C) Ai cos(2k1x – 1t) (D) None
3
PQ = (0.2 kg, 3 m), QS = (0.15 kg, 4m) respectively.
All the strings are under same tension. Wave-1 is
incident at P. It is partly reflected (wave-2) and Section D – Reflection and transmission
partly transmitted (wave-3). Now wave-3 is incident between 2 string
at Q. It is again partly transmitted (wave-5) and
17. A composition String is made up by joining two
partly reflected (wave-4). Phase difference
strings of different masses per unit length
between wave-1 and wave
  and 4. the composite string is under the same
1 3 5
tension. A transverse wave pulse: Y = (6mm) sin
P Q
(5t+40x), Where ‘t’ is in seconds and ‘x’ in meters,
R S
is sent along the lighter string towards the joint. The
2 4 joint is at x=0. The equation of the wave pulse
(A) 2 is  (B) 4 is zero reflected from the joint is
(C) both (a) and (b) are correct (A) (2mm) sin (5t-40x) (B) (4 mm) sin (40x-5t)
(D) both (a) and (b) are wrong (C) –(2 mm) sin (5t-40x) (D) (2 mm) sin (5t-10x)
18. In the previous question, the percentage of power 22. Under what conditions 75% of incident
transmitted to the heavier string through the joint is energy is transmitted
approximately v1 1 v1 1
(A) 33% (B) 89% (A) v  2 (B) v  3
2 2
(C) 67% (D) 75%
v1 1 v1 2
19. A wave moving with constant speed on a uniform (C) v  4 (D) v  3
2 2
string passes the point x = 0 with amplitude A0,
angular frequency 0 and average rate of energy
transfer P0. As the wave travels down the string it Section E – Equation of standing wave
gradually loses energy and at the point x = , the (Stationary waves)
P
average rate of energy transfer becomes 0 . At 20
2 23. A Standing Wave y  A sin(  x ) cos (1000 t ) is
the point x = , angular frequency and amplitude 3
are respectively. maintained in a taut string where y and x are
expressed in meters. The distance between the
(A) 0 and A0 / 2
successive points oscillating with the amplitude A/2
(B) 0/ y  Asin  ωt kx  . and A0 across a node is equal to
(C) less than 0 and A0 (A) 2.5 cm (B) 25 cm
(D) 0/ 2 and A0 / 2 (C) 5 cm (D) 10 cm

20. A metallic Wire of length L is fixed between two 24. A wave represented by the equation
rigid supports. If the wire is cooled through a y = a cos (kx – t) is superposed with another wave
temperature difference T (Y = young’s modulus,  to form a stationary wave such that the point x = 0
= density,  = coefficient of linear expansion) is a node. The equation for other wave is :
then the frequency of transverse vibration is (A) a sin (kx + t) (B) – a cos (kx + t)
proportional to: (C) – a cos (kx – t) (D) – a sin (kx – t)
 Y
(A)  Y (B) 25. A taut string at both ends vibrates in its nth overtone.
 The distance between adjacent Node and antinode
 is found to be ‘d’ If the length of the string is L, then

(C) (D) (A) L=2d(n+1) (B) L=d(n+1)
Y Y
(C) L=2dn (D) L = 2d(n – 1)

Passage Q. No. 21 to 22 26. A standing wave pattern of amplitude A in a string

In the shown figure answer the following two of length L shows 2 nodes (plus those at two ends).
questions. If one end of the string corresponds to the origin
and v is the speed of progressive wave, the
Transmitted
disturbance in the string, could be represented (with
incident
appropriate phase) as:
1 2
 2x   2 vt 
(A) y(x, t) = A sin   cos  
Reflected  L   L 

21. If P i , P r and P t are powers of incident,  3x   2 vt 

(B) y(x, t) = A cos   sin  
reflected and transmitted waves and ii, Ir and It the  L   L 
corresponding intensities, then
 4x   4 vt 
(A) Pi = Pr +Pt (C) y(x, t) = A cos   cos  
 L   L 
(B) Ii = Ir + It
(C) both (a) and (b) are correct  3x   3vt 
(D) both (a) and (b) are wrong (D) y(x, t) = A sin   cos  
 L   L 
27. The equation of a wave disturbance is given as : 29. A string of length 1m and linear mass density 0.01
kgm–1 is stretched to a tension of 100N.when both
 
y  0.02cos   50t  cos(10x) , where x and y ends of the string are fixed, the three lowest
2 
frequencies for standing wave are f1, f2 and f3. when
are in meters and t in seconds. Choose the wrong only one end of the string is fixed, the three lowest
statement frequencies for standing wave are n1, n2 and n3. Then
(A) Antinode occurs at x = 0.3 m (A) n3 = 5n1 = f3 = 125 Hz
(B) The wavelength is 0.2 m (B) f3 = 5f1 = n2 = 125 Hz
(C) The speed of the constituent waves is 4m/s (C) f3 = n2 = 3f1 = 150 Hz
(D) Node occurs at x = 0.15 m
f1  f 2
(D) n 2   75 Hz
2
Section F – Stationary waves in strings,
vibration in string wave, sono
meter wire 30. A string is fixed at both ends vibrates in a resonant
mode with a separation 2.0 cm between the
28. The frequency of a sonometer wire is f, but when consecutive nodes. For the next higher resonant
the weights producing the tensions are completely frequency, this separation is reduced to 1.6 cm. The
immersed in water the frequency becomes f/2 and length of the string is
on immersing the weights in a certain liquid the (A) 4.0 cm (B) 8.0 cm
frequency becomes f/3. The specific gravity of the
(C) 12.0 cm (D) 16.0 cm
liquid is:

4 16 31. Figure, shows a stationary wave between two fixed

(A) (B)
3 9 points P and Q.

15 32 Which point(s) of 1, 2 and 3 are in phase with the

(C) (D) point x?
12 27

P X 1 23 Q

(A) 1, 2 and 3 (B) 1 and 2 only

(C) 2 and 3 only (D) 3 only
Exercise - 2 (Level-II) Multiple Correct | JEE Advanced

Section A – Euqation of W ave, Particle 4. The points moving with maximum speed is/are
Velocity and Acceleration (A) b (B) c
1. A wave equation which gives the displacement (C) d (D) h
along the Y direction is given by
Y = 10–4 sin (60t + 2x) 5. The points moving upward is/are
where x and y are in metres and t is time in seconds. (A) a (B) c
This represents a wave (C) f (D) g
(A) travelling with a velocity of 30 m/s in the negative
x direction.
6. The points moving downwards is/are
(B) of wavelength  metre
(A) o (B) b
(C) of frequency 30/ hertz
(C) d (D) h
(D) of amplitude 10–4 metre travelling along the
negative x direction.
7. A perfectly elastic uniform string is suspended
2. The displacement of a particle in a medium due to vertically with its upper end fixed to the ceiling and
a wave travelling in the x-direction through the lower end loaded with the weight. If a transverse
the medium is given by y = A sin (t – x), where wave is imparted to the lower end of the string, the
t = time, and  and  are constants : pulse will
(A) the frequency of the wave is  (A) not travel along the length of the string
(B) the frequency of the wave is /2 (B) travel upwards with increasing speed
(C) the wavelength is 2/ (C) travel upwards with decreasing speed
(D) the velocity of the wave is /
(D) travelled upwards with constant acceleration

Section B – Speed of transverse wave on

8. One end of a string of length L is tied to the ceiling
string, energy in waves
of a lift accelerating upwards with an acceleration
Question No. 3 to 6 (4 questions)
2g. The other end of the string is free. The linear
The figure represents the instantaneous picture of
mass density of the string varies linearly from 0 to
a transverse harmonic wave traveling along the
 from bottom to top.
negative x-axis. Choose the correct alternative(s)
related to the movement of the nine points shown (A) The velocity of the wave in the string will be 0.
in the figure. (B) The acceleration of the wave on the string will
y b be 3g/4 every where.
a c h (C) The time taken by a pulse to reach from bottom
o x
d
e g to top will be 8L / 3g .
f
3. The stationary points is/are (D) The time taken by a pulse to reach from bottom

(A) o (B) b to top will be 4L / 3g .

(C) f (D) h
Section D – Reflection and transmission be- Section F – Stationary waves in strings,
tween 2 string vibration in string wave, sono
meter wire

 x
9. A plane wave y = A sin   t   undergo a normal 12. A clamped string is oscillating in nth harmonic, then
 v
(A) total energy of oscillations will be n2 times that
incidence on a plane boundary separating medium of fundamental frequency
M1 and M2 and splits into a reflected and transmitted
(B) total energy of oscillations will be
wave having speeds v1 and v2 then
(n–1)2 times that of fundamental frequency
(A) for all values of v1 and v2 the phase of
(C) average kinetic energy of the string over a
transmitted wave is same as that of incident wave
complete oscillations is half of that of the total
(B) for all values of v1 and v2 the phase of reflected energy of the string.
wave is same as that of incident wave
(D) none of these.
(C) the phase of transmitted wave depends upon
v1 and v2
13. In a stationary wave,
(D) the phase of reflected wave depends upon v1
and v2 (A) all the particles of the medium vibrate in phase
(B) all the antinodes vibrate in phase
(C) the alternate antinodes vibrate in phase
Section E – Equation of standing wave
(Stationary waves) (D) all the particles between consecutive nodes
vibrate in phase
10. The vibration of a string fixed at both ends are
described by Y = 2 sin(x)sin (100t) where Y is in
14. Two waves of equal frequency f and velocity v travel
mm, x is in cm, t in sec then
in opposite directions along the same path. The
(A) Maximum displacement of the particle at waves have amplitudes A and 3A. Then :
x = 1/6 cm would be 1mm.
(A) the amplitude of the resulting wave varies with
(B) velocity of the particle at x = 1/6 cm at time position between maxima of amplitude 4A and
t = 1/600 sec will be 1573 mm/s minima of zero amplitude
(C) If the length of the string be 10 cm, number of (B) the distance between a maxima and adjacent
loop in it would be 5 minima of amplitude is V/2f
(D) None of these (C) at point on the path the average displacement
is zero
11. In a standing wave on a string. (D) the position of a maxima or minima of amplitude
(A) In one time period all the particles are does not change with time
simultaneously at rest twice.
(B) All the particles must be at their positive
extremes simultaneously once in one time period.
(C) All the particles may be at their positive
extremes simultaneously once in a time period.
(D) All the particles are never at rest simultaneously.
Exercise - 3 | Level-I Subjective | JEE Advanced

Section A – Equation of W ave, Particle Section B – Speed of transverse wave on

Velocity and Acceleration string, energy in waves

4. Two strings A and B with  = 2 kg/m and

1. Consider the wave y = (5 mm) sin (1 cm–1) x – (60
 = 8 kg/m respectively are joined in series and
s–1)t] Find (a) the amplitude (b) the wave number,
kept on a horizontal table with both the ends fixed.
(c) the wavelength, (d) the frequency, (e) the time
The tension in the string is 200 N. If a pulse of
period and (f) the wave velocity.
amplitude 1 cm travels in A towards the junction,
then find the amplitude of reflected and transmitted
pulse.
2. The wave function for a traveling wave on a taut
string is (in SI unit) 5. A parabolic pulse given by equation y (in cm) = 0.3
y(x, t) = (0.350 m) sin (10 t – 3x + /4) – 0.1 (x – 5t)2 ( y  0) x in meter and t in second
travelling in a uniform string. The pulse passes
(a) What are the speed and direction of travel of
through a boundary beyond which its velocity
the wave ?
becomes 2.5 m/s. What will be the amplitude of
(b) What is the vertical displacement of the string pulse in this medium after transmission ?
at t = 0, x = 0.100 m ?
6. In the arrangement shown in figure, the string has
(c) What are wavelength and frequency of the mass of 4.5 g. How much time will it take for a
wave ? transverse disturbance produced at the floor to
(d) What is the maximum magnitude of the reach the pulley? Take g = 10 m/s2
transverse speed of a particle of the string ?

25cm
3. The string shown in figure is driven at a frequency
of 5.00 Hz. The amplitude of the motion is 12.0 cm, 2.0m
2kg
and the wave speed is 20.0 m/s. Furthermore, the
wave is such that y = 0 at x = 0 and t = 0. Determine
(a) the angular frequency and (b) wave number for
this wave. (c) Write an expression for the wave
Section C – Super position principle
function. Calculate (d) the maximum transverse
interference of waves
speed and (e) the maximum transverse acceleration
of a point on the string. 7. Two waves are described by
y1 = 0.30 sin [(5x – 200)t] and y2 = 0.30 sin [(5x
y – 200t) + /3]
where y1, y2 and x are in meters and t is in seconds.
x
When these two waves are combined, a traveling
wave is produced. What are the (a) amplitude, (b)
x=0
wave speed, and (c) wave length of that traveling
wave ?
8. A particle on stretched string supporting a travelling Section E – Equation of standing wave
wave, takes 5.0 ms to move from its mean position (Stationary waves)

to the extreme position. The distance between two 12. A nylon guitar string has a linear density of 7.20
consecutive particles, which are at their mean g/m and is under a tension of 150 N. The fixed
position, is 2.0 cm. Find the frequency, the supports are distance D = 90.0 cm apart. The string
wavelength and the wave speed. is oscillating in the standing wave pattern shown in
figure. Calculate the (a) speed wavelength, and (c)
frequency of the traveling waves whose
9. A 200 Hz wave with amplitude 1 mm travels on a superposition gives this standing wave.
long string of linear mass density 6 g/m kept under D
a tension of 60 N. (a) Find the average power
transmitted across a given point on the string. (b)
Find the total energy associated with the wave in a
2.0m long portion of the string.
13. A string oscillates according to the equation

Section D – Reflection and transmission   –1  

y=(0.50 cm) sin  cm  x  cos [(40  s–1)t]
between 2 string  3  

10. A travelling wave of amplitude 5 A is partially What are the (a) amplitude and (b) speed of the
reflected from a boundary with the amplitude 3 A. two waves (identical except for direction of travel)
whose superposition gives this oscillation ? (c) What
Due to superposition of two waves with different
is the distance between nodes ? (d) What is the
amplitudes in opposite directions a standing wave
transverse speed of a particle of the string at the
pattern is formed. Determine the amplitude at node
position x = 1.5 cm when t = 9/8 s ?
and antinodes.

Section F – Stationary waves in strings,

11. The equation of a plane wave travelling along vibration in string wave, sono
meter wire
2
positive direction of x-axis is y = a sin (vt – x)
 14. A string vibrates in 4 loops with a frequency of
When this wave is reflected at a rigid surface and 400 Hz.
its amplitude becomes 80%, then find the equation (a) What is its fundamental frequency ?
of the reflected wave (b) What is frequency will cause it to vibrate into
7 loops.

15. A sonometer wires resonates with a given tuning

fork forming standing waves with five antinodes
between the two bridges when a mass of 9 kg is
suspended from the wire. When this mass is
replaced by M, the wire resonates with the same
tuning fork forming three antinodes for the same
position of bridges. Find the value of M.
Exercise - 3 | Level-II Subjective | JEE Advanced

Section A – Equation of W ave, Particle Section C – Super position principle

Velocity and Acceleration interference of waves

1. The figure shows a snap photograph of a

5. A string that is stretched between fixed supports
vibrating string at t = 0. The particle P is
separated by 75.0 cm has resonant frequencies of
observed moving up with velocity 20 cm/s. The
420 and 315 Hz with no intermediate resonant
angle made by string with x-axis at P is 6°.
frquencies. What are
y
–3 (a) the lowest resonant frequencies and (b) the wave
(in10 m)
4 P speed ?
x
0 1.5 3.5 5.5 7.5
–2
(in10 m) 6. A string fixed at both ends is vibrating in the lowest
(a) Find the direction in which the wave is moving mode of vibration for which a point at quarter of its
(b) the equation of the wave lengths from one end is a point of maximum
displacement. The frequency of vibration in this
(c) the total energy carried by the wave per cycle
mode is 100 Hz. What will be the frequency emitted
of the string, assuming that , the mass per unit
length of the string = 50 gm/m. when it vibrates in the next mode such that this
point is again a point of maximum displacement.

2. A uniform rope of length L and mass m is held at

one end and whirled in a horizontal circle with Section D – Reflection and transmission
angular velocity . Ignore gravity. Find the time between 2 string
required for a transverse wave to travel from one
end of the rope to the other. 7. In a stationary wave pattern that forms as a result
of reflection of waves from an obstacle the ratio of
the amplitude at an antinode and a node is  = 1.5.
Section B – Speed of transverse wave on
What percentage of the energy passes across the
string, energy in waves
obstacle?
3. A symmetrical triangular pulse of maximum height
0.4m and total length 1 m is moving in the positive
8. A 6.00 m segment of a long string has a mass of
x-direction on a string on which the wave speed is
180 g. A high-speed photograph shows the at
24 m/s. At t = 0 the pulse is entirely located between
segment contains four complete cycles of wave.
x = 0 and x = 1 m. Draw a graph of the transverse
velocity of particle of string versus time at x = +1m. The string is vibrating sinusoidally with a frequency
of 50.0 Hz and a peak=to-valley displacement of
15.0 cm. (The "peak-to-valley" displacement is the
4. A uniform rope of length 12 m and mass 6 kg hange
vertical distance from the farthest positive
vertically from a rigid support. A block of mass
displacement to the farthest negative displacement.
2kg is attached to the free end of the rope. A
transverse pulse of wavelength 0.06 m is produced (a) Write the function that describes this wave
at the lower end of the rope. What is the wavelength traveling in the positive x direction. (b) Determine
of the pulse when it reaches the top of the rope ? the power being supplied to the string.
Section E – Equation of standing wave Section F – Stationary waves in strings,
(Stationary waves) vibration in string wave, sono
meter wire
9. What are (a) the lowest frequency, (b) the second
lowest frequency, and (c) the third lowest frequency
11. A steel wire of length 1 m and density
for standing waves on a wire that is 10.0 m long
8000 kg m–3 is stretched tightly between two rigid
has a mass of 100 g. and is stretched under a tension
supports. When vibrating in its fundamental mode,
of 250 N which is fixed at both ends ?
its frequency is 200 Hz.
(a) What is the velocity of transverse waves along
10. In an experiment of standing waves, a string 90 cm this wire ?
long is attached to the prong of an electrically driven
(b) What is the longitudinal stress in the wire?
tuning fork that oscillates perpendicular to the length
of the string at a frequency of 60 Hz. The mass of (c) If the maximum acceleration of the wire is 800
the string is 0.044 kg. What tension must the string ms–2, what is the amplitude of vibration at the mid-
be under (weights are attached to the other end) if point ?
it is to oscillate in four loops ?
12. A guitar string is vibrating in its fundamental mode,
with nodes at each end. The length of the segment
of the string that is free to vibrate is 0.386 m. The
maximum transverse acceleration of a point at the
middle of the segment is 8.40 × 103 m/s2 and the
maximum transverse velocity is 3.80 m/s.
a) What is the amplitude of this standing wave?
b) What is the wave speed for the transverse
traveling waves on this string?
Exercise - 4 | Level-I Previous Year | JEE Main

1. A string is stretched between fixed points separated 5. A travelling wave represented by

by 75.0 cm. It is observed to have resonant y  Asin  ωt  kx  is superimposed on another
frequencies of 420 Hz and 315 Hz. There are no wave represented by y  Asin  ωt  kx  . The
other resonant frequencies between these two. resultant is [AIEEE 2011]
Then, the lowest resonant frequency for this (A) A standing wave having nodes at
string is [AIEEE 2006]  1λ
(A) 105 Hz (B) 1.05 Hz x   n   , n  0,1,2
 22
(C) 1050 Hz (D) 10.5 Hz (B) A wave travelling along + x direction
(C) A wave travelling along - x direction
2. A wave travelling along the x-axis is described by nλ
(D) A standing wave having nodes at x  ; n=0, 1, 2
the equation y(x,t) = 0.005 cos  αx  βt  . If the 2

wavelength and the time period of the wave are

6. Statement I Two longitudinal waves given by
0.08 m and 2.0s, respectively. then α andβ in
equations- y1  x, t   2asin  ωt  kx  and
appropriate units are [AIEEE 2008]
y 2  x, t   asin  2ωt  2 kx  will have equal
0.08 2.0 intensity. [AIEEE 2011]
(A) α  25.00π,β  π (B) α  ,β 
π π Statements II Intensity of waves of given
0.04 1.0 π frequency in same medium is proportional to square
(C) α  ,β  (D) α  12.50π,β  fo amplitude only.
π π 2.0
(A) Statement I is true, Statement II is true
(B) Statement I is true, Statement II is false
3. The equation of a wave on a string of linear mass (C) Statement I is true, Statement II true; Statement
density 0.04 kg m–1 is given by II is the correct explanation of Statement I
(D) Statement I is true, Statement II is true; Statement
  t x 
y = 0.02 (m) sin  2 π    II is not correct explanation of Statement I
  0.04(s) 0.50(m) 
The tension in the string is [AIEEE 2009] 7. A uniform string of length 20 m is suspended from
(A) 4.0 N (B) 12.5 N a rigid support. A short wave pulse is introduced at
(C) 0.5 N (D) 6.25 N its lowest end. It starts moving up the string. The
time taken to reach the support is :
4. The transverse displacement y (x, t) of awave on a (take g = 10 ms–2) [AIEEE 2016]


 ax 2  bt 2  2 ab xt  (A) 2 s (B) 2 2 s
string is given by y  x, t   e . This
represents a [AIEEE 2011] (C) 2 s (D) 2 2 s
b
(A) wave moving in-x direction with speed
a 8. A pipe open at both ends has a fundamental
frequency f in air. The pipe is dipped vertically in
(B) standing wave of frequency b water so that half of it is in water. The fundamental
1 frequency of the air column is now :[AIEEE 2016]
(C) standing wave of frequency 3f
b (A) (B) 2f
4
a f
(D) wave moving in + x direction with speed (C) f (D)
b 2
Exercise - 4 | Level-II Previous Year | JEE Advanced

1. A transverse harmonic disturbance is produced in 5. When two progressive waves y1 = 4 sin (2x – 6t)
a string. The maximum transverse velocity is 3 m/s
and maximum transverse acceleration is 90 m/s2. If  
and y2 = 3 sin  2x  6t   are superimposed, the
the wave velocity is 20 m/s then find the waveform.  2
[JEE-2005] amplitude of the resultant wave is : [JEE 2010]

2. A massless rod is suspended by two identical strings 6. A horizontal stretched string, fixed at two ends, is
AB and CD of equal length. A block of mass m is vibrating in its fifth harmonic according to the
suspended from point O such that BO is equal to equation, y(x,t) = (0.01 m) sin [(62.8 m-1)x] cos
‘x’. Further, it is observed that the frequency of 1st
[ 628s 1  t] Assuming   = 3.14, the correct
harmonic (fundamental frequency) in AB is equal
to 2nd harmonic frequency in CD. Then, length of statement (s) is (are) [JEE-2013]
BO is [JEE-2006] (A) The number of nodes is 5.
L (B) The length of the string is 0.25 m.
(A)
5 (C) The maximum displacement of the midpoint of
C
A the string, from its equilibrium position is 0.01 m.
L
(B) (D) The fundamental frequency is 100 Hz.
4

4L B D 7. One end of a taut string of length 3m along the x

(C)
5 axis is fixed at x = 0. The speed of the waves in the
L
x string is 100 ms-1. The other end of the string is
3L m vibrating in the direction so that stationary waves
(D)
4 are set up in the string. the possible waveform of
these stationary waves is (are)
3. A transverse sinusoidal wave moves along a string [JEE Advanced 2014]
in the positive x-direction at a speed of 10 cm/s. x 50 t
The wavelength of the wave is 0.5 m and its ampli- (A) y (t) = A sin cos
6 3
tude is 10 cm. At a particular time t, the snap-shot
of the wave is shown in figure. The velocity of point x 100t
P when its displacement is 5 cm is Figure : (B) y (t) = A sin cos
3 3
y
3 5x 250t
(A) ĵ m/s P (C) y (t) = A sin cos
50 6 3
x
3 ˆ 5x
(B) – j m/s (D) y (t) = A sin cos 250 t
50 2
3 3 ˆ
(C) î m/s (D) – i m/s
50 50 8. Four harmonic waves of equal frequencies and equal
[JEE-2008] intensities I0 have phase angles 0, /3, 2/3 and .
when they are superposed, the intensity of the
4. A 20 cm long string, having a mass of 1.0 g, is fixed resulting wave is nI0. The value of n is -
at both the ends. The tension in the string is 0.5 N.
[JEE-2015]
The string is set into vibrations using an external
vibrator of frequency 100 Hz. Find the separation
(in cm) between the successive nodes on the string.
[JEE 2009]
Exercise - 1 Objective Problems | JEE Main
1. B 2. C 3. B 4. A 5. D
6. A 7. D 8. D 9. C 10. C
11. C 12. B 13. D 14. C 15. D
16. D 17. C 18. A 19. A 20. D
21. C 22. A 23. D 24. D 25. B
26. A 27. B 28. D 29. A 30. B
31. D 32. C 33. C 34. A 35. A

Exercise - 2 (Leve-I) Objective Problems | JEE Main

1. C 2. A 3. D 4. B 5. C
6. D 7. A 8. C 9. C 10. C
11. B 12. B 13. C 14. B 15. C
16. C 17. C 18. B 19. A 20. B
21. A 22. B 23. C 24. B 25. A
26. D 27. C 28. D 29. D 30. B
31. C

Exercise - 2 (Level-II) Multiple Correct | JEE Advanced

1. A,B,C,D 2. B,C,D 3. B,C 4. C,D 5. A,D
6. C 7. B,D 8. B,C 9. A,D 10. A,B
11. A,C 12. A,C 13. C,D 14. C,D

Exercise - 3 | Level-I Subjective | JEE Advanced

2
1. (a) amplitude A = 5 mm ; (b) wave number k = 1 cm–1 ; (c) wavelength  = = 2 cm
k

 60 1 
(d) frequency v = = Hz ; (e) time period T = = s
2 2 v 30
(f) wave velocity u = n = 60 cm/s
10 
2. (a) im / s (b) –5.48 cm (c) 0.667 m, 5.00 Hz (d) 11.0 m/s
3
3. (a) 10  rad/s (b) /2 rad/m (c) y = (0.120m) sin (1.57x – 31.4 t) (d) 1.2  m/s
(e) 118 m/s2
1 2
4. Ar = – cm, At = cm 5. 0.2 cm 6. 0.02 s
3 3
7. (a) 0.52 m ; (b) 40 m/s ; (c) 0.40 m 8. 50 Hz, 4.0 cm, 2.0 m/s
 2 
9. (a) 0.47 W, (b) 9.4 mJ 10. 2 A, 8 A 11. y = 0.8 a sin ( vt  x  )
 2
12. (a) 144 m/s ; (b) 60.0 cm ; (c) 241 Hz
13. (a) 0.25 cm (b) 1.2 × 102 cm/s; (c) 3.0 cm; (d) 0
14. (a) 100 Hz (b) 700 Hz 15. 25 kg

Exercise - 3 | Level-II Subjective | JEE Advanced

 1 
1. (a) negative x; (b) y = 4 × 10 –3 sin 100  3 t  0.5 x   (x, y in meter) ;
 400 
vp
19.2 m/s

(c) 144  × 10–5 J 2.
2
3. 1/48 sec1/24 sec t
–19.2m/s
4. 0.12 m 5. (a) 105 Hz ; (b) 158 m/s
6. 300 Hz 7. 4. 96% 8. (a) y=(7.50 cm) sin (4.19 x–314 t) (b) 625 W

5 10 5 10
9. (a) Hz ; (b) 5 10 Hz ; (c) Hz 10. 36 N
2 2

11. (a) C = 400 ms–1 ; (b) stress =1.28 × 109 Nm–2 ; (c) a = 0.02/42

12. a. 1.72 × 10–3 m. ; b. 272 m/s.

Exercise - 4 | Level-I Previous Year | JEE Main

1. A 2. A 3. D 4. A 5. A
6. B 7. B 8. C

Exercise - 4 | Level-II Previous Year | JEE Advanced

3
1. y = (10 cm) sin (30 t ± x + f) 2. A 3. A 4. 5
2
5. 5 6. B, C 7. A,C,D 8. 3