Wave on a String

WAVE ON A STRING



WAVES
Wave motion is the phenomenon that can be observed almost everywhere around us, as well it
appears in almost every branch of physics. Surface waves on bodies of matter are commonly observed.
Sound waves and light waves are essential to our perception of the environment. All waves have a
similar mathematical description, which makes the study of one kind of wave useful for the study of
other kinds of waves. In this chapter, we will concentrate on string waves, which are type of a
mechanical waves. Mechanical waves require a medium to travel through. Sound waves, water waves
are other examples of mechanical waves. Light waves are not mechanical waves, these are
electromagnetic waves which do not require medium to propagate.
Mechanical waves originate from a disturbance in the medium (such as a stone dropping in a pond) and
the disturbance propagates through the medium. The forces between the atoms in the medium are
responsible for the propagation of mechanical waves. Each atom exerts a force on the atoms near it,
and through this force the motion of the atom is transmitted to the others. The atoms in the medium do
not, however, experience any net displacement. As the wave passes, the atoms simply move back and
forth. Again for simplicity, we concentrate on the study of harmonic waves (that is those that can be
represented by sine and cosine functions).

TYPES OF MECHANICAL WAVES

Mechanical waves can be classified according to the physical properties of the medium, as well as in
other ways.
1. Direction of particle motion :
Waves can be classified by considering the direction of motion of the particles in the medium as wave
passes. If the disturbance travels in the x direction but the particles move in a direction, perpendicular
to the x axis as the wave passes it is called a transverse wave. If the motion of the particles were
parallel to the x axis then it is called a longitudinal wave. A wave pulse in a plucked guitar string is a
transverse wave. A sound wave is a longitudinal wave.

2. Number of dimensions :
Waves can propagate in one, two, or three dimensions. A wave moving along a taut string is a one
dimensional wave. A water wave created by a stone thrown in a pond is a two dimensional wave. A
sound wave created by a gunshot is a three-dimensional wave

3. Periodicity :
A stone dropped into a pond creates a wave pulse, which travels outward in two dimensions. There
may be more than one ripple created, but there is still only one wave pulse. If similar stones are
dropped in the same place at even time intervals, then a periodic wave is created.
4. Shape of wave fronts : The ripples created by a stone dropped into a pond are circular in shape. A
sound wave propagating outward from a point source has spherical wavefronts. A plane wave is a three
dimensional wave with flat wave fronts.
(Far away from a point source emitting spherical waves, the waves appear to be plane waves.)
A solid can sustain transverse as well as longitudinal wave. A fluid has no well-defined form or structure
to maintain and offer far more resistance to compression than to a shearing force. Consequently, only
longitudinal wave can propagate through a gas or within the body of an ideal (non viscous) liquid.
However, transverse waves can exist on the surface of a liquid. In the case of ripples on a pond, the
force restoring the system to equilibrium is the surface tension of the water, whereas for ocean waves,
it is the force of gravity.
Also, if disturbance is restricted to propagate only in one direction and there is no loss of energy during
propagation, then shape of disturbance remains unchanged.

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 Wave on a String

DESCRIBING WAVES :
Two kinds of graph may be drawn - displacement-distance and displacement-time.
A displacement - distance graph for a transverse mechanical wave shows the displacement y of the
vibrating particles of the transmitting medium at different distance x from the source at a certain instant
i.e. it is like a photograph showing shape of the wave at that particular instant.
The maximum displacement of each particle from its undisturbed position is the amplitude of the wave.
In the figure 1, it is OA or OB.

One wavelength

Displacement

O
Distance x
B

Crest Trough Crest Trough

The wavelength  of a wave is generally taken as the distance between two successive crests or two
successive trough. To be more specific, it is the distance between two consecutive points on the wave
which have same phase.
A displacement-time graph may also be drawn for a wave motion, showing how the displacement of
one particle at a particular distance from the source varies with time. If this is simple harmonic variation
then the graph is a sine curve.
WAVE LENGTH, FREQUENCY, SPEED
If the source of a wave makes f vibrations per second, so too will the particles of the transmitting
medium. That is, the frequency of the waves equals frequency of the source.
When the source makes one complete vibration, one wave is generated and the disturbance spreads
out a distance  from the source. If the source continues to vibrate with constant frequency f, then f
waves will be produced per second and the wave advances a distance f  in one second. If v is the
wave speed then
v = 
This relationship holds for all wave motions.
Travelling wave :
Imagine a horizontal string stretched in the x direction. Its equilibrium shape is flat and straight. Let y
measure the displacement of any particle of the string from its equilibrium position, perpendicular to the
string. If the string is plucked on the left end, a pulse will travel to the right. The vertical displacement y
of the left end of the string (x = 0) is a function of time.
i.e. y (x = 0, t) = f(t)
If there are no frictional losses, the pulse will travel undiminished, retaining its original shape. If the
pulse travels with a speed v, the ‘position’ of the wave pulse is x = vt. Therefore, the displacement of
x
the particle at point x at time t was originated at the left end at time t – . [y, (x, t) is function of both x
v
x x
and t]. But the displacement of the left end at time t is f(t) thus at time t – , it is f(t – ).
v v
Therefore :
x x
y(x, t) = y (x = 0, t – ) = f (t – )
v v
This can also be expressed as

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 Wave on a String

f f
 (vt – x)  – (x – vt) y (x, t) = g(x – vt)
v v
using any fixed value of t (i.e. at any instant), this shows shape of the string.
If the wave is travelling in – x direction, then wave equation is written as
x
y(x, t) = f(t + )
v
The quantity x – vt is called phase of the wave function. As phase of the pulse has fixed value
x – vt = const.
dx
Taking the derivative w.r.t. time =v
dt
where v is the phase velocity although often called wave velocity. It is the velocity at which a particular
phase of the disturbance travels through space.
In order for the function to represent a wave travelling at speed v, the three quantities x, v and t must
appear in the combination (x + vt) or (x – vt). Thus (x – vt)2 is acceptable but x2 – v2 t2 is not.

TRAVELLING SINE WAVE IN ONE DIMENSION (WAVE ON STRING) :
 x
The wave equation y  f  t   is quite general. If holds for arbitrary wave shapes, and for transverse
 v
as well as for longitudinal waves.
A complete description of the wave requires specification of f(x). The most important case, by far, in
physics and engineering is when f(x) is sinusoidal, that is, when the wave has the shape of a sine or
cosine function. This is possible when the source, that is moving the left end of the string, vibrates the
left end x = 0 in a simple harmonic motion. For this, the source has to continuously do work on the
string and energy is continuously supplied to the string.
The equation of motion of the left end may be written as
f (t) = A sin t
where A is amplitude of the wave, that is maximum displacement of a particle in the medium from its
equilibrium position  is angular frequency, that is 2f where f is frequency of SHM of the source.
The displacement of the particle at x at time t will be
 x  x
y  f t   or y = A sin   t   y = A sin (t – kx)
 v  v
 2 2 1
where k =  is called wave number. T = = is period of the wave, that is the time it takes
v   f
to travel the distance between two adjacent crests or through (it is wavelength ).
The wave equation y = A sin (t – kx) says that at x = 0 and t = 0, y = 0. This is not necessarily the
case, of source. For the same condition, y may not equal to zero. Therefore, the most general
expression would involve a phase constant , which allows for other possibilities,
y = A sin (t – kx + )
A suitable choice of  allows any initial condition to be met. The term kx – wt +  is called the phase of
the wave. Two waves with the same phase (on phase differing by a multiple of 2) are said to be “in
phase”. They execute the same motion at the same time.
The velocity of the particle at position x and at time t is given by
y
 A cos (t – kx + )
t
The wave equation has been partially differentiated keeping x as constant, to specify the particle. Note
dx
that wave velocity is different from particle velocity while waves velocity is constant for a medium
dt
and it along the direction of string, whereas particle velocity is perpendicular to wave velocity and is
dependent upon x and t.

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 Wave on a String

Example 1. A sinusoidal wave traveling in the positive x direction has an amplitude of 15 cm, wavelength
40 cm and frequency 8 Hz. The vertical displacement of the medium at t = 0 and x = 0 is also
15 cm, as shown.
y(cm)
40

15

x(cm)

(a) Find the angular wave number, period, angular frequency and speed of the wave.
(b) Determine the phase constant , and write a general expression for the wave function.
2 2 rad 
Solution : (a) k= = = rad/cm
 40 cm 20
1 1
T= = s  = 2f = 16 s–1
f 8
v = f = 320 cm/s
(b) It is given that A = 15 cm
and also y = 15 cm at x = 0 and t = 0
then using y = A sin (t – kx + )
15 = 15 sin   sin  = 1

or  = rad.
2
Therefore, the wave function is

y = A sin (t – kx + )
2
   rad  
= (15 cm) sin (16s1 )t    .x  
  20 cm  2

Example 2. A sinusoidal wave is travelling along a rope. The oscillator that generates the wave completes
60 vibrations in 30 s. Also, a given maximum travels 425 cm along the rope in 10.0 s. What is
the wavelength?
425 60 v
Solution : v= = 42.5 cm/s. f= = 2 Hz = = 21.25 cm.
10 30 f


THE LINEAR WAVE EQUATION :
By using wave function y = A sin (t – kx + ), we can describe the motion of any point on the string.
Any point on the string moves only vertically, and so its x coordinate remains constant. The transverse
velocity vy of the point and its transverse acceleration ay are therefore
 dy  y
vy =    = A cos (t – kx + ) ....(1)
 dt  x constant t
 dv y  v y  2 y
ay =     2 = –2 A sin (t – kx + ) ....(2)
 dt  x constant t t
and hence vy, max = A
ay, max = 2A

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 Wave on a String

The transverse velocity and transverse acceleration of any point on the string do not reach their
maximum value simultaneously. Infact, the transverse velocity reaches its maximum value (A) when
the displacement y = 0, whereas the transverse acceleration reaches its maximum magnitude (2A)
when y = ± A
 dy  y
further  dx   = –kA cos (t – kx + ) ....(3)
  t constant x
2y
= = – k2 A sin (t – kx + ) ....(4)
x 2
y  y
From (1) and (3) =–
t k x
 vP = – vw × slope
i.e. if the slope at any point is negative, particle velocity is positive and vice-versa, for a wave moving
along positive x axis i.e. vw is positive.
For example, consider two points A and B on the y-x curve for a wave, as shown. The wave is moving
along positive x-axis.
y

A
B

x
x1 x2

Slope at A is positive therefore at the given moment, its velocity is negative. That means it is coming
downward. Reverse is the situation for particle at point B.
Now using equation (2) and (4)
2 y k2 2 y 2 y 1 2 y
  
x 2 2 t 2 x 2 v 2 t 2
This is known as the linear wave equation or differential equation representation of the travelling wave
model. We have developed the linear wave equation from a sinusoidal mechanical wave travelling
through a medium, but it is much more general. The linear wave equation successfully describes waves
on strings, sound waves and also electromagnetic waves.

Example 3. Verify that wave function

2
y=
(x  3t)2  1
is a solution to the linear wave equation. x and y are in cm.
Solution : By taking partial derivatives of this function w.r.t. x and to t
2 y 12(x  3t)2  4
 ,
x2 [(x  3t)2  1]3
2 y 108(x  3t)2  36
and 
t 2 [(x  3t)2  1]3

2 y 1 2 x
or 
x 2 9 t 2
Comparing with linear wave equation, we see that the wave function is a solution to the linear
wave equation if the speed at which the pulse moves is 3 cm/s. It is apparent from wave
function therefore it is a solution to the linear wave equation.

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 Wave on a String


THE SPEED OF TRANSVERSE WAVES ON STRINGS
The speed of a wave on a string is given by
T
v

where T is tension in the string (in Newtons) and  is mass per unit length of the string (kg/m).
It should be noted that v is speed of the wave w.r.t. the medium (string).
In case the tension is not uniform in the string or string has nonuniform linear mass density then v is
speed at a given point and T and  are corresponding values at that point.

Example 4. Find speed of the wave generated in the string as in the situation shown. Assume that the
tension is not affected by the mass of the cord.

500 gm/m

20 kg
Solution : T = 20 × 10 = 200 N
200
v  20m / s
0.5

Example 5. A uniform rope of mass m and length L hangs from a ceilling. (a) Show that the speed of a
transverse wave on the rope is a function of y, the distance from the lower end, and is given by
v = gy . (b) Show that the time a transverse wave takes to travel the length of the rope is
given by t = 2 L / g .
Solution :

M, i
Y

(a) As mas per unit length

m
µ=
i
T
   V=  Tension at P = yg 

yg
    V= = yg

dy
(b) Now = yg
dt
t
dy

O
y
=

g dt
o

t=2 /g

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 Wave on a String

Example 6. A taut string having tension 100 N and linear mass density 0.25 kg/m is used inside a cart to
generate a wave pulse starting at the left end, as shown. What should be the velocity of the cart
so that pulse remains stationary w.r.t. ground.

T
Solution : Velocity of pulse =  20m / s

Now vPG  vPC  vCG
0 = 20 i + vCG
vCG = – 20i m/s

POWER TRANSMITTED ALONG THE STRING BY A SINE WAVE
When a travelling wave is established on a string, energy is transmitted along the direction of
propagation of the wave, in form of potential energy and kinetic energy
Average Power P = 22 f2 A2 v
Derivation of average power :

suppose equation of the wave is :

x
y = A sin ( (t – ))
v
Power = Fy Vy
y y x
P = (F sin ) = F sin  
t x t
Here F is tension in the string F = v2 and
y x
For small angle sin   tan  = and as y = A sin ( (t – )) so
x v
y A x
=– cos ( (t – )) and
x v v
y x
= A cos ( (t – ))
t v
y y x
putting the values P = v2
x x t
2
x  y 
= v so P = v3  
t  x 
2
 A  x   x 
P = v3  – cos  (t – )   or P = vA 2 2 cos2  (t – ) 
 v  v   v 
 x  1
 Average value of cos2  (t – )  = so
 v  2

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 Wave on a String

1
Average power P = vA 2 2
2
As  = 2 f
so Average Power P = 22 f2 A2 v

t
Energy Transferred =  0
Pav dt

Energy transferred in one time period = Pav T

This is also equal to the energy stored in one wavelength.

Intensity : Energy transferred per second per unit cross sectional area is called intensity of the wave.
Power P 1
=   = 2 A2v
Cross sectional area s 2
This is average intensity of the wave.
Energy density of a wave is energy per unit volume.
Pdt 
= 
svdt v

Example 7. A string with linear mass density m = 5.00 × 10–2 kg/m is under a tension of 80.0 N. How much
power must be supplied to the string to generate sinusoidal waves at a frequency of 60.0 Hz
and an amplitude of 6.00 cm?
Solution : The wave speed on the string is
1/ 2
T  80.0N 
v= =   = 40.0 m/s
  
 5.00  10 kg/ m 
2

Because  = 60 Hz, the angular frequency  of the sinusoidal waves on the string has the value
 = 2f = 2(60.0 Hz) = 377 s–1
Using these values in following Equation for the power, with A = 6.00 × 10 –2 m,. gives
1
p= 2A2 v
2
1
= (5.00 × 10–2 kg/m) (377s–1)2 × (6.00 × 10–2 m)2 (40.0 m/s)
2
= 512 W.

Example 8. Two waves in the same medium are represented by y-t curves in the figure. Find ratio of their
average intensities ?
y

5
wave 1

2 wave 2
t

1 12 A12 f 2  A2 1 25 25
Solution :  2 2 = 12 12 . = =
2 2 A 2 f2  A 2 4 4 16

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 Wave on a String


THE PRINCIPLE OF SUPERPOSITION
When two or more waves simultaneously pass through a point, the disturbance at the point is given by
the sum of the disturbances each wave would produce in absence of the other wave(s).
In general, the principle of superposition is valid for small disturbances only. If the string is stretched too
far, the individual displacements do not add to give the resultant displacement. Such waves are called
nonlinear waves. In this course, we shall only be talking about linear waves which obey the
superposition principle.
To put this rule in a mathematical form, let y1(x, t) and y2(x, t) be the displacements that any element of
the string would experience if each wave travelled alone. The displacement y(x, t) of an element of the
string when the waves overlap is then given by
y(x, t) = y1(x, t) + y2(x, t)
The principal of superposition can also be expressed by stating that overlapping waves algebraically
add to produce a resultant wave. The principle implies that the overlapping waves do not in any way
alter the travel of each other.
If we have two or more waves moving in the medium the resultant waveform is the sum of wave
functions of individual waves.
Fig: A sequence of pictures showing two pulses travelling in opposite directions along a stretched
string. When the two disturbances overlap they give a complicated pattern as shown in (b). In (c), they
have passed each other and proceed unchanged.

(a)

(b)

(c)

An Illustrative examples of this principle is phenomena

of interference and reflection of waves.

Example 9. Two waves passing through a region are represented by

y = (1.0 m) sin [(3.14 cm–1) x – (157 s–1) t]
and y = (1.5 cm) sin [(1.57 cm –1)x – (314 s–1) t].
Find the displacement of the particle at x = 4.5 cm at time t = 5.0 ms.
Solution : According to the principle of superposition, each wave produces its disturbance independent of
the other and the resultant disturbance is equal to the vector sum of the individual disturbance.
The displacements of the particle at x = 4.5 cm at time t = 5.0 ms due to the two waves are,
y1 = (1.0 cm) sin [(3.14 cm–1) (4.5 cm)– (157 s–1) (5.0 × 10–3 s)]
    1.0 cm
= (1.0 cm) sin  4.5    = (1.0 cm) sin  4   =
 4   4  2
and y2 = (1.5 cm) sin [(1.57 cm–1)(4.5 cm) – (314 s–1) (5.0 × 10–3 s)]
     1.5 cm
= (1.5 cm) sin  2.25    = (1.5 cm ) sin  2   = (1.5 cm) sin =–
 2  4 4 2
0.5 cm
The net displacement is : y = y1 + y2 = = – 0.35 cm.
2

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 Wave on a String


INTERFERENCE OF WAVES GOING IN SAME DIRECTION
Suppose two identical sources send sinusoidal waves of same angular frequency  in positive
x-direction. Also, the wave velocity and hence, the wave number k is same for the two waves. One
source may be situated at different points. The two waves arriving at a point then differ in phase. Let the
amplitudes of the two waves be A1 and A2 and the two waves differ in phase by an angle . Their
equations may be written as
y1 = A1 sin (kx – t) and y2 = A2 sin (kx – t + ).
According to the principle of superposition, the resultant wave is represented by
y = y1 + y2 = A1 sin (kx – t) + A2 sin (kx – t + ).
we get y = A sin (kx – t + )
where, A= A12  A22  2A1A2 cos  (A is amplitude of the resultant wave)
A 2 sin 
Also, tan  = ( is phase difference of the resultant wave with the first wave)
A1  A 2 cos 
Constructive and Destructive Interference
Constructive Interference :
When resultant amplitude A is maximum
A = A1 + A2
when cos  = + 1 or  = 2n
where n is an integer.
Destructive interference :
When resultant amplitude A is minimum
or A = |A1 – A2| When cos  = – 1 or  = (2n + 1)
where n is an integer.

Example 10. Two sinusoidal waves of the same frequency travel in the same direction along a string. If
A1 = 3.0 cm, A2 = 4.0 cm, 1 = 0, and 2 = /2 rad, what is the amplitude of the resultant wave?

Solution : Resultant amplitude = 32  42  2  3  4  cos90º = 5 cm.


REFLECTION AND TRANSMISSION OF WAVES
A travelling wave, at a rigid or denser boundary, is reflected with a phase reversal but the reflection at
an open boundary (rarer medium) takes place without any phase change. The transmitted wave is
never inverted, but propagation constant k is changed.

INCIDENT WAVE INCIDENT WAVE

yi = Ai sin(t – k1x) yi = Ai sin(t – k1 x)

TRANSMITTED WAVE TRANSMITTED WAVE

yt = At sin(t – k2x) yt = At sin(t – k2x)

REFLECTED WAVE REFLECTED WAVE

yr = Ar sin(t + k1x + ) yr = Ar sin(t + k1x)

Fig. : Reflection at denser boundary Fig. : Reflection at rarer boundary

Amplitude of reflected and transmitted waves :

v1 and v2 are speeds of the wave in incidenting and reflecting mediums respectively then
v  v1 2v 2
Ar = 2 Ai At = .A i
v1  v 2 v1  v 2
Ar is positive if v2 > v1, i.e., wave is reflected from a rarer medium.

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Example 11. A harmonic wave is travelling on string 1. At a junction with string 2 it is partly reflected and
partly transmitted. The linear mass density of the second string is four times that of the first
string, and that the boundary between the two strings is at x = 0. If the expression for the
incident wave is, yi = Ai cos (k1 x – 1t)
What are the expressions for the transmitted and the reflected waves in terms of Ai, k1 and 1?
Solution : Since v = T/ , T2 = T1 and 2 = 41
v1
we have, v2 = ... (i)
2
The frequency does not change, that is,
1 = 2 ....(ii)
Also, because k = /v, the wave numbers of the harmonic waves in the two strings are related
by,
 1 
k2 = 2 = = 2 1 = 2k1 ....(iii)
v2 v1 / 2 v1
The amplitudes are,
 2v 2   2(v1 / 2)  2
At =   Au =   Ai = Ai ....(iv)
v
 1  v 2  v
 1  (v1 / 2)  3
 v  v1   (v1 / 2)  v1  Ai
and Ar =  2  Au =   Ai = ....(v)
 v1  v 2   v1  (v1 / 2)  3
Now with equation (ii), (iii) and (iv), the transmitted wave can be written as,
2
yt = A cos (2k1 x – 1t) Ans.
3 i
Similarly the reflected wave can be expressed as,
A
= i cos (k1x + 1t + ) Ans.
3

STANDING WAVES :
Suppose two sine waves of equal amplitude and frequency propagate on a long string in opposite
directions. The equations of the two waves are given by
y1 = A sin (t – kx) and y2 = A sin (t + kx + ).
These waves interfere to produce what we call standing waves. To understand these waves, let us
discuss the special case when  = 0.
The resultant displacements of the particles of the string are given by the principle of superposition as
y = y1 + y2
= A [sin (t – kx) + sin(t + kx)]
= 2A sin t cos kx
or, y = (2A cos kx) sin t. ....

This is the required result and from this it is clear that :

1. As this equation satisfies the wave equation,
2 y 1 2 y

x 2
v2  t2
it represents a wave. However, as it is not of the form f(ax ± bt), the wave is not travelling and
so is called standing or stationary wave.

2. The amplitude of the wave

As = 2A cos kx
is not constant but varies periodically with position (and not with time as in beats).

NEET_XI_PHYSICS (WAVES FLUIDS & PROPERTIES OF MATTER) # 80

Wave on a String

3. The points for which amplitude is minimum are called nodes and for these
 3 5
cos kx = 0, i.e., kx = , ,
2 2 2
 3 5  2 
i.e., x= , , , .... as k   
4 4 4  
i.e., in a stationary wave, nodes are equally spaced.
4. The points for which amplitude is maximum are called antinodes and for these,
cos kx = ± 1, i.e., kx = 0, , 2, 3, ......
 2 3  2 
i.e., x = 0, , , ,....  as k 
2 2 2   
i.e., like nodes, antinodes are also equally spaced with spacing (/2) and Amax = ± 2A.
Furthermore, nodes and antinodes are alternate with spacing (/4).
5. The nodes divide the medium into segments (or loops). All the particles in a segment vibrate in
same phase, but in opposite phase with the particles in the adjacent segment. Twice in one
period all the particles pass through their mean position simultaneously with maximum velocity
(As), the direction of motion being reversed after each half cycle.
Node

t=0

t = T/8

t = T/4
t = 3T/8

t = T/2

t = 5T/8

t = 3T/4

t = 7T/8

t=T

Antinode
(a)
6. Standing waves can be transverse or longitudinal, e.g., in strings (under tension) if reflected
wave exists, the waves are transverse-stationary, while in organ pipes waves are longitudinal-
stationary.
7. As in stationary waves nodes are permanently at rest, so no energy can be transmitted across
them, i.e., energy of one region (segment) is confined in that region. However, this energy
oscillates between elastic potential energy and kinetic energy of the particles of the medium.
When all the particles are at their extreme positions KE is minimum while elastic PE is
maximum (as shown in figure A), and when all the particles (simultaneously) pass through their
mean position KE will be maximum while elastic PE minimum (Figure B). The total energy
confined in a segment (elastic PE + KE), always remains the same.
v=0 v = max. = As string
string
v = min = 0
y y

x x

Elastic PE = max = E Elastic PE = min = 0

Kinetic energy = min = 0 Kinetic energy = max = E
(A) (B)

NEET_XI_PHYSICS (WAVES FLUIDS & PROPERTIES OF MATTER) # 81

 Wave on a String

Example 12. Two waves travelling in opposite directions produce a standing wave. The individual wave
functions are
y1 = (4.0 cm) sin(3.0x – 2.0t)
y2 = (4.0 cm) sin (3.0x + 2.0t)
where x and y are in centimeter.
(a) Find the maximum displacement of a particle of the medium at x = 2.3 cm.
(b) Find the position of the nodes and antinodes.
Solution : (a) When the two waves are summed, the result is a standing wave whose mathematical
representation is given by Equation, with A = 4.0 cm and k = 3.0 rad/cm;
y = (2A sin kx) cos t = [(8.0 cm) sin 3.0 x] cos 2.0 t
Thus, the maximum displacement of a particle at the position x = 2.3 cm is
ymax = [(8.0 cm) sin 3.0x]x = 2.3 cm
= (8.0 m) sin (6.9 rad) = 4.6 cm
(b) Because k = 2/ = 3.0 rad/cm, we see that  = 2/3cm. Therefore, the antinodes are
located at
  
x=n   cm (n = 1, 3, 5, .....)
 6.0 
and the nodes are located at
   
x=n   cm (n = 1, 2, 3, .....)
2  3.0 

Example 13. Two travelling waves of equal amplitudes and equal frequencies move in opposite direction
along a string. They interfere to produce a standing wave having the equation.
y = A cos kx sin t
in which A = 1.0 mm, k = 1.57 cm –1 and  = 78.5 s–1. (a) Find the velocity and amplitude of the
component travelling waves. (b) Find the node closest to the origin in the region x > 0. (c) Find
the antinode closest to the origin in the region x > 0. (d) Find the amplitude of the particle at x =
2.33 cm.
Solution : (a) The standing wave is formed by the superposition of the waves
A A
y1 = sin (t – kx) and y2 = sin (t + kx).
2 2
The wave velocity (magnitude) of either of the waves is
 78.5s1
v= = = 50 cm/s; Amplitude = 0.5 mm.
k 1.57cm1
(b) For a node, cos kx = 0.
The smallest positive x satisfying this relation is given by
 3.14
kx = /2 or, x= = = 1 cm
2k 2  1.57 cm 1
(c) For an antinode, |cos kx| = 1.
The smallest positive x satisfying this relation is given by

kx =  or, x= = 2 cm
k
(d) The amplitude of vibration of the particle at x is given by | A cos kx |. For the given point,
7 
kx = (1.57 cm –1) (2.33 cm) = =+ .
6 6
Thus, the amplitude will be
3
(1.0 mm) | cos ( + /6) | = mm = 0.86 mm.
3


NEET_XI_PHYSICS (WAVES FLUIDS & PROPERTIES OF MATTER) # 82
 Wave on a String


VIBRATION OF STRING :
(a) Fixed at both ends :
Suppose a string of length L is kept fixed at the ends x = 0 and x = L. In such a system suppose we
send a continuous sinusoidal wave of a certain frequency, say, toward the right. When the wave
reaches the right end. It gets reflected and begins to travel back. The left-going wave then overlaps the
wave, which is still travelling to the right. When the left-going wave reaches the left end, it gets reflected
again and the newly reflected wave begins to travel to the right. Overlapping the left-going wave. This
process will continue and, therefore, very soon we have many overlapping waves, which interfere with
one another. In such a system, at any point x and at any time t, there are always two waves, one
moving to the left and another to the right. We, therefore, have
y1(x, t) = ym sin (kx – t) (wave travelling in the positive direction of x-axis)
and y2(x, t) = ym sin (kx + t) (wave travelling in the negative direction of x-axis).
The principle of superposition gives, for the combined wave
y’(x, t) = y1(x, t) + y2(x, t)
= ym sin (kx – wt) + ym sin (kx + t) = (2ym sin kx) cos t
It is seen that the points of maximum or minimum amplitude stay at one position.
Nodes : The amplitude is zero for values of kx that give sin kx = 0 i.e. for,
kx = n, for n = 0, 1, 2, 3,.....
Substituting k = 2 in this equation, we get

x=n , for n = 0, 1, 2, 3,.....
2

The positions of zero amplitude are called the nodes. Note that a distance of or half a wavelength
2
separates two consecutive nodes.
Antinodes :
The amplitude has a maximum value of 2ym, which occurs for the values of kx that give |sin kx| = 1.
Those values are
kx = (n + 1/2) for n = 0, 1, 2, 3,....
Substituting k = 2 in this equation, we get.

x = (n + 1/2) for n = 0, 1, 2, 3,....
2
as the positions of maximum amplitude. These are called the antinodes. The antinodes are separated
by 
/2 and are located half way between pairs of nodes.
For a stretched string of length L, fixed at both ends, the two ends of the ends is chosen as position
x = 0, then the other end is x = L. In order that this end is a node; the length L must satisfy the condition

L=n , for n = 1, 2, 3,....
2
This condition shows that standing waves on a string of length L have restricted wavelength given by
2L
= , for n = 1, 2, 3,.....
n
The frequencies corresponding to these wavelengths follow from Eq. as
v
f=n , for n = 1, 2, 3,.....
2L

NEET_XI_PHYSICS (WAVES FLUIDS & PROPERTIES OF MATTER) # 83

 Wave on a String

where v is the speed of travelling waves on the string. The set of frequencies given by equation are
called the natural frequencies or modes of oscillation of the system. This equation tells us that the
v
natural frequencies of a string are integral multiples of the lowest frequency f = , which corresponds
2L
to n = 1. The oscillation mode with that lowest frequency is called the fundamental mode or the first
harmonic. The second harmonic or first overtone is the oscillation mode with n = 2. The third harmonic
and second overtone corresponds to n = 3 and so on. The frequencies associated with these modes
are often labeled as 1, 2, 3 and so on. The collection of all possible modes is called the harmonic
series and n is called the harmonic number.

Some of the harmonic of a stretched string fixed at both the ends are shown in figure.
A
(a)
Fundamental
or first harmonic
f0 = v/2L
A

A A (b)
second harmonic
or first overtone =
N f1 = 2f0 = 2v/2L
A A
(c)
A A A third harmonic
or 2nd
N N overtone =
A A A f2 = 3f0, = 3v/2L
(d)
A A A A
fourth harmonic
or 3rd
N N N overtone =
A A A A
f3 = 4f0, = 4v/2L

Example 14. A middle C string on a piano has a fundamental frequency of 262 Hz, and the A note has
fundametal frequency of 440 Hz. (a) Calculate the frequencies of the next two harmonics of the
C string. (b) If the strings for the A and C notes are assumed to have the same mass per unit
length and the same length, determine the ratio of tensions in the two strings.
Solution :. (a) Because f1 = 262 Hz for the C string, we can use Equation to find the frequencies f 2 and f3;
f2 = 2f1 = 524 Hz
f3 = 3f1 = 786 Hz
Using Equation for the two strings vibrating at their fundamental frequencies gives
1 TA 1 TC
f1A =  f1C =
2L  2L 
2 2
f1A TA TA f   440Hz 
 =  =  1A  =   = 2.82. Ans.
f1C TC TC  f1C   262Hz 

NEET_XI_PHYSICS (WAVES FLUIDS & PROPERTIES OF MATTER) # 84

 Wave on a String

Example 15. A wire having a linear mass density 5.0 × 10–3 kg/m is stretched between two rigid supports
with a tension of 450 N. The wire resonates at a frequency of 420 Hz. The next higher
frequency at which the same wire resonates is 490 Hz. Find the length of the wire.
Solution : Suppose the wire vibrates at 420 Hz in its nth harmonic and at 490 Hz in its (n + 1)th harmonic.
n F
420 s–1 = ....(i)
2L 
(n  1) F
and 490 s–1 = ....(ii)
2L 
490 n 1
This gives = or n = 6.
420 n
Putting the value in (i),
6 450N 900
420 s–1 = = m/s
2L 5.0  10 3 kg/ m L
900
or L= m = 2.1 m
420


(b) Fixed at one end :
Standing waves can be produced on a string which is fixed at one end and whose other end is free to
move in a transverse direction. Such a free end can be nearly achieved by connecting the string to a
very light thread.
If the vibrations are produced by a source of “correct” frequency, standing waves are produced. If the
end
x = 0 is fixed and x = L is free, the equation is again given by
y = 2A sin kx cos t
with the boundary condition that x = L is an antinode. The boundary condition that x = 0 is a node is
automatically satisfied by the above equation. For x = L to be an antinode,
sin kL = ± 1
 1 2L  1
or, kL =  n    or, = n  
 2   2
1
n
2Lf 1  1 v 2
or, =n+ or, f = n   = T /  .....
v 2  2  2L 2L
These are the normal frequencies of vibration. The fundamental frequency is obtained when n = 0, i.e.,
f0 = v/4L

(a) Fundamental
N
A
The overtone frequencies are
3v
f1 = = 3f0
4L
First
(b) Overtone
N N
A A
5v
f2 = = 5f0
4L
Second
(c)
Overtone
A N N
A A
We see that all the harmonic of the fundamental are not the allowed frequencies for the standing
waves. Only the odd harmonics are the overtones. Figure shows shapes of the string for some of the
normal modes.

NEET_XI_PHYSICS (WAVES FLUIDS & PROPERTIES OF MATTER) # 85

Wave on a String


SONOMETER :
Sonometer consists of a hollow rectangular box of light wood. One end of the experimental wire is
fastened to one end of the box. The wire passes over a frictionaless pulley at the other end of the box.
The wire is stretched by a tension T.

The box serves the purpose of increasing the loudness of the sound produced by the vibrating wire. If
the length the wire between the two bridges is , then the frequency of vibration is
l T
f =
2 m
To test the tension of a tuning fork and string, a small paper rider is placed on the string. When a
vibrating tuning fork is placed on the box, and if the length between the bridges is properly adjusted,
then when the two frequencies are exactly equal, the string quickly picks up the vibrations of the fork
and the rider is thrown off the wire.
Comment :
mass of wire r 2 d
m = = = r2d
length of wire
where r is the radius of the wire and d is the density of the material of the wire. Thus the frequency of
vibration of a given string under tension is
1 1
f  Thus f  (for same material wires)
r 2d r
1
and f  (for different material wires of same radius).
d
MELDE'S EXPERIMENT :
In Melde's experiment, one end of a flexible piece of thread is tied to the end of a tuning fork. The other
end passed over a smooth pulley carries a pan which can be suitable loaded.

Case 1. In a vibrating string of fixed length, the product of number of loops in a vibrating string and
square root of tension is a constant or
n T = constant.
Case 2. When the tuning fork is set vibrating as shown in fig. then the prong vibrates at right angles to
the thread. As a result the thread is set into motion. The frequency of vibration of the thread (string) is
equal to the frequency of the tuning fork. If length and tension are properly adjusted then, standing
waves are formed in the string. (This happens when frequency of one of the normal modes of the string
matched with the frequency of the tuning fork). Then, if n loops are formed in the thread, then the
frequency of the tuning fork is given by
n T
f =
2 m

NEET_XI_PHYSICS (WAVES FLUIDS & PROPERTIES OF MATTER) # 86

 Wave on a String

Case 3. If the tuning fork is turned through a right angle, so that the prong vibrates along the length of
the thread, then the string performs only a half oscillation for each complete vibrations of the prong.
This is because the thread sags only when the prong moves towards the pulley i.e. only once in a
vibration.

The thread performs sustained oscillations when the natural frequency of the given length of the thread
under tension is half that of the fork. Thus if n loops are formed in the thread, then the frequency of the
tuning fork is
2n T
f =
2 m

Problem 1. A wave pulse moving along the x axis is represented by the wave function
2
y(x, t) =
 x  3t 2  1
where x and y are measured in cm and t is in seconds.
(i) In which direction is the wave moving?
(ii) Find speed of the wave.
2
Solution : y =
 x  3t 2  1
(i) As wave is moving in +ve x direction because
y = (x, t) = f(t – x/v) = f/v(vt – x)
(ii) Now x – v t is compared with x – 3t
 v = 3 cm/sec.
Ans. (i) Positive x axis (ii) 3 cm/s.
6
Problem 2. At t = 0, a transvalues wave pulse in a wire is described by the function y = where
x 3
3

x and y are in metres write the function y(x, t) that describes this wave if it is travelling in the
positive x direction with a speed of 4.5 m/s.
6 6 6
Solution : y= 3 = f(x) As y(x, t) = f(x – vt) = Ans.
x 3 (x  4.5t)  3  x  4.5t 2  3
2

Problem 3. The wave function for a travelling wave on a string is given as


y (x, t) = (0.350 m) sin (10 t – 3x + )
4
(a) What are the speed and direction of travel of the wave?
(b) What is the vertical displacement of the string at t = 0, x = 0.1 m?
(c) What are wavelength and frequency of the wave?

Solution : Y (x, t) = (0.350m) sin (10t – 3  x + )
4
comparing with equation ;

Y = A sin (wt – kx + ) w = 10  , k = 3, f =
4
w 10
(a) speed = = = 3.33 m/sec and along +ve x axis
k 3

NEET_XI_PHYSICS (WAVES FLUIDS & PROPERTIES OF MATTER) # 87

 Wave on a String

   3 
(b) y (0.1, 0) = 0.35 sin (10  x O – 3  (0.1) + ) = 0.35 sin    = – 5.48 cm
4  4 10 
2 2  10 / 3
(c) k= = 3  =)   cm = 0.67 cm and ¦= = = 5 Hz.
 3  2/3
Problem 4. Show that the wave function y = eb(x – vt) is a solution of the linear wave equation.
y y
Solution : Y = eb(x–v t) = beb(x–v t) and = (bv)eb(xv– v t)
x t
2y 2y
= b2 eb(x–v t) and = (bv)2 eb(x – v t)
x 2 t 2
2y 1 2y
obviously ; = which is a Linear wave equation.
x 2 v 2 t 2
Problem 5. A transverse wave of amplitude 0.50 mm and frequency 100 Hz is produced on a wire
stretched to a tension of 100 N. If the wave speed is 100 m/s, what average power is the
source transmitting to the wire?
Solution : Average Power;
< P > = 2 2 2 A2µV
100
= 22 (100)2 (0.5 × 10–3)3 = 49 mW Ans 49 mW
(100)2
Problem 6. Two sinusoidal waves of the same frequency are to be sent in the same direction along a taut
string. One wave has an amplitude of 5.0 mm, the other 8.0 mm. (a) What phase difference 1
between the two waves results in the smallest amplitude of the resultant wave? (b) What is that
smallest amplitude? (c) What phase difference 2 results in the largest amplitude of the
resultant wave? (d) What is that largest amplitude? (e) What is the resultant amplitude if the
phase angle is (1 – 2)/2?
Solution : (a) For smallest amplitude ;
AR = | A1 – A2 | and that is possible when 1 =  between A1 and A2
(b) AR = | A1 – A2 | = 3 mm
(c) for largest amplitude ;
AR = | A1 + A2 | and that is possible when  2 = 0 between A1 and A2
(d) AR = | A1 + A2 | = 13 mm
1  2 o  
(e) when  = = =  AR = [A12 + A22 + 2A1 A2 cos ]1/2 = 9.4 mm
2 2 2 2
Ans. (a)  rad; (b) 3.0 mm; (c) 0 rad; (d) 13 mm; (e) 9.4 mm

Problem 7. A string fixed at both ens is 8.40 m long and has a mass of 0.120 kg. It is subjected to a
tension of 96.0 N and set oscillating. (a) What is the speed of the waves on the string? (b) What
is the longest possible wavelength for a standing wave? (c) Give the frequency of the wave.
T 96
Solution : (a) V = = = 82 m/sec.
  0.12 
 8.4 
 
(b) for longent possible warelength ;


2
 = 2 = 2 × 8.4 = 16.8 m
v 82
(c) V =  = ) ¦ = = = 4.88 HZ. Ans. (a) 82.0 m/s, (b) 16.8 m, (c) 4.88 Hz.
 16.8

NEET_XI_PHYSICS (WAVES FLUIDS & PROPERTIES OF MATTER) # 88