CBSE Class 11 Physics Mechanical Wave in Strings Theory and Examples

Mechanical waves in strings are an important part of the Waves chapter in CBSE Class 12 Physics. This notes and solved examples page covers the introduction to waves, the classification of waves, the equation of a plane progressive wave, terms related to wave motion, the important forms of the wave equation, interference and superposition, reflection and transmission, standing waves in strings, the laws of vibration of a stretched string, and Melde's experiment, with all the solved examples and answers.

Topics Covered

1. Introduction
2. Classification of waves
3. Equation of a plane progressive wave
4. Various terms related to wave motion
5. Important forms of the equation of wave
6. Interference and principle of superposition of waves
7. Reflection and transmission of waves
8. Standing waves in strings and normal modes of vibration
9. Laws of vibration in a stretched string
10. Experiment of Melde's

1. INTRODUCTION

There are two modes of energy transfer:
(a) Particle motion: When a particle is transferred from one part of space to the other, it carries energy with it.
(b) Wave motion: When no particle is transferred from one part of space to the other, although energy is transported. For example, the vibration of a string, or the surface waves produced on the surface of a solid and a liquid.

2. CLASSIFICATION OF WAVES

Waves may be of the following two types:

2.1 Electromagnetic waves (E.M. waves) (medium is not necessary for propagation)
(I) Periodic changes take place in electric and magnetic fields, hence it is called an electromagnetic wave.
(II) In vacuum, E.M. waves travel with the velocity of light.
(III) E.M. waves can be polarised.
(IV) E.M. waves are transverse in nature.
(V) Medium is not required for the propagation of E.M. waves.
(VI) E.M. waves have momentum.
Examples: radio waves, light waves, thermal radiation.

2.2 Mechanical waves (medium is essential)
(I) Method of energy propagation in which a disturbance propagates with a definite velocity without changing its form is called a mechanical wave.
(II) Energy and momentum propagate by the motion of particles of the medium. But the medium remains at its previous position. Mass transfer does not take place.
(III) Propagation is possible due to the properties of the medium, viz. elasticity and inertia. Mechanical waves may be of two types.

2.2.1 Transverse waves
(a) The particles of the medium vibrate in a direction perpendicular to the direction of propagation of the wave.
(b) Crests and troughs are formed.
Examples: the vibration of a string, and the surface waves produced on the surface of a solid and a liquid.
Note: Polarisation of transverse waves is possible, but it is not possible for longitudinal waves.

2.2.2 Longitudinal waves
(a) The vibration of the particles of the medium is in the direction of wave propagation.
(b) The wave proceeds in the form of compression (C) and rarefaction (R).
(c) At places of compression the pressure and density are maximum, while at places of rarefaction they are minimum.
Examples: sound waves, waves in gases.

Ex.1 The sound wave produced in a gas is always -
(A) longitudinal
(B) Transverse
(C) Stationary
(D) Electro magnetic

Answer: (A) longitudinal

1. In gases, only longitudinal waves can propagate.
2. Hence the correct answer is (A).

Teacher's Note:
a) Transverse waves need a medium that can support shear, and gases cannot do this.
b) Sound in air is a compression wave, so the particles move along the direction of travel.

3. EQUATION OF A PLANE PROGRESSIVE WAVE

(1) When a particle executes S.H.M., the displacement is given by y = A sin(ωt + φ), where φ is the phase difference.

(2) In a progressive wave, each particle performs S.H.M. and transfers its energy and momentum to the nearest particle. Let point P be the reference point, and let it oscillate in simple harmonic manner: y = A sin ωt. This is the oscillation equation of point P.

Let the wave travel in the positive direction with constant velocity v. Point Q is at a distance x from point P. At time t, the displacement of point Q will be the same as the displacement of point P at time (t − x/v). Therefore, the displacement of point Q at time t is:
y = A sin ω(t − x/v) ... (a)

[Figure: Position of different particles at some instant in a string, showing a sine curve with the wavelength 2π/k marked along the x-axis. Points A, B, C, D, E, F, G, H and I are labelled on the curve. Arrows show the direction of propagation of the wave. Along the y-axis, the displacement is shown.]

(i) Equation of progressive wave in the positive x direction:
\( y = A \sin ω\left(t − \frac{x}{v}\right) \) ... (1)
where x is the coordinate of point P with respect to the origin.

(ii) Equation of progressive wave in the negative x direction:
\( y = A \sin ω\left(t + \frac{x}{v}\right) \) ... (2)
where x is the coordinate of point Q with respect to the origin.

Note:
(1) Equations (1) and (2) represent the displacement of a particle from equilibrium at time t. It means that y is the displacement of point Q from the mean position.
(2) The periodic motion of point Q can be expressed by y = A sin ωt, in the same way as that of point P. In equations (1) and (2), it is supposed that point P is the reference point, that is why the sign of x is opposite. Otherwise, equations (1) and (2) are the same.
(3) At time t, the velocity of a particle is \( \frac{dy}{dt} \), and the wave velocity is \( \frac{dx}{dt} \).
(4) Equation (a) represents the displacement of a particle in the direction perpendicular to the wave motion with respect to the mean position. By this equation we can calculate the displacement of a particle in the direction perpendicular to the wave motion. Equation (a) is called the equation of a plane progressive wave.

4. VARIOUS TERMS RELATED TO WAVE MOTION

(1) Amplitude (A): The maximum displacement of a vibrating particle of the medium from the mean position. 'A' is the amplitude in y = A sin ωt.
(2) Periodic time (T): The time taken to complete one oscillation, denoted by T. \( T = \frac{1}{n} \). n is called the frequency.
(3) Wave frequency (n): The number of vibrations made per second by the particles, denoted by n. \( n = \frac{1}{T} \), unit: Hz.
Angular frequency: ω = 2πn, unit: rad/sec.
(4) Wavelength (λ): The distance between two consecutive particles in the same phase, or the distance travelled by the wave in one periodic time, denoted by λ.

[Figure: Displacement against distance graph of a wave. Points A, B, C, D, E, F, G, H and I are marked on the curve, with the wavelength λ marked between two crests B and F. The direction of propagation of the wave is shown along the x-axis, and the displacement is shown on the vertical axis.]

Wave velocity: \( V = nλ = \frac{λ}{T} = \frac{ω}{2π} · λ = \frac{ω}{k} \)
where k = 2π/λ is defined as the wave number.

[Figure: Displacement against distance graph showing one complete wave with a wavelength of 2π/k marked along the x-axis. The displacement is shown as the vertical y-axis and the position of the particles at a constant time t.]

(5) Intensity of wave: In a medium, the propagated energy perpendicular to unit area per second is called the intensity of wave.
I = 2π²n²a²ρv
where ρ = density of the medium, v = wave velocity, a = amplitude, and n = wave frequency. The unit is W/m².
If v and ρ are constant, then I ∝ (frequency)² and I ∝ (amplitude)².

(6) Energy density: Energy per unit volume is called energy density.
\( E = \text{Energy density} = \frac{\text{Energy intensity}}{\text{Energy velocity}} = \frac{2π²n²a²ρv}{v} \), so E = 2π²n²a²ρ

(7) Wave pressure: When a wave is incident on a surface, it produces pressure on it. This is called wave pressure.

5. IMPORTANT FORMS OF EQUATION OF WAVE

(I) y = A sin(ωt − kx)
(II) y = A sin(ωt − (2π/λ)x)
(III) y = A sin(ωt − (ωxT/λ))
(IV) y = A sin 2π(nt − x/λ)
(V) y = A sin (2π/T)(t − xT/λ)
(VI) y = A sin (2π/λ)(vt − x)
(VII) y = A sin 2π(nt − nx/v)
(VIII) y = A sin ω(t − x/v)

[Figure: Two sine graphs. The upper graph shows particle displacement against time for a particle at different times t, with T/4, T/2, 3T/4, T, 3T/2 and 2T marked on the time axis. The lower graph shows particle displacement against distance for various particles at a constant time t, with λ/4, λ/2, 3λ/4, λ, 3λ/2 and 2λ marked on the distance axis.]

[Figure: Table-style diagram. With reference to point O, for motion in the positive x direction: far away from point O, y = A sin ω(t − x/v); towards point O, y = A sin ω(t + x/v). For motion in the negative x direction: far away from point O, y = A sin ω(t + x/v); towards point O, y = A sin ω(t − x/v). Point O is considered as reference point.]

Ex.2 A progressive wave of frequency 500Hz is travelling with a velocity of 360 meter/sec. How far apart are two points 60° out of phase?
(A) 0.12 cm
(B) 1.2 m
(C) 0.06 m
(D) 0.12 m

Answer: (D) 0.12 m

1. v = 360 m/sec and n = 500 Hz
2. Wavelength λ = v/n = 360/500 = 0.72 m
3. If Δx is the minimum distance between two points, then the phase difference is Δφ = (2π/λ) × Δx
4. Here Δφ = 60° = π/3
5. Path difference Δx = (λ/2π) × Δφ = \( \frac{0.72}{2π} × \frac{π}{3} \) = 0.12 m
6. Hence the correct answer is (D).

Teacher's Note:
a) Convert the phase difference to radians before using it.
b) The wavelength is found from v = nλ, and it is the first step.
c) Check: the phase difference of 2π corresponds to one full wavelength, 0.72 m.

Ex.3 The equation of a simple harmonic progressive wave is y = 8 sin(0.628x − 12.56t), where y and x are in cm and t in second. The amplitude, frequency and speed of the wave and the phase difference between two particles at a distance of 2.0 cm apart at any instant will be.
(A) 8cm, 2 per sec, 10 cm/sec, 52°
(B) 4cm, 4 per sec, 20 cm/sec, 52°
(C) 8cm, 2 per sec, 20 cm/sec, 72°
(D) 8cm, 2 per sec, 10 cm/sec, 52°

Answer: (C) 8cm, 2 per sec, 20 cm/sec, 72°

1. y = 8 sin 2π(0.1x − 2t) cm
2. Comparing with the standard equation y = a sin 2π(x/λ − t/T), amplitude a = 8 cm
3. Time period T = 1/2 sec, so frequency n = 1/T = 2 per sec
4. Wavelength λ = 1/0.1 = 10 cm
5. Wave speed v = nλ = 2 × 10 = 20 cm/sec
6. Phase difference Δφ = (2π/λ) × Δx = (2π/10) × 2.0 = 2π/5 radian
7. Δφ = (2/5) × 180° = 72°
8. Hence the correct answer is (C).

Teacher's Note:
a) Compare the given equation with the standard form to read the amplitude, period and wavelength.
b) Speed is the product of frequency and wavelength, not the wavelength alone.
c) Options (A), (B) and (D) all give the wrong phase difference of 52°, and option (D) also gives the wrong speed.

Ex.4 The disturbance of a wave propagating in the positive x-direction at t = 0 is y = \( \frac{1}{(1 + x²)} \), and at t = 2 s it becomes y = \( \frac{1}{1 + (x − 1)²} \). Then the phase velocity of the wave will be -
(A) \( \frac{1}{2} \) m/s
(B) \( \frac{1}{4} \) m/s
(C) \( \frac{1}{6} \) m/s
(D) \( \frac{1}{8} \) m/s

Answer: (A) \( \frac{1}{2} \) m/s

1. Displacement of the wave in the x direction is Δx = x − (x − 1) = 1 m
2. Time interval Δt = 2 − 0 = 2 s
3. Phase velocity = \( \frac{Δx}{Δt} = \frac{1}{2} \) m/s
4. Hence the correct answer is (A).

Teacher's Note:
a) The pulse shape is the same at both times, and it has only moved by 1 m.
b) Phase velocity is the distance moved divided by the time taken.

Ex.5 The distance between two particles on a string is 10 cm. If the frequency of the wave propagating in it is 400 Hz and its speed is 100 m/s, then the phase difference between the particles will be -
(A) 0.8 π Radian
(B) 0.4 π Radian
(C) 0.2 π Radian
(D) π Radian

Answer: (A) 0.8 π Radian

1. Phase difference δ = (2π/λ) × path difference
2. Wavelength λ = v/n = 100/400 = 0.25 m = 25 cm
3. Phase difference = (2π/λ) × Δx = (2π × 400/100) × 0.1 = 0.8 π radian
4. Hence the correct answer is (A).

Teacher's Note:
a) Use the distance in metres, which is 0.1 m, in this formula.
b) Phase difference is 2π times the number of wavelengths between the points.

Ex.6 An observer standing at the sea-coast observes 54 waves reaching the coast per minute. If the wavelength of the wave is 10 m, find the velocity. What type of waves did he observe? Explain.

Answer:
1. Frequency n = 54/60 = 0.9 Hz
2. Velocity v = nλ = (54/60) × 10 m/sec = 9 m/sec
3. The waves observed are transverse, because on the water surface transverse waves are formed.

Teacher's Note:
a) Convert waves per minute to waves per second before multiplying by the wavelength.
b) Surface waves on water are the standard example of transverse waves.

Ex.7 Write the equation for a wave travelling in the negative direction along the x-axis and having an amplitude 0.010 metre, a frequency 550 vib./sec, and a speed 330 metres/sec.

Answer:
1. Amplitude a = 0.01 m
2. Angular frequency ω = 2πn = 2π × 550 = 1100π
3. Wavelength λ = v/n = 330/550 = 0.6 m
4. Wave number k = 2π/λ = 2π/0.6 = \( \frac{10π}{3} \)
5. For a wave travelling in the negative x direction, y = a sin(ωt + kx)
6. Hence y = 0.01 sin(1100πt + \( \frac{10π}{3} \)x)

Teacher's Note:
a) A wave moving in the negative direction has a plus sign between ωt and kx.
b) The printed solution carries an additional initial phase constant φ, which is zero here.

6. INTERFERENCE AND PRINCIPLE OF SUPERPOSITION OF WAVES

6.1 Principle of superposition of waves
(1) The superposition principle states that the net disturbance at a given place and time caused by a number of waves in the same space is the vector sum of the disturbances which would have been produced by each wave independently of the others.
(2) Resultant displacement: y = y₁ + y₂ + y₃ + ...
(3) The value of the resultant displacement depends on: (a) amplitude of waves, (b) phase difference and path difference of waves, (c) frequency of waves, and (d) direction of wave motion.
The phenomena produced due to superposition are interference and the related effects described below.

6.2 Interference
When waves of equal frequency and nearly equal amplitude are superimposed, interference occurs.

6.2.1 Mathematical interference
At time t, at point x, two waves of equal frequency y₁ = a₁ sin(ω₁t − k₁x + φ₁) and y₂ = a₂ sin(ω₂t − k₂x + φ₂) are superimposed. Then the amplitude A and the intensity I of the resultant wave are:
A² = a₁² + a₂² + 2a₁a₂ cos Δφ
I = I₁ + I₂ + 2√(I₁I₂) cos Δφ
Δφ = (ω₂ − ω₁)t − (k₂ − k₁)x + φ₂ − φ₁
The intensity of the resultant wave changes periodically from minimum to maximum, and from maximum to minimum, from one point to another point.

6.2.2 Constructive interference
Phase difference Δφ = 2πn
Path difference Δx = nλ
I_max ∝ (a₁ + a₂)²

6.2.3 Destructive interference
Phase difference Δφ = (2n − 1)π
Path difference Δx = (n − 1/2)λ = \( \left(n − \frac{1}{2}\right)λ \)
I_min ∝ (a₁ − a₂)²

Ex.8 Two waves of the same frequency but of amplitude in the ratio 1 : 3 are superposed. What is the ratio of maximum to minimum intensity?

Answer: 4 : 1

1. Let the two amplitudes be a and 3a.
2. Maximum amplitude = a + 3a = 4a, so I_max = (4a)² = 16a²
3. Minimum amplitude = 3a − a = 2a, so I_min = (2a)² = 4a²
4. I_max / I_min = \( \frac{16a²}{4a²} \) = \( \frac{4}{1} \)

Teacher's Note:
a) Intensity is proportional to the square of the amplitude, so amplitudes 4a and 2a give intensities 16a² and 4a².
b) The ratio simplifies to 4 : 1, which the marking usually expects in this form.

Ex.9 Two waves of the same frequency and amplitudes 2a and 3a are super-imposed on each other. (i) For what values of the phase difference φ the amplitude A of the resultant wave will be maximum and for what values minimum? (ii) Calculate the maximum and minimum intensities of the resultant wave.

Answer:
1. (i) The intensity is maximum for φ = 0, 2π, 4π, ...
2. The intensity is minimum for φ = π, 3π, 5π, ...
3. (ii) I_max = (a₁ + a₂)² = (2a + 3a)² = 25a²
4. I_min = (a₁ − a₂)² = (3a − 2a)² = a²

Teacher's Note:
a) Maximum intensity occurs when the waves are in phase, and minimum when they are out of phase.
b) Subtract the amplitudes, not the intensities, when finding I_min.
c) The resultant amplitude is 5a at the maximum and a at the minimum.

Ex.10 Consider interference between waves from two sources of intensities I and 4I. Find the intensities at points where the phase difference is (i) π/2 (ii) π.

Answer:
1. Resultant intensity I_R = I₁ + I₂ + 2√(I₁I₂) cos δ = I + 4I + 4I cos δ
2. (i) For δ = π/2: I_R = 5I + 4I cos(π/2) = 5I
3. (ii) For δ = π: I_R = 5I + 4I cos π = I
4. So the intensities are 5I for (i) and I for (ii).

Teacher's Note:
a) At φ = π/2 the cosine term is zero, so the result is simply the sum of intensities.
b) At φ = π, the cosine is −1, so the intensity falls to I.

Ex.11 Two sources of intensities I and 4I are used in an interference experiment. Find the intensity at points where the waves from the sources superimpose with a phase difference of (a) zero (b) π/2 (c) π.

Answer:
1. (a) I_R = I + 4I + 2√(4I²) cos 0 = I + 4I + 4I = 9I
2. (b) I_R = I + 4I + 4I cos(π/2) = 5I
3. (c) I_R = I + 4I + 4I cos π = I

Teacher's Note:
a) The maximum intensity for these two sources is 9I, when the waves are in phase.
b) The minimum intensity is I, which occurs when the phase difference is π.

7. REFLECTION AND TRANSMISSION OF WAVES

[Figure: Two diagrams. The left diagram is labelled Reflection with inversion, and the right diagram is labelled Reflection without inversion, each showing an incident pulse, a reflected pulse and a transmitted pulse.]

Reflection with inversion and transmission, and transmission and reflection without inversion:
(a) If a wave enters a region where the wave velocity is smaller, the reflected wave is inverted.
(b) If a wave enters a region where the wave velocity is larger, the reflected wave is not inverted.
(c) The transmitted wave is never inverted.

8. STANDING WAVES IN STRINGS AND NORMAL MODES OF VIBRATION

Let a harmonic wave be set up on a string of length L, fixed at the two ends x = 0 and x = L. This wave gets reflected from the two fixed ends of the string continuously, and as a result of superposition of these waves, standing waves are formed on the string.

Let the wave pulse moving on the string from left to right (along the positive direction of the x axis) be represented by:
\( y_1 = r \sin \frac{2π}{λ}(vt − x) \) ... (1)

As there is a phase change of π radian on reflection at the fixed end of the string, the reflected wave pulse travelling from right to left on the string is represented by:
\( y_2 = r \sin\left[\frac{2π}{λ}(vt − x) + π\right] = −r \sin \frac{2π}{λ}(vt − x) \) ... (2)

Note that here x is the distance from the origin in the direction of the wave (from left to right). It is often convenient to take the origin (x = 0) at the interface (the site of reflection), on the right fixed end of the string. In that case, the sign of x is reversed, because it is measured from the interface in a direction opposite to the incident wave. The equation of the incident wave may, therefore, be written as:
\( y_1 = r \sin \frac{2π}{λ}(vt + x) \)

According to the superposition principle, the resultant displacement y at time t and position x is:
y = y₁ + y₂ = r sin \( \frac{2π}{λ}(vt + x) − r \sin \frac{2π}{λ}(vt − x) \) ... (3)
\( y = r\left[\sin\frac{2π}{λ}(vt + x) − \sin\frac{2π}{λ}(vt − x)\right] \)

Using the relation sin C − sin D = 2 cos \( \frac{C + D}{2} \) sin \( \frac{C − D}{2} \), we get:
\( y = 2r \cos \frac{2π}{λ} vt \sin \frac{2π}{λ} x \) ... (4)

As the arguments of the trigonometric functions involved in (4) do not have the form (vt ± x), therefore, it does not represent a moving harmonic wave. Rather, it represents a new kind of wave called a standing or stationary wave.

At one end of the string, where x = 0, from (4):
y = 2r cos \( \frac{2π}{λ} vt \) sin 0° = 0 (since sin 0° = 0)

At the other end of the string, where x = L, from (4):
\( y = 2r \cos \frac{2π}{λ} vt \sin \frac{2π}{λ} L \) ... (5)

As the other end of the string is fixed, y = 0 at this end. For this, from (5):
\( \sin \frac{2π}{λ} L = 0 = \sin nπ \), where n = 1, 2, 3, ...
\( \frac{2πL}{λ} = nπ \) or \( λ = \frac{2L}{n} \)

Note: n = 1, 2, 3, ... correspond to the 1st, 2nd, 3rd, ... normal modes of vibration of the string.

9. LAWS OF VIBRATION IN A STRETCHED STRING

(1) Law of length: n ∝ 1/L, so n₁/n₂ = L₂/L₁, where T and m are constant.
(2) Law of tension: n ∝ √T, so n₁/n₂ = √(T₁/T₂), where L and m are constant.
(3) Law of mass: n ∝ 1/√m, so n₁/n₂ = √(m₂/m₁), where L and T are constant.
(4) Law of radius: n ∝ 1/r, so n₁/n₂ = r₂/r₁, where d, L and T are constant.
(5) Law of density: n ∝ 1/√d, so n₁/n₂ = √(d₂/d₁), where r, T and L are constant.

General formula:
n = \( \frac{P}{2L} \sqrt{\frac{T}{m}} \)
Here T is the tension in the string and m is the mass per unit length of the string. Also m = πr²d, so:
n = \( \frac{P}{2Lr} \sqrt{\frac{T}{πd}} \) = \( \frac{P}{2Lr} \sqrt{\frac{Mg}{πd}} \)

The speed of a transverse wave in a string is given by v = √(T/m).

Change in frequency with tension: From n = k√T, log n = log k + ½ log T. Differentiating, \( \frac{dn}{n} = \frac{1}{2} \frac{dT}{T} \), so the percentage change in frequency is \( \frac{dn}{n} × 100 = \frac{1}{2} \left( \frac{dT}{T} × 100 \right) \).

Ex.12 The mass of a 4.0 meter wire is 0.01 kg, and it is stretched by a force of 400 N. What is the speed of the transverse wave in the wire?
(A) 400 m/sec
(B) 200 m/sec
(C) 800 m/sec
(D) None

Answer: (A) 400 m/sec

1. Speed of the transverse wave is v = √(T/m)
2. Here T = 400 N and m = \( \frac{0.01 \text{ kg}}{4.0 \text{ m}} \) = 0.0025 kg/m
3. v = √(400 × 4.0 / 0.01) = √160000 = 400 m/sec
4. Hence the correct answer is (A).

Teacher's Note:
a) Mass per unit length is the total mass divided by the length, not the mass alone.
b) Multiply the tension by the length and divide by the total mass, which is the quick route here.

Ex.13 A string of linear density 0.03 g/cm vibrates in three loops to emit a frequency of 600 Hz while it is under a tension of 480 N. Find the length of the string.

Answer:
1. Linear density m = 0.03 g/cm = 0.003 kg/m
2. For the third harmonic (three loops), n₃ = \( \frac{3}{2l} \sqrt{\frac{T}{m}} \)
3. 600 = \( \frac{3}{2l} \sqrt{\frac{480}{0.003}} \)
4. \( \sqrt{\frac{480}{0.003}} \) = √160000 = 400
5. 600 = \( \frac{3}{2l} × 400 \), so l = 1 m

Teacher's Note:
a) Convert the linear density from g/cm to kg/m before using it.
b) Three loops means the number of loops, p = 3, appears in the formula.

Ex.14 A stretched wire 75 cm long emits a fundamental frequency of 300 Hz. By how much the length is changed so that it emits a fundamental note of 400 Hz?

Answer:
1. Using the formula n₁l₁ = n₂l₂
2. l₂ = \( \frac{n_1 l_1}{n_2} \) = \( \frac{300 × 75}{400} \) = \( \frac{225}{4} \) = 56.25 cm
3. Change in length Δl = l₁ − l₂ = 75 − 56.25 = 18.75 cm

Teacher's Note:
a) The string must be shortened, since the frequency is increased.
b) For a fixed tension and mass, the frequency is inversely proportional to the length.

Ex.15 If the tension of a string be increased by 2%, what would be the % change in its frequency?

Answer:
1. n = k√T, so log n = log k + ½ log T
2. Differentiating, \( \frac{dn}{n} = \frac{1}{2} \frac{dT}{T} \)
3. % change in frequency = \( \frac{dn}{n} × 100 = \frac{1}{2} × 2\% = 1\% \)

Teacher's Note:
a) Frequency changes by half the percentage change in tension.
b) A 2% rise in tension gives only a 1% rise in frequency.

Ex.16 Two wires of the same material and radii r and 2r respectively are welded together end to end. The combination is used as a sonometer wire and kept under tension T. The welded point is midway between the two bridges. When the stationary waves are set up in the composite wire, the joint is a node. What is the ratio of the number of loops formed in the thinner to thicker wire?

Answer: 1 : 2

1. Let p₁ and p₂ be the numbers of loops in the two wires. They are connected in series, so n₁ = n₂
2. \( \frac{p_1}{2l_1} \sqrt{\frac{T}{m_1}} = \frac{p_2}{2l_2} \sqrt{\frac{T}{m_2}} \) ... (1)
3. The welded point is midway, so l₁ = l₂
4. m₁ = πr²d and m₂ = π(2r)²d = 4πr²d, so \( \frac{m_1}{m_2} = \frac{1}{4} \)
5. From equation (1), \( \frac{p_1}{p_2} = \frac{l_1}{l_2} \sqrt{\frac{m_1}{m_2}} = 1 × \sqrt{\frac{1}{4}} = \frac{1}{2} \)
6. So p₁ : p₂ = 1 : 2

Teacher's Note:
a) The thicker wire has the larger mass per unit length, and it carries twice as many loops.
b) Area goes as the square of the radius, so the mass ratio is 1 : 4 here.

10. EXPERIMENT OF MELDE'S

(1) This is an experimental demonstration of a transverse stationary wave.
(2) The unknown frequency of a tuning fork can be determined by this experiment.
(3) The laws of transverse vibration in a stretched string can be proved by this experiment.
(4) One stretched string is tied to the arm of a tuning fork. Now it is vibrated.

There are two arrangements:

[Figure: Two diagrams of the Melde's apparatus. The left diagram shows a tuning fork with the string attached to one prong, with the prongs vibrating up and down at right angles to the string, and a pulley at the right end. The right diagram shows the tuning fork with the string attached so that the prongs vibrate along the length of the string, and a pulley at the right end.]

Ex.17 The mathematical form of Melde's law is -
(A) p√T = constant
(B) pT = constant
(C) P/√T = constant
(D) P/T = constant

Answer: (A) p√T = constant

1. From Melde's law, if L and n are constant, then P√T = constant.
2. Hence the correct answer is (A).

Teacher's Note:
a) Melde's law links the number of loops with the square root of the tension.
b) The number of loops is directly proportional to the square root of the tension, which makes option (A) correct.