HG T KJ

HG T KJ

and

y(t + T) = y(t).

Obtain the Fourier series expansion of y (t).

9. A displacement curve is given by

f (t) = sin ωt for 0 < t < T/2,

f (t) = –sin ωt for T/2 < t < T

2π where T =

and f (t + T) = f (t).

ω Give a rough sketch of the displacement curve and obtain the Fourier series expansion for f(t).

FOURIER ANALYSIS 201

10. A square wave is defined as

f (t) = B for –

<< t

TT = 0 elsewhere in the range − to ,

where 0 < α < 1 and f (t) is periodic with period T = . Obtain the Fourier series

of the function.

11. For a periodic function f (t) of fundamental frequency f 0 (i.e., f 0 = , where T is the T

period) we have the Fourier series expansion

∑ b a n cos 2 π nft 0 + b n sin 2 π nft 0 g 2 ,

f (t) = 0 +

F or, − << t I and f(t + T) = f(t).

T where f(t) is defined for the time –

<< t

2 f 0 2 f 0 HG 2 2 KJ

Show that the Fourier coefficients are given by

a n = 2 f 0 ft bgb cos 2 π nf t dt 0 g , n = 012 , , , ...

b n = 2 f 0 ft bgb sin 2 π nf t dt 0 g , n = 123 , , , ... .

and

12. Show that for values of x between 0 and π, the function π(π – 2x) can be expanded

in the cosine series

13. Find a Fourier sine series corresponding to the function

f (x) = cos x, 0 < x < π.

14. Expand f (x) = x, 0 < x < 2 in a half range in (a) sine series (b) cosine series.

15. (a) Prove that for 0 ≤ x ≤ π,

cos 2 x cos 4 x cos 6

x(π – x) =

2 + NM ...

1 2 3 QP

(b) Show that

bg

(i) ∑ =

(ii)

202 WAVES AND OSCILLATIONS

16. Show that when 0 ≤ x ≤ π,

1 2 2 cos 3 x cos 5 x

ππ b − 2 x ge π + 2 π x − 2 x j = cos x + 4 + 4 + ...

17. Find the Fourier cosine and sine transforms of exp (– at), t > 0. By taking inverse Fourier transform show that

d e − (ii) ax ω = , x > 0

18. (a) Find the Fourier transform of the rectangular pulse

R 0 for || x > X

f (x) = S| 1 for || x < X T| 2 X

(b) By taking Fourier inverse transform show in the limit X → 0 that

2π −∝ z

19. By taking the Fourier transform of the differential equation

dy 2

2 + xy =0 dx

obtain the differential equation satisfied by g (ω), and find its solution. Inverting the transform show that

2 π −∝ z N MM HG 3 KJ Q PP

y(x) =

d ω exp − i ω x −

[The integral is known as Airy integral ].

20. Using the definition of momentum function g(p) [see problem 23] find the momentum function representation for the one-dimensional Schrödinger equation for harmonic oscillator:

− h 2 d 2 ψ bg

1 kx + 2

ψ bg x = Eψ(x).

2m dx 2

21. A linear quantum oscillator is its ground state has the normalised wave function

ψ(x) = a − 12 π − 14 exp − x 2 e 2 a 2 j .

FOURIER ANALYSIS 203

Show that the corresponding momentum function is

− 12 π − 14 h − 12 − 2 2 g(p) = 2 a exp e ap 2 h j . L

−∝ z

e − β x dx j = exp β 4 MM α

Use the relation exp − α x 2 π

PP Q

22. Using the relation

< p 2 >= 2 g * bgbg p p g p dp ,

Find < p 2 > in the ground state of quantum harmonic oscillator of problem 2l.

23. Squaring both sides of the expression of problem 16 of supplementary problems and integrating x from 0 to π show that

[Hints: cos b 2 n + 1 g x cos b 2 m + 1 g x dx = δ z mn

Vibrations of Strings and Membranes

8.1 TRANSVERSE VIBRATION OF A STRING FIXED AT TWO ENDS

The normal mode frequencies of transverse waves on a string of length l which is under tension T and is fixed at two ends are given by

, n = 1, 2, 3,....,

where µ is the mass per unit length of the string. For the fundamental mode, n = 1 and higher values of n correspond to higher harmonics (overtones).

8.2 PLUCKED STRING

A string which is fixed at both ends and is under some tension, is plucked at some point to

a small height and then released from rest. The string then executes small transverse vibration.

8.3 STRUCK STRING

A perfectly flexible string which is fixed at both ends and is under some tension, is struck by a hammer at a point, the time of contact between the string and the hammer being very very small. The force given by the hammer is of the nature of an impulse and it imparts initially (t = 0) a velocity to the point struck but all other points have zero velocity. We investigate the motion of the string at later times.

8.4 BOWED STRING

We study the motion of the violin string when bowed at some point. The string is fixed at both ends and is under some tension.

Characteristics of a bowed string: For maintained vibration of the bowed string, Helmholtz observed the following characteristics: (i) The bowed point moves with the same velocity as that of the bow.

VIBRATIONS OF STRINGS AND MEMBRANES 205

(ii) All points of the string vibrate in a plane at y any instant. The motion of any point on the string consists of an ascent with uniform forward velocity followed by a descent with another uniform backward velocity. The two velocities are equal in magnitude at the middle point of the string.

The displacement-time graph of a point on the

string can be represented as two step zig-zag straight lines (Fig. 8.1). Here, the point under observation moves forward with constant velocity for the time

AB = T 1 and moves backward with another constant

velocity for the time BC = T 2 . The time period of

vibration τ = T

1 + T 2 = AC.

Fig. 8.1

8.5 TRANSVERSE VIBRATION OF MEMBRANES

A perfectly flexible thin membrane is stretched and then fixed along its entire boundary in the xy-plane. The tension T per unit length caused by the stretching of the membrane is same in all directions. The deflection u (x, y, t) for small transverse vibration of the membrane satisfies the two-dimensional wave equation

2 =v 2 ∇ 2 u

2 ∂ 2 ∂ where v = 2 Tσσ , being mass per unit area of the membrane and ∇ =

SOL SOL SOL SOL SOLVED PR VED PROBLEMS VED PR VED PR VED PR OBLEMS OBLEMS OBLEMS OBLEMS

1. Derive the formula given in Eqn. (8.1).

Solution

Since the string is fixed at its ends, each end must be stationary and therefore nodes are

n=1

produced at the two ends. Again, the string must

be an integral number of half-wavelengths in length (Fig. 8.2): n=2

l=n n , n = 1, 2, 3, ...

where λ n is the wavelength of the nth normal

n=3

mode. The frequency of the nth mode is

Fig. 8.2

where v = Tµ is the velocity of propagation of transverse wave along the string.

206 WAVES AND OSCILLATIONS

2. A string of length l = 0.5 m and mass per unit length 0.01 kgm –1 has a fundamental frequency of 250 Hz. What is the tension in the string?

Solution

We have

1 T ν= 2l µ

or T = 4l 2 ν 2 µ = 4 × (0.5) 2 (250) 2 × 0.01

= 625 N.

3. Two wires of radii r and 2r respectively are welded together end to end. This combination is used as a sonometer wire kept under tension T. The welded point is midway between the two bridges. What would be the ratio of the number of loops formed in the wires such that the joint is a node when stationary vibrations are set up in the wires.

(I.I.T. 1976)

Solution

Let n 1 and n 2 be the number of loops formed in the wires of radii r and 2r respectively. Then

and ν 2 =

Since the welded wire is continuous, ν 1 =ν 2 and

1 π r 2 ×× 1 ρ

2 = n 2 µ 2 MM N π bg 2 r ×× 1 ρ Q PP 2

where ρ = density of the material of the wires.

4. A metal wire of diameter 1 mm is held on two knife edges separated by a distance of

50 cm. The tension in the wire is 100 N. The wire vibrating with its fundamental frequency and a vibrating tuning fork together produce 5 beats/sec. The tension in the wire is then reduced to 81 N. When the two are excited, beats are heard at the same rate. Calculate (i) the frequency of the fork and (ii) the density of the material of wire.

(I.I.T. 1980)

Solution

Let the frequency of the tuning fork be n. When tension in the wire is 100 N, the fundamental frequency of the wire is n + 5, and when T = 81 N, the fundamental frequency of the wire is n – 5, so that 5 beats/sec are produced in both the cases. Thus,

On solving these two equations, we get

n = 95 Hz, µ = 0.01 kg/m.

VIBRATIONS OF STRINGS AND MEMBRANES 207

Now, ρ = Density of the material of wire

π 2 r × 1 π × . 0 5 10 × − 32 e j

= 12732.4 kg/m 3 .

5. A string 25 cm long and having a mass of 2.5 g is under tension. A pipe closed at one end is 40 cm long. When the string is vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats per second are heard. It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in air is 320 m/s, find the tension in the string.

(I.I.T. 1982)

Solution

The fundamental frequency of the closed pipe is

The frequency of the vibrating string is 208 Hz so that with decrease in tension of the string the frequency of the string decreases and the beat frequency also decreases.

The first overtone of the string is

= 208 × 208 × l 2 × (m/l)

where

l = 25 cm = 0.25 m

T = 27.04 N.

6. A sonometer wire fixed at one end has a solid mass M hanging from its other end to produce tension in it. It is found that a 70 cm length of the wire produces a certain fundamental frequency when plucked. When the same mass M is hanging in water, completely submerged in it, it is found that the length of the wire has to be changed by 5 cm in order that it will produce the same fundamental frequency. Calculate the density of the material of the mass M hanging from the wire.

(I.I.T 1972)

Solution

The fundamental frequency of the sonometer wire is

When M is submerged in water completely, the tension in the wire decreases to [M – M/ρ] g, where ρ is the density of the material of the mass M. The frequency remains the same for a length of 65 cm. Hence,

1 M − M ρ g ν=

208 WAVES AND OSCILLATIONS

From these two equations, we have

HG 65 KJ

M − M /ρ

or

ρ = 7.26 g/cm 3 = 7.26 × 10 3 kg/m 3 .

7. A steel wire of length 1 m, mass 0.1 kg and uniform cross sectional area 10 –6 m 2 is rigidly fixed at both ends. The temperature of the wire is lowered by 20°C. If transverse waves are set up by plucking the string in the middle, calculate the frequency of the fundamental mode of vibration.

Young’s modulus of steel = 2 × 10 11 N/m 2 ,

Coefficient of linear expansion of steel = 1.21 × 10 –5 /°C. (I.I.T. 1984)

Solution

Change in length dl of the wire when the temperature is lowered by 20°C is dl = l × α × 20 = 1 × 1.21 × 10 –5 × 20

= 2.42 × 10 –4 m. TA

or T = YA dl/l = 2 × 10 11 × 10 –6 × 2.42 × 10 –4

= 48.4 N.

The fundamental frequency is

O = 11 Hz.

1 L 48.4 2

2 NM 0.1 QP

8. If y be the displacement at x of an elementary segment δx of a uniform string under tension T at any instant, then show that the kinetic energy (K.E.) and potential energy (P.E.) of the element at that instant are given by

K.E. =

2 HG ∂ t KJ

P.E. =

2 HG ∂ x KJ

where µ is the mass per unit length of the string.

ds dy

Solution

The mass of the element δx is µδx and the velocity is

. Thus, the kinetic energy is µδ x F I .

2 HG ∂ t KJ

dx

Fig. 8.3

Let δS be the element in the displaced position (Fig. 8.3). The work done against the tension T when the element is stretched is the potential energy of the element.

Since,

(δS) 2 = (δx) 2 + (δy) 2 ,

212 L / F

MM HG δ N x KJ PP Q

δS = δx

VIBRATIONS OF STRINGS AND MEMBRANES 209

≈ δx 1 + F I in the limit δx → 0.

N MM 2 HG ∂ x KJ Q PP

We assume that the y-displacement of the string is small so that

HG is small. ∂ x KJ

I δ. x

Thus,

P.E. =

2 HG ∂ x KJ

Here, we neglect the increase in tension of the string when it is stretched.

9. If a wave of the form y = f (u) with u = x – vt moves along the string under tension T with velocity v, show that the instantaneous power passing any position x is given by

P = vT [f′ (u)] 2 .

Solution

At any position and time, the kinetic energy density (K.E. per unit length of the string)

I = µ vfu 2 ′

bg

2 HG ∂ t KJ 2 HG ∂ u ∂ t KJ 2

Tfu ′ bg .

= Tfu ' bg

The potential energy density = T

2 HG ∂ x KJ 2 Total energy density = E 1 = T[f ′(u)] 2 .

Since the wave moves with velocity v, the instantaneous power passing any position x is

P = vE 1 = vT [f ′(u)] 2 .

10. Using the following general expression for the power that passes any position x along the string

P = FV, ∂ y

where F = y – component of the force = –T

V = the transverse velocity =

and the general form of the travelling wave y = f(x – vt) = f(u) with u = x – vt,

show that (i)

F =µ T

(ii) P = vT [f ′(u)] 2 .

Solution

We have,

= f ′(u) and

=− vf u ′ ( ).

Thus

= µT

210 WAVES AND OSCILLATIONS

[Note : The ratio F/V = µT is called the wave impedance or, the characteristic impedance (z) of a transverse wave on the string. This expression is analogous with the

electrical impedance = voltage/current, while electric power = voltage × current. Voltage and current are the electrical analogous of the mechanical force and displacement velocity.]

11. Show that the mean power required to maintain a travelling wave of amplitude A and angular frequency ω on a long string is

P=

µ vA 2 ω 2 = zA 2 ω 2 ,

where µ is the mass per unit length of the string, v is the speed of transverse waves on the string and z (= µT ) is the characteristic impedance of the string for transverse waves.

Solution

For the displacement of the string let us take y = A sin (kx – ωt) = A sin ku = f(u).

where u = x – vt. The instantaneous power passing any position is

P = vT[f ′(u) 2 ] = vT A 2 k 2 cos 2 ku.

2 Since the average value of cos 1 ku is , the average power is

vTA k = µ v A 2 2 ω 1 zA 2 ω. P 2 = =

12. A long string of mass per unit length 0.1 kgm –1 is stretched to a tension of 250 N. Find the speed of transverse waves on the string and the mean power required to maintain

a travelling wave of amplitude 5mm and wavelength 0.5 m.

Solution

v= T/µ= 50 ms − 1

µ vA 2 ω 2 = × . 0 1 50 × × × 5 10 32 × F I

Mean power =

HG = 24.67 W.

05 . KJ

13. Consider the motion of the transverse

waves on a long string consisting of two parts.

The left part has a linear mass density µ 1 and

the right part a different linear mass density µ 2 with both parts under the same tension T Fig. 8.4

For convenience, we place the x-origin at the discontinuity. Suppose that a source of

sinusoidal waves on the negative x-axis is sending waves toward the discontinuity and that the waves continue past it are absorbed with no reflection by a distant sink. Find the power

reflection and transmission coefficients at the point of discontinuity.

VIBRATIONS OF STRINGS AND MEMBRANES 211 Solution

There are two independent boundary conditions at the point of discontinuity where the two strings are joined;

(i) Continuity of the displacement of the string: y left =y right at x = 0 for all times.

(otherwise they would not be joined together) (ii) Continuity of the transverse force in the string: If the force is not continuous at the

boundary, an infinitesimal mass therefore would be subject to a finite force, resulting in an infinite acceleration, which is not possible. Thus, we have

at x = 0 for all times,

∂ x left

∂ x right

or,

at x = 0 for all times,

∂ x left

∂ x right

Let us take the incident wave coming from the left to be the real part of y 1 =A 1 exp [i(k 1 x – ωt)], –∝ < x < 0.

which has the amplitude A 1 and the velocity v 1 = ω/k 1 = T/ µ 1 . The wave transmitted past the discontinuity is assumed to be the real part of

y 2 =A 2 exp [i(k 2 x – ωt)], 0 < x < ∝ ,

which has the amplitude A 2 and the velocity v 2 = ω/k 2 = Tµ 2 .

Both the waves must necessarily have the same frequency. There must exist a third wave that is reflected from the boundary. We assume that the

reflected wave travelling to the left is the real part of y′ 1 =B 1 exp [i(–k 1 x – ωt)], –∝ < x < 0, which is moving towards the negative direction. The reflected wave has the amplitude B 1 and

the wave number k 1 which is appropriate to the string on the left side of the boundary. The boundary conditions (i) and (ii) now give

A 1 + B 1 =A 2 –T(ik 1 A 1 ) + T(ik 1 B 1 ) = –T(ik 2 A 2 )

From these two equations, we get

and

Thus, we have

Amplitude reflection coefficient

=R

Amplitude transmission coefficient = T

212 WAVES AND OSCILLATIONS

The characteristic impedances of the two parts of the string are:

z 1 =µ 1 v 1 = Tµ 1 z 2 =µ 2 v 2 = Tµ 2

and the wave numbers k 1 and k 2 are given by

R = 1 − z 2 µ−µ 1 a 2 =

Thus, we get

z 1 + z 2 µ+µ 1 2

z 1 + z 2 µ+µ 1 2

We find that both R a and T a are real. If µ 1 >µ 2 , R a is positive which implies that the reflected wave has the same phase as the incident wave. If µ 1 <µ 2 , R a is negative showing that the reflected and incident waves are 180° out of phase. T a is always positive showing that the transmitted wave has the same phase as the incident wave.

Next, we define the power reflection coefficient R p and power transmission coefficient

2 p . The power carried by a travelling wave is µvω A 2 2 . Thus, we may define

2 1 1 Power reflection coefficient = R 1

1 HG z 1 + z 2 KJ

Power transmission coefficient = T

2 µ vA 2 22 2 4 zz 12

11 vA 1 b z 1 + z 2 g

We see that R p +T p = 1, showing that the incident power equals the reflected power plus the transmitted power. Since R p and T p depend only on the properties of the string and not on the frequency and amplitude of the waves, the expressions (8.3) and (8.4) must hold for waves of arbitrary shape.

VIBRATIONS OF STRINGS AND MEMBRANES 213

14. A perfectly elastic string of length l which is under tension T and is fixed at both ends, has the linear mass density (i.e., mass per unit length) µ. The string is given initial deflection and initial velocity at its various points and is released at time t = 0. The string executes small transverse vibrations. The initial deflection and the initial velocity of the string at any point x are denoted by h (x) and V (x) respectively. Find the different normal modes of vibrations and the deflection of the string at any point x and at any time t > 0.

Solution

We have to solve the wave equation

under the following boundary conditions: (i) y(0, t) = 0

…(8.6) (ii) y(l, t) = 0

…(8.7) (iii) y(x, 0) = h(x)

…(8.8) ∂ y

(iv) ∂ x t=0 = V(x) …(8.9) (v) |y(x, t)| < M

…(8.10) where M is a fixed number i.e. the motion is bounded. Solution of Eqn. (8.5) by the method of separation of variables: We assume that y (x,

t) can be written as

y(x, t) = F(x) G(t)

where F (x) is a function of x only and G (t) is a function of t only. Substituting in Eqn. (8.5), we get

vG 2 dt 2 F dx 2 …(8.11) L.H.S. of Eqn. (8.11) is a function of t only and the R.H.S of Eqn. (8.11) is a function of x only. Since Eqn. (8.11) is true for all values of x and t, the two sides of Eqn. (8.11) must

be equal to a constant, independent of x and t. This constant should also be negative on physical grounds, |y(x, t)|< M (i.e., the displacement is bounded). If it were a positive

constant q, then G (t) would be G (t) ~ exp (± vqt 2 ). The positive sign in the exponential is not allowed since it would mean growing displacement and the negative sign is not

acceptable since there is no damping force in the system. Thus, we have

where –p 2 is the separation constant. The solution for F(x) is

F(x) = A cos px + B sin px.

214 WAVES AND OSCILLATIONS

The boundary conditions of Eqns. (8.6) and (8.7) give

A = 0 and B sin pl = 0,

or

pl = nπ, n = 1, 2, 3,...

or

p = nπ/l

For any particular n, we have

F n (x) = B n sin(nπx/l).

For different values of n we obtain different solutions. In fact there are infinitely many solutions. For a particular n, the differential equation for G (t) is

dG 2 n

dt 2

where ω n = Nπv/l. The general solution of this equation is

G n (t) = D′ n cos ω n t + E′ n sin ω n t.

Thus, the displacement for the nth mode is

nx π

y n (x, t) = F n (x)G n (t) = B n sin

n cos ω n t+E n sin ω n t) …(8.12) where

We have obtained the solutions y n (x, t) of the partial differential Eqn. (8.5) satisfying the boundary conditions (8.6), (8.7) and (8.10). The functions y n (x, t) are called the eigenfunctions or characteristic functions and the values ω n = nπv/l are called the eigen frequencies or characteristic frequencies of the vibrating string. Each y n represents a harmonic motion having the angular frequency ω n = 2πν n, where

nv

The motion is called the nth normal mode of the string. The first normal mode (n = 1) is known as the fundamental mode, and higher modes (n = 2, 3, 4, ...) as overtones. Form Eqn. (8.12), we have y n (x, t) = 0 for all time when

x=k , k = 0, 1, 2,....., n

These are the points of the string which do not move (nodes). For k = 0, n we have x = 0, l which are the two fixed end points of the string.

When n = 1, the nodes are at x = 0, l (Fig. 8.5). The fundamental frequency is

Tµ

2l

Fig. 8.5

VIBRATIONS OF STRINGS AND MEMBRANES 215

and the fundamental wavelength is

=2. l

When n = 2, the nodes are at x = 0, l/2,

l (Fig. 8.6). The corresponding frequency and

wavelength are

Fig. 8.6

ν 2 = 2ν 1 and λ 2 =l=λ 1 /2. l 2 l

When n = 3, the nodes are at x = 0, , ,

l (Fig. 8.7). The corresponding frequency and

wavelength are

1 Fig. 8.7

ν 3 = 3ν 1 and λ 3 = λ 1 .

Equation (8.12) gives the nth normal mode solution of Eqn. (8.5) satisfying the boundary conditions (8.6), (8.7) and (8.10). The sum of infinitely many solutions y n (x, t) is also a solution. Therefore, the general solution is

y(x, t) = ∑ b D n cos ω n t + E n sin ω n t g sin nx π . …(8.13)

n = 1 l The boundary condition (8.8) gives

y(x, 0) = ∑ D n sin = hx bg

nx π

which is the Fourier sine series of h (x). Thus, we have

hx bg sin dx , n = 1, 2, 3, ...

2 l nx π

By applying the boundary condition (8.9), we get

bg

Vx bg sin dx

2 l nx π

or

nv π z 0 l

Vx bg sin dx , n = 1, 2, 3,... …(8.15)

2 l nx π

or

The deflection of the string at any point x and at any time t is given by Eqn. (8.13) where the coefficients D n and E n are obtained from Eqns. (8.14) and (8.15).

15. (a) A string of length l = π which is under tension T and is fixed at both ends has mass per unit length µ. The initial deflection at any point x is given by

h(x) = 0.01 x(π – x).

216 WAVES AND OSCILLATIONS

The initial velocity is zero at any point x. Find the deflection y(x, t) of the string at point x and at any time t > 0.

(b) What is the ratio of the amplitudes of the fundamental mode and the next non-zero overtone (i.e., D 1 /D 3 )?

(c) Find the ratio

1 e D 1 + D 3 + D 5 + .... j .

Solution (a) The deflection y(x, t) of the string is given by

D n cos ω n t ∑ sin

nx π

y(x, t) =

where

hx bg sin nx dx

. 0 01 x b π − x g sin nx dx

. 0 04 =– n

3 [ bg − 1 − 1 π ] n

Thus,

D 2 =D 4 = D 6 = ... = 0.

(b) D 1 /D 3 = 27 (c) The deflection at time t = 0 is

0.01x ( π – x) =

sin

Squaring this expression and integrating from 0 to π , we get

. 0 01 x b π − x g dx = ∑ ∑

2 b . 0 08 g

× sin b 2 k + 1 g x sin b 2 l + 1 g x dx

k = 0 b 2 k + 1 g 960

GG ∑

Now,

k = 0 b 2 k + 1 g H JJ K

D 2 1 2 + D 3 + D 5 2 + ...

VIBRATIONS OF STRINGS AND MEMBRANES 217

16. A perfectly elastic string of length l y which is under tension T and is fixed at both ends, has mass per unit length µ. It is plucked at the point x = a to a height h (Fig. 8.8) and then released from rest. The string executes small transverse vibration. Find the different

normal modes of vibrations and the deflection h of the string at any point x at any later time.

Solution

a Initially we have triangular deflection: x l

hl bg − x

For a ≤ x ≤ l,

, or y =

Thus, at time t = 0, the deflection of the string is given by

R xh for 0≤ x ≤ a || h(x) = a

S hl − x

a ≤ x ≤ T l || b g la −

for

We have also ∂ y

Thus, from Eqn. (8.13), we obtain

y(x, t) = ∑ D n cos sin

n vt π

nx π

hx bg sin dx

2 l nx π

where

l MM z

z a la − l PP

hl bg − x nx π O

b sin

ala 2 2

For nth harmonic we have the vibration mode

2 hl 2 L 1 na π nx π n vt π O

y n (x, t) =

π 2 al b − a g 2 NM n l l

l QP

nx π

At the antinode of this particular mode, sin =1 and the maximum amplitude of

vibration for the nth mode is

na π 2 hl 2 sin

218 WAVES AND OSCILLATIONS

The amplitude of higher harmonics decreases very fast due to appearance of n 2 in the denominator.

l When a = (plucked at the mid-point of the string)

8 h L 1 n π nx π n vt π O

y n (x, t) =

NM cos

sin

sin

l QP

and

8 h L 1 π x π v t 1 3 π x 3 π vt

y(x, t) =

2 2 sin cos

− sin cos

π NM 1 l l 3 2 l l

− ... O

+ 2 sin

cos

QP

We see that the 2nd, 4th, 6th and all the even harmonics are absent. l

If a = ,

l we see that 3rd, 6th, 9th, ..., harmonics will be absent. In general, if a = ,

3 p where p is an integer, pth, (2p)th, (3p)th, ..., harmonics will be absent in the vibrations.

17. A string of length l which is fixed at both ends is under tension T. It is plucked at the point x = a to a height h (Fig. 8.8) and then released from rest. The string executes small transverse vibrations.

µvhl 22 (i) Show that the initial potential energy of the string is

2 al b − a g

(ii) Find the total energy for the nth harmonic of the string. (iii) Show that the total energy is the sum of the energies of the harmonics. (iv) Show that the total energy at any instant is equal to the initial potential energy

of the string.

Solution

(i) At time t = 0, the deflection of the string is given by, (see problem 16),

R xh for 0 ≤ x ≤ a , y(x) = || a S

hl bg − || x for a ≤ x ≤ T l

la −

From problem 8, we have for the total potential energy of the string,

z 0 2 HG ∂ x KJ

Total P.E. =

dx

2 z H a K z H a la − K

dx + F I dx

N MM 0 PP Q

µv

µ vhl 22

2 ala bg −

VIBRATIONS OF STRINGS AND MEMBRANES 219

(ii) For the nth harmonic the total energy is given by

+ v F n I n O dx (see problems 8 and 16)

2 z MM HG ∂ t KJ HG ∂ x KJ

PP Q

π alan b − 2 g l

After performing the integrations, we get

µ vhl 223

1 2 na π

ala bg −

sin

π 22 2 n 2

Thus, the energy of higher harmonics decreases very fast with increase in n. (iii) The total energy of the string is given by

2 z HG ∂ t KJ 2 z HG ∂ x KJ

∂ y I dx µ v 2 F ∂ y

I dx

E=

where y = ∑ y n is the deflection of the string.

∂ y I nv π mv π nx π mx π

∂ t KJ ∑ ∑

Now,

HG =

Since sin

An I 2222 µ

2 z HG 0 ∂ t KJ n 4 l l

2 F ∂ y 2 I µ An 2222 n π v 2 n vt Similarly, π v dx = cos .

2 z HG ∂ x KJ ∑ 4 l l

E= ∑

22 2 2 π sin ala bg − ∑

2 π 2 ala b − g l

220 WAVES AND OSCILLATIONS

[See Supplementary problem 11]

2 ala b − g

18. A perfectly flexible string of length l and linear mass density µ , which is fixed at both ends and is under tension T, is struck by a pointed hammer at the point x = a, the time of

contact between the string and the hammer being very very small. Write the wave equation and the proper boundary conditions of this problem of struck string. Find the deflection of the string at any point x at a later time.

Solution Since the string is perfectly flexible, the point, say x = a, where the hammer strikes is

not at rest, though other points are at rest initially i.e., F ∂ y I = y 0 ≠ 0 at x = a and 0 =0

HG ∂ t KJ b t = 0 g

when x ≠

a. In this case, there is initial motion, but no initial displacement i.e., y (x, 0) = 0. Thus, we have to solve the wave equation.

under the following boundary conditions: (i) y (0, t) = 0 (ii) y (l, t) = 0 (iii) |y (x, t)| < M (iv) y(x, 0) = 0 (v) 0 = 0 (x) δ (x – a)

where M is a fixed number i.e., the motion is bounded, v = T µ is the velocity of propagation of transverse wave on the string and δ (x – a) is the Dirac delta function having the definition

δ (x – a) = 0 when x ≠ a

0 when x = a.

The general equation for deflection of the vibrating string satisfying the boundary conditions (i), (ii) and (iii) is [see problem 14].

A B I cos π sin ∑ sin n + n

∝ F n vt π

n vt π nx

y(x, t) =

n = 1 HG l

l KJ l

From the condition (iv) we have

nx y(x, 0) = ∑ A n sin π= 0

or, A n = 0, since the above equation is true for all values of x. Thus,

y(x, t) = ∑ B n sin sin

n vt π

nx π

F ∝ ∂ y I nv π nx π

and

HG sin ∂

t KJ

VIBRATIONS OF STRINGS AND MEMBRANES 221

which is nothing but Fourier sine series with coefficients

B n = 2 nx yx π 0 bgb δ x − a g sin dx

nv π z 0 l

or

ya 0 bg sin

2 ya ∝ 0 bg 1 na π nx π n vt π

y(x, t) =

l The amplitude of the nth mode of vibration is given by

2 ya 0 bg 1 na π nx π

sin

sin

which decreases as

. When a =

, 2nd, 4th, 6th, . . . harmonics are absent. When

a = , 3rd, 6th, 9th, . . . harmonics are absent.

3 Note: In practice, the results for the vibration of struck string are to be modified because

of finite time of contact between the hammer and the string and also because the area under the hammer is finite.

19. A violin string of length l and linear mass density µ is fixed at both ends and is under tension T. The string is bowed at some point. It is observed that a point x on the string

has a constant forward velocity v 1 from t = 0 to t = T 1 and a constant backward velocity v 2 (in magnitude) from t = T 1 to t = τ where τ is the period of vibration. Show that the deflection of the string is given by

1 + v 2 g 1 n vT π nv π

y(x, t) =

sin

sin

l HG 2 KJ

2l

where v = T µ = velocity of transverse wave along the string. Solution

We have to solve the wave equation ∂ 2 y

under the following boundary conditions:

(i) y (0, t) = 0

(ii) y (l, t) = 0 (iii) |y(x, t)| < M

R v 1 for 0 << t T 1

(iv) ∂ = t

S|

T| − v 2 for T 1 << t τ

222 WAVES AND OSCILLATIONS

We know that the general solution for finite displacement of a string fixed at x = 0 and x = l is

n vt π nx π y(x, t) = ∑ A n cos + B n sin I sin ,

n vt π

n = 1 HG l

l KJ l

where the boundary conditions (i), (ii) and (iii) are satisfied. The above equation can be rewritten as

y(x, t) = ∑ b A n cos nt ω + B n sin nt ω g sin

nx π

…(8.16) n = 1 l

where

ω=π v and τ = = .

nx π Now,

∑ b − nA ω n sin ntnB ω + ω n cos nt ω g sin

n = 1 l According to the Fourier series, we have

τ z 0 HG ∂ t KJ

τ 0 z HG ∂ t KJ

which gives

L T 1 nx τ π

τ MM

v 1 sin n t dt ω + bg − v 2 sin n t dt ω

PP

–nωA n sin

τ MM z

nx π

v 1 cos n t dt ω + bg − v 2 cos n t dt ω

PP

Thus, we get

y(x, t) = ∑ 2 gb cos nT ω 1 − 1 g cos nt ω

n = 1 πω n

+ sin nωT 1 sin nωt]

VIBRATIONS OF STRINGS AND MEMBRANES 223

b v 1 + v 2 g nT ω

∑ 2 2 − 2 sin 2 1 cos nt ω