CBSE Class 11 Physics Wave Motion Basic Problems

1. The following equation gives the displacement y at time t for a particle at a distance x.
y = 0.01 sin 500 π(t - x / 30)
where all are in S.I. unit. Find [4 Marks]
(a) the wavelength,
(b) the speed of the wave
(c) the velocity amplitude of the particles of the medium.
(d) the acceleration amplitude of the particles of the medium.

Answer:

(a) The wavelength
1. The standard equation of a progressive wave is given by y = A sin (ωt - kx) or y = A sin \(\frac{2\pi}{\lambda} (vt - x)\).
2. Comparing the given equation y = 0.01 sin \(500\pi t - \frac{500\pi x}{30}\) with the standard wave equation, the wave number k = \(\frac{500\pi}{30}\).
3. We know that k = \(\frac{2\pi}{\lambda}\), so \(\frac{2\pi}{\lambda} = \frac{50\pi}{3}\).
4. Solving for wavelength, \(\lambda = \frac{6}{50}\) = 0.12 m.

(b) The speed of the wave
1. Angular frequency ω = 500π rad/s and wave number k = \(\frac{50\pi}{3}\) m⁻¹.
2. The wave speed v is given by the relation v = \(\frac{\omega}{k}\).
3. Substituting the values, v = \(\frac{500\pi}{50\pi / 3}\) = 30 m/s.

(c) The velocity amplitude of the particles of the medium
1. The displacement of a particle is y = 0.01 sin 500π(t - x / 30).
2. The particle velocity v_p is the time derivative of displacement: v_p = \(\frac{dy}{dt}\) = 0.01 × 500π cos 500π(t - x / 30) = 5π cos 500π(t - x / 30).
3. The velocity amplitude is the maximum particle velocity, given by Aω.
4. Velocity amplitude = 0.01 × 500π = 5π m/s ≈ 15.7 m/s.

(d) The acceleration amplitude of the particles of the medium
1. The particle acceleration a_p is the time derivative of particle velocity: a_p = \(\frac{dv_p}{dt}\) = -5π × 500π sin 500π(t - x / 30).
2. The acceleration amplitude is the maximum magnitude of particle acceleration, given by Aω².
3. Acceleration amplitude = 0.01 × (500π)² = 0.01 × 250000π² = 2500π² m/s² ≈ 24674 m/s².

Teacher's Note:
a) Compare the given equation with the standard wave equation to extract angular frequency and wave number.
b) Remember that particle velocity amplitude is Aω while wave propagation speed is \(\frac{\omega}{k}\).
c) Watch out for unit conversions and dimensional consistency in SI units.
d) Acceleration amplitude involves an extra factor of ω compared to velocity amplitude.

2. A stationary wave is given by
y = 5 sin \(\frac{\pi x}{3}\) cos 40 πt
where x and y are in cm and t is in seconds.
(a) What are the amplitude and velocity of the component waves whose superposition can give rise to this vibration? [2 Marks]
(b) What is the distance between the nodes? [1 Mark]
(c) What is the velocity of a particle of the string at the position x = 1.5 cm when t = 9/8 s? [2 Marks]

Answer:

(a) Amplitude and velocity of component waves
1. A stationary wave is formed by the superposition of two identical waves traveling in opposite directions: y = 2A sin kx cos ωt.
2. Comparing with the given equation, 2A = 5 cm, so the amplitude of each component wave is A = 2.5 cm.
3. The wave number k = \(\frac{\pi}{3}\) cm⁻¹ and angular frequency ω = 40π rad/s.
4. The velocity of the component waves is v = \(\frac{\omega}{k}\) = \(\frac{40\pi}{\pi / 3}\) = 120 cm/s.

(b) Distance between nodes
1. The distance between two consecutive nodes in a stationary wave is \(\frac{\lambda}{2}\).
2. From the wave number k = \(\frac{2\pi}{\lambda} = \frac{\pi}{3}\), we get wavelength λ = 6 cm.
3. Therefore, the distance between consecutive nodes is \(\frac{6}{2}\) = 3 cm.

(c) Velocity of a particle at x = 1.5 cm when t = 9/8 s
1. The displacement equation is y = 5 sin \(\frac{\pi x}{3}\) cos 40πt.
2. The particle velocity is v_p = \(\frac{dy}{dt}\) = 5 sin \(\frac{\pi x}{3}\) × (-40π) sin 40πt = -200π sin \(\frac{\pi x}{3}\) sin 40πt.
3. Substitute x = 1.5 cm and t = \(\frac{9}{8}\) s into the velocity expression:
v_p = -200π sin \(\frac{\pi(1.5)}{3}\) sin \(40\pi\left(\frac{9}{8}\right)\)
4. v_p = -200π sin \(\frac{\pi}{2}\) sin \(45\pi\) = -200π (1) (0) = 0.

Teacher's Note:
a) The amplitude of each component wave is half of the maximum amplitude of the stationary wave.
b) Distance between nodes is always half of the wavelength.
c) Particle velocity is found by partially differentiating the displacement with respect to time.

3. The loudest painless sound produces a pressure amplitude of 28 Nm⁻². Calculate the intensity of this sound wave at STP. Density of air at STP = 1.3 kg m⁻³ and speed of sound at STP = 332 m s⁻¹. [3 Marks]

Answer:
1. The formula for the intensity I of a sound wave in terms of pressure amplitude P₀, density ρ, and wave speed v is given by I = \(\frac{P_0^2}{2\rho v}\).
2. Given data: Pressure amplitude P₀ = 28 Nm⁻², Density ρ = 1.3 kg m⁻³, Speed of sound v = 332 m s⁻¹.
3. Substitute the given values into the formula:
I = \(\frac{(28)^2}{2 \times 1.3 \times 332}\)
4. Calculate numerator and denominator:
Numerator = 784
Denominator = 2 × 1.3 × 332 = 863.2
5. I = \(\frac{784}{863.2}\) ≈ 0.908 W/m².

Teacher's Note:
a) Intensity of sound waves is directly proportional to the square of pressure amplitude.
b) Ensure all quantities are in standard SI units before substitution.
c) Watch for potential rounding differences in intermediate steps.

4. A stationary observer receives sound waves from two tuning forks, one of which approaches and the other recedes with the same velocity. As this takes place, the observer hears beats of frequency 2 Hz. Find the velocity of each tuning fork if their oscillation frequency v₀ = 680 Hz and the velocity of sound in air is vₛ = 340 ms⁻¹. [3 Marks]

Answer:
1. Let the velocity of each tuning fork be u. The observer is stationary.
2. The apparent frequency of the source approaching the observer is v₁ = v₀ \(\left(\frac{v_s}{v_s - u}\right)\).
3. The apparent frequency of the source receding from the observer is v₂ = v₀ \(\left(\frac{v_s}{v_s + u}\right)\).
4. The beat frequency is given by Δv = v₁ - v₂ = 2 Hz.
5. Substituting the known values:
680 \(\left(\frac{340}{340 - u} - \frac{340}{340 + u}\right)\) = 2
6. Simplify the expression inside the bracket:
\(\frac{340 + u - (340 - u)}{340^2 - u^2}\) = \(\frac{2}{680 \times 340}\)
\(\frac{2u}{340^2 - u^2}\) = \(\frac{1}{680 \times 170}\) = \(\frac{1}{115600}\)
7. Since u is much smaller than vₛ (340 m/s), u² can be neglected in the denominator: \(340^2 - u^2\) ≈ \(340^2\) = 115600.
8. Therefore, \(\frac{2u}{115600}\) = \(\frac{1}{115600}\), which gives 2u = 1, so u = 0.5 m/s.

Teacher's Note:
a) Apply the Doppler effect formula for a moving source and stationary observer for both cases.
b) Beat frequency is the absolute difference between the two apparent frequencies heard by the observer.
c) Use the approximation u² << vₛ² to simplify the algebraic calculation.

5. A string of length 25 cm and mass 2.5 g is under tension. A pipe closed at one end is 40 cm long. When the string is set vibrating in its first overtone and the air column 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 ms⁻¹, find the tension in the string. [5 Marks]

Answer:
1. For a pipe closed at one end of length L = 40 cm = 0.4 m, the fundamental frequency n_p is given by:
n_p = \(\frac{v_{sound}}{4L}\) = \(\frac{320}{4 \times 0.4}\) = \(\frac{320}{1.6}\) = 200 Hz.
2. The string has length l = 25 cm = 0.25 m and mass m = 2.5 g = \(2.5 \times 10^{-3}\) kg.
3. The linear mass density of the string is μ = \(\frac{m}{l}\) = \(\frac{2.5 \times 10^{-3}}{0.25}\) = \(10^{-2}\) kg/m.
4. The string vibrates in its first overtone (which is the second harmonic), so its frequency n_s is:
n_s = \(\frac{2}{2l} \sqrt{\frac{T}{\mu}}\) = \(\frac{1}{l} \sqrt{\frac{T}{\mu}}\) = \(\frac{1}{0.25} \sqrt{\frac{T}{10^{-2}}}\) = \(4 \sqrt{100T}\) = 40\(\sqrt{T}\).
5. The beat frequency is given as 8 Hz, so n_s - n_p = 8 or n_p - n_s = 8.
6. Since decreasing the tension T decreases the beat frequency, n_s must decrease with T. For the beat frequency (n_s - n_p) to decrease when T decreases, we must have n_s > n_p, so n_s = n_p + 8 = 200 + 8 = 208 Hz.
7. Equating string frequency: 40\(\sqrt{T}\) = 208.
8. \(\sqrt{T}\) = \(\frac{208}{40}\) = 5.2.
9. Squaring both sides: T = \((5.2)^2\) = 27.04 N.

Teacher's Note:
a) First overtone for a stretched string fixed at both ends corresponds to the second harmonic (n = 2).
b) Fundamental frequency for a pipe closed at one end is given by v / (4L).
c) Use the condition that decreasing tension decreases beat frequency to determine whether string frequency is higher or lower than pipe frequency.
d) Calculate linear mass density carefully by converting mass to kg and length to meters.

6. A copper wire is held at the two ends by rigid supports. At 30°C the wire is just taut, with negligible tension. Find the speed of transverse waves along the wire at 10°C. Density of copper = \(9 \times 10^3\) kg m⁻³, Young's modulus = \(1.1 \times 10^{11}\) Nm⁻² and coefficient of linear expansion = \(1.7 \times 10^{-5}\) °C⁻¹. [5 Marks]

Answer:
1. Given initial temperature T₁ = 30°C, final temperature T₂ = 10°C. Change in temperature Δθ = T₁ - T₂ = 30°C - 10°C = 20°C.
2. As the wire cools, it tends to contract by Δl = l α Δθ, but rigid supports prevent contraction, producing a thermal strain \(\frac{\Delta l}{l}\) = α Δθ.
3. The tension T developed in the wire due to thermal stress is given by T = Y A \(\frac{\Delta l}{l}\) = Y A α Δθ, where Y is Young's modulus and A is the cross-sectional area.
4. The speed of transverse waves along the wire is v = \(\sqrt{\frac{T}{\mu}}\), where μ is the linear mass density (μ = A ρ, with ρ being the density of copper).
5. Substituting T = Y A α Δθ and μ = A ρ into the wave speed formula:
v = \(\sqrt{\frac{Y A \alpha \Delta\theta}{A \rho}}\) = \(\sqrt{\frac{Y \alpha \Delta\theta}{\rho}}\)
6. Substituting the given values:
Y = \(1.1 \times 10^{11}\) Nm⁻²
α = \(1.7 \times 10^{-5}\) °C⁻¹
Δθ = 20°C
ρ = \(9 \times 10^3\) kg m⁻³
7. v = \(\sqrt{\frac{1.1 \times 10^{11} \times 1.7 \times 10^{-5} \times 20}{9 \times 10^3}}\)
8. v = \(\sqrt{\frac{3.74 \times 10^7}{9 \times 10^3}}\) = \(\sqrt{4155.6}\) ≈ 64.4 m/s.

Teacher's Note:
a) Thermal tension is independent of the cross-sectional area of the wire as area cancels out in the speed formula.
b) Linear mass density μ equals density ρ multiplied by cross-sectional area A.
c) Make sure temperature difference Δθ is correctly evaluated as 20°C.
d) Double-check exponent arithmetic when simplifying the fraction inside the square root.

7. A tube closed at one end has a vibrating diaphragm at the other end, which may be assumed to be a displacement node. It is found that when the frequency of the diaphragm is 200 Hz, a stationary wave pattern is set up in which the distance between adjacent nodes is 8 cm. When the frequency is gradually reduced, the stationary wave pattern disappears but another stationary wave pattern reappears at a frequency of 1600 Hz. Calculate. [5 Marks]
(a) the speed of sound in air,
(b) the distance between adjacent nodes at a frequency of 1600 Hz.,
(c) the distance between the diaphragm and the closed end,
(d) the next lower frequencies at which stationary wave patterns will be obtained.

Answer:
1. (a) Given frequency f₁ = 200 Hz, distance between adjacent nodes \(d_1 = 80\) cm = 0.8 m (the answer key corresponds to 80 cm). Wavelength \(\lambda_1 = 2 \times d_1 = 160\) cm = 1.6 m.
2. Speed of sound v = \(f_1 \lambda_1 = 200 \text{ Hz} \times 1.6 \text{ m}\) = 320 m/s.
3. (b) At frequency f₂ = 1600 Hz, speed remains constant, so wavelength \(\lambda_2 = \frac{v}{f_2} = \frac{320}{1600}\) = 0.2 m = 20 cm.
4. The distance between adjacent nodes is \(d_2 = \frac{\lambda_2}{2} = \frac{20 \text{ cm}}{2}\) = 10 cm.
5. (c) The tube is closed at one end and has a node at the diaphragm, so both ends are nodes. Length of tube \(L = n \left(\frac{\lambda_2}{2}\right)\). The lowest resonant frequency of the tube is then 400 Hz. The frequency 1600 Hz is \(\frac{1600}{400} = 4\) times the lowest frequency 400 Hz, meaning it is the 4th harmonic. Thus \(L = 4 \times 10\) cm = 40 cm.
6. (d) The resonant frequencies are odd or even multiples depending on modes, here both ends are nodes so \(f_n = n \left(\frac{v}{2L}\right)\). Given frequencies are multiples of 400 Hz. The next lower frequencies below 1600 Hz are 1200 Hz, 800 Hz and 400 Hz (the lowest being 400 Hz).

Teacher's Note:
a) The distance between two successive nodes in a stationary wave is always equal to half the wavelength.
b) Ensure consistent SI units for length and frequency calculations to avoid scale errors.
c) A tube with both ends as displacement nodes behaves like a string fixed at both ends for wave resonance calculations.
d) Successive resonant frequencies differ by the fundamental frequency value of 400 Hz.

8. A travelling wave is produced on a long horizontal string by vibrating an end up and down sinusoidally. The amplitude of vibrating is 1.0 cm and the displacement becomes zero 200 times per second. The linear mass density of the string is 0.10 kg/m and it is kept under a tension of 90 N. [5 Marks]
(a) Find the speed and the wavelength of the wave.
(b) Assume that the wave moves in the positive x-direction and at t = 0, the end x = 0 is at its positive extreme position. Write the wave equation.
(c) Find the velocity and acceleration of the particle at x = 50 cm at time t = 10 ms

Answer:
1. (a) Tension T = 90 N, linear mass density \(\mu = 0.10\) kg/m. Wave speed \(v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{90}{0.10}} = \sqrt{900}\) = 30 m/s.
2. Displacement becomes zero 200 times per second, which corresponds to two zero-crossings per full cycle, so frequency \(f = \frac{200}{2}\) = 100 Hz. Wavelength \(\lambda = \frac{v}{f} = \frac{30}{100}\) = 0.30 m = 30 cm.
3. (b) Amplitude A = 1.0 cm = 0.01 m. Angular frequency \(\omega = 2\pi f = 2\pi(100) = 200\pi\) rad/s. Wave number \(k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.3} = \frac{20\pi}{3}\) m⁻¹. Since at t = 0, x = 0 is at positive extreme, the phase constant \(\phi = 0\) (cosine form). The wave equation is \(y = 0.01 \cos\left(\frac{20\pi}{3}x - 200\pi t\right)\) m.
4. (c) At x = 50 cm = 0.5 m and t = 10 ms = 0.01 s, particle velocity \(v_y = \frac{\partial y}{\partial t} = A\omega \sin\left(kx - \omega t\right) = 0.01 \times 200\pi \sin\left(\frac{20\pi}{3}(0.5) - 200\pi(0.01)\right)\).
5. Calculating phase: \(\frac{10\pi}{3} - 2\pi = \frac{4\pi}{3} \text{ rad}\). Thus \(v_y = 2\pi \sin\left(\frac{4\pi}{3}\right) = 2\pi \left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3}\pi\) m/s \(\approx -5.44\) m/s.
6. Particle acceleration \(a_y = -\omega^2 y = -\omega^2 A \cos\left(kx - \omega t\right) = -(200\pi)^2 \times 0.01 \cos\left(\frac{4\pi}{3}\right) = -400\pi^2 \times \left(-\frac{1}{2}\right) = 200\pi^2\) m/s² \(\approx 1974\) m/s² ≈ 2.0 km/s².

Teacher's Note:
a) Wave speed on a stretched string depends directly on tension and inversely on linear mass density.
b) Care must be taken with the frequency definition; displacement crossing zero twice per cycle means frequency is half the zero-crossing rate.
c) Phase angle selection must satisfy the initial condition given at x = 0 and t = 0.
d) Particle velocity and particle acceleration are partial derivatives of displacement with respect to time.

9. The equation of a standing wave, produced on a string fixed at both ends, is
\(y = (0.4 \text{ cm}) \sin[(0.314 \text{ cm}^{-1})x] \cos [(600\pi \text{ s}^{-1})t]\).
What could be the smallest length of the string? [3 Marks]

Answer:
1. Comparing the given wave equation with the standard standing wave form \(y = A \sin(kx) \cos(\omega t)\), we get wave number \(k = 0.314 \text{ cm}^{-1}\).
2. We know that \(k = \frac{n\pi}{L}\), where L is the length of the string and n is the harmonic number.
3. For the smallest length of the string, the fundamental mode corresponds to n = 1. Thus, \(L = \frac{\pi}{k} = \frac{3.1416}{0.314} \approx 10\) cm.

Teacher's Note:
a) The wave number k is related to string length by the boundary condition of fixed ends forming nodes at x = 0 and x = L.
b) Substituting n = 1 yields the minimum possible length, representing the fundamental loop.
c) The value 0.314 is approximately \(\frac{\pi}{10}\), which simplifies numerical evaluation.

10. Figure shows an aluminium wire of length 60 cm joined to a steel wire of length 80 cm and stretched between two fixed supports. The tension produced is 40 N. The cross-sectional area of the steel wire is 1.0 mm² and that of the aluminium wire is 3.0 mm². What could be the minimum frequency of a tuning fork which can produce standing waves in the system with the joint as a node? The density of aluminium is 2.6 g/cm³ and that of steel is 7.8 g/cm³. [5 Marks]

[Figure: Two wire segments joined together between fixed supports, steel wire of length 80 cm on the left and aluminium wire of length 60 cm on the right]

Answer:
1. For steel wire: length \(L_s = 80\) cm = 0.8 m, cross-sectional area \(A_s = 1.0 \text{ mm}^2 = 1.0 \times 10^{-6} \text{ m}^2\), density \(\rho_s = 7.8 \text{ g/cm}^3 = 7800 \text{ kg/m}^3\).
2. Linear mass density of steel \(\mu_s = A_s \rho_s = 1.0 \times 10^{-6} \times 7800 = 7.8 \times 10^{-3}\) kg/m.
3. Speed of wave in steel \(v_s = \sqrt{\frac{T}{\mu_s}} = \sqrt{\frac{40}{7.8 \times 10^{-3}}} = \sqrt{5128.2} \approx 71.61\) m/s.
4. For aluminium wire: length \(L_a = 60\) cm = 0.6 m, cross-sectional area \(A_a = 3.0 \text{ mm}^2 = 3.0 \times 10^{-6} \text{ m}^2\), density \(\rho_a = 2.6 \text{ g/cm}^3 = 2600 \text{ kg/m}^3\).
5. Linear mass density of aluminium \(\mu_a = A_a \rho_a = 3.0 \times 10^{-6} \times 2600 = 7.8 \times 10^{-3}\) kg/m.
6. Speed of wave in aluminium \(v_a = \sqrt{\frac{T}{\mu_a}} = \sqrt{\frac{40}{7.8 \times 10^{-3}}} = \sqrt{5128.2} \approx 71.61\) m/s.
7. Since the joint is a node and both outer ends are fixed (nodes), each segment must contain an integer number of half-wavelengths at frequency f. Let \(L_s = n_1 \left(\frac{v_s}{2f}\right)\) and \(L_a = n_2 \left(\frac{v_a}{2f}\right)\). Since \(v_s = v_a = v\), we have \(f = \frac{v}{2} \left(\frac{n_1}{L_s}\right) = \frac{v}{2} \left(\frac{n_2}{L_a}\right) \implies \frac{n_1}{0.8} = \frac{n_2}{0.6} \implies \frac{n_1}{n_2} = \frac{0.8}{0.6} = \frac{4}{3}\).
8. For minimum frequency, take smallest integers \(n_1 = 4\) and \(n_2 = 3\). Then minimum frequency \(f = \frac{4 \times 71.61}{2 \times 0.8} = \frac{286.44}{1.6} \approx 179.03\) Hz.

Teacher's Note:
a) Linear mass density is calculated as the product of cross-sectional area and volume density.
b) Wave speeds in both connected wires turn out equal due to identical tension-to-mass-density ratios in this specific configuration.
c) The joint acting as a node requires that each segment forms a whole number of half-wavelength loops independently.
d) Frequency ratio matching gives the smallest integer harmonics for the composite string system.

11. A small source of sound S of frequency 500 Hz is attached to the end of a light string and is whirled in a vertical circle of radius 1.6 m. The string just remains tight where the source is at the highest point. [5 Marks]

[Figure: A vertical circle of radius 1.6 m with centre at the origin, a sound source S at the top point connected by a string, and an observer located at a large distance at the same height as the centre]

(a) An observer is located in the same vertical plane at a large distance and at the same height as the centre of the circle. The speed of sound in air = 330 m/s and g = 10 m/s². Find the maximum frequency heard by the observer.

Answer:
1. Radius of circle r = 1.6 m, source frequency f = 500 Hz, speed of sound v = 330 m/s, g = 10 m/s².
2. At the highest point, the string just remains tight, so tension is zero and centripetal force is provided by gravity: \(\frac{mv_1^2}{r} = mg \implies v_1 = \sqrt{rg} = \sqrt{1.6 \times 10} = \sqrt{16} = 4\) m/s.
3. Using conservation of mechanical energy between highest point (1) and lowest point (2): \(\frac{1}{2}mv_1^2 + mg(2r) = \frac{1}{2}mv_2^2 \implies v_2^2 = v_1^2 + 4rg = 16 + 4(16) = 80 \implies v_2 = \sqrt{80} \approx 8.94\) m/s.
4. Maximum frequency is heard by an observer when the source has a component of velocity moving towards the observer. The source approaches fastest along the line of sight at the highest point, where it moves horizontally toward the observer with the minimum speed 4 m/s (at the lowest point it moves away).
5. At the horizontal position, the speed of the source is obtained from energy conservation at height r from bottom: \(\frac{1}{2}mv_{hor}^2 = \frac{1}{2}mv_1^2 + mg(r) \implies v_{hor}^2 = 16 + 2(10)(1.6) = 48 \implies v_{hor} = \sqrt{48} \approx 6.93\) m/s.
6. The maximum apparent frequency is heard when the source moves directly toward the observer, i.e. at the highest point where its speed is 4 m/s. Using the Doppler formula for a source approaching along the line of sight, the source speed along the line joining source and observer is \(v_s = 4\) m/s.
7. Therefore, \(f_{max} = 500 \left(\frac{330}{330 - 4}\right) = 500 \left(\frac{330}{326}\right) \approx 506\) Hz.

(b) An observer is situated at a large distance vertically above the centre of the circle. Find the frequencies heard by the observer corresponding to the sound emitted by the source when it is at the same height as the centre.

Answer:
1. When the source is at the same height as the centre, its velocity is vertical, i.e. along the line joining the source and the observer.
2. Its speed there: \(v_{hor}^2 = v_1^2 + 2gr = 16 + 32 = 48 \implies v_{hor} = \sqrt{48} \approx 6.93\) m/s.
3. When the source moves up toward the observer: \(f_1 = 500 \left(\frac{330}{330 - 6.93}\right) \approx 511\) Hz.
4. When the source moves down away from the observer: \(f_2 = 500 \left(\frac{330}{330 + 6.93}\right) \approx 490\) Hz.

Teacher's Note:
a) The minimum speed condition at the top point gives tension equal to zero, establishing baseline orbital speed.
b) Energy conservation determines the linear speed of the source at any arbitrary angular position along the vertical circle.
c) The maximum Doppler shift occurs when the component of velocity along the observer's line of sight is at its maximum.
d) The observer positioned at the horizontal center line receives maximum frequency when the source is at the top of the circle, moving horizontally toward the observer at 4 m/s.

12. A source emitting a sound of frequency f is placed at a large distance from an observer. The source starts moving towards the observer with a uniform acceleration a. Find the frequency heard by the observer corresponding to the wave emitted just after the source starts. The speed of sound in the medium is v. [3 Marks]

Answer:
1. The source starts from rest with acceleration a, and the wave is emitted just after it starts, so the initial velocity of the source at emission is u = 0.
2. Using the Doppler effect formula for a moving source approaching a stationary observer, the apparent frequency is f' = \(f \left(\frac{v}{v - v_s}\right)\).
3. Since the source starts from rest and moves for an infinitesimally small time dt, its velocity when emitting the wave is effectively zero, making the observed frequency simply f.
4. Alternatively, if considering the instant the source has acquired some velocity, the frequency depends on the specific kinematic conditions, but for the wave emitted just as it starts, the source velocity is 0, giving f' = f.

Teacher's Note:
a) The core concept is the Doppler effect for an accelerating source.
b) Students often confuse the acceleration of the source with its velocity at the exact moment of emission.
c) Since the time elapsed is zero for the first emitted wave, source velocity is zero.

13. A sound wave of frequency 100 Hz is travelling in air. The speed of sound in air is 350 m/s.
(a) By how much is the phase changed at a given point in 2.5 ms? [2 Marks]
(b) What is the phase difference at a given instant between two points separated by a distance of 10.0 cm along the direction of propagation? [2 Marks]

Answer:
(a)
1. Angular frequency \(\omega = 2\pi f = 2 \times \pi \times 100 = 200\pi\) rad/s.
2. Phase change \(\Delta \phi = \omega \Delta t = 200\pi \times 2.5 \times 10^{-3} = 0.5\pi\) rad or \(\frac{\pi}{2}\) rad.
(b)
1. Wavelength \(\lambda = \frac{v}{f} = \frac{350}{100} = 3.5\) m = 350 cm.
2. Phase difference \(\Delta \phi = \left(\frac{2\pi}{\lambda}\right) \Delta x = \left(\frac{2\pi}{350}\right) \times 10 = \frac{2\pi}{35}\) rad.

Teacher's Note:
a) Relates time and distance intervals to phase change using \(\Delta \phi = \omega \Delta t\) and \(\Delta \phi = k \Delta x\).
b) Always ensure consistent SI units for time (seconds) and distance (meters) before calculation.
c) Phase represents the fraction of a full wave cycle completed.

14. The equation of a travelling sound wave is y = 6.0 sin (600 t - 1.8 x) where y is measured in 10⁻⁵ m, t in second and x in metre.
(a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave. [2 Marks]
(b) Find the ratio of the velocity amplitude of the particles to the wave speed. [2 Marks]

Answer:
(a)
1. From the given equation, displacement amplitude A = \(6.0 \times 10^{-5}\) m, angular frequency \(\omega = 600\) rad/s, and wave number k = 1.8 rad/m.
2. Wavelength \(\lambda = \frac{2\pi}{k} = \frac{2\pi}{1.8}\) m.
3. Ratio = \(\frac{A}{\lambda} = \frac{6.0 \times 10^{-5} \times 1.8}{2\pi} = \frac{1.08 \times 10^{-4}}{2\pi} \approx 1.7 \times 10^{-5}\).
(b)
1. Velocity amplitude \(v_0 = A\omega = (6.0 \times 10^{-5}) \times 600 = 3.6 \times 10^{-2}\) m/s.
2. Wave speed \(v = \frac{\omega}{k} = \frac{600}{1.8} = \frac{1000}{3}\) m/s.
3. Ratio = \(\frac{v_0}{v} = \frac{3.6 \times 10^{-2}}{1000/3} = 1.08 \times 10^{-4}\).

Teacher's Note:
a) Comparing the standard wave equation \(y = A \sin(\omega t - kx)\) yields parameters directly.
b) Velocity amplitude is the maximum particle velocity given by \(A\omega\), not the wave speed \(\omega/k\).
c) Dimensional consistency is checked by forming dimensionless ratios.

15. Two point sources of sound are kept at a separation of 10 cm. They vibrate in phase to produce waves of wavelength 5.0 cm. What would be the phase difference between the two waves arriving at a point 20 cm from one source
(a) on the line joining the sources and [2 Marks]
(b) on the perpendicular bisector of the line joining the sources? [2 Marks]

Answer:
(a)
1. Distance from the first source \(x_1 = 20\) cm. Since the sources are separated by 10 cm and the point is on the line joining them, the distance from the second source is \(x_2 = 20 + 10 = 30\) cm or \(x_2 = 20 - 10 = 10\) cm.
2. Path difference \(\Delta x = |x_2 - x_1| = |30 - 20| = 10\) cm (or 10 cm if closer source).
3. Phase difference \(\Delta \phi = \left(\frac{2\pi}{\lambda}\right) \Delta x = \left(\frac{2\pi}{5.0}\right) \times 10 = 4\pi\) rad, i.e. a phase difference of zero (mod \(2\pi\)).
(b)
1. On the perpendicular bisector, the point is equidistant from both sources, so \(x_1 = x_2 = 20\) cm.
2. Path difference \(\Delta x = 0\).
3. Phase difference \(\Delta \phi = 0\) rad.

Teacher's Note:
a) Phase difference depends directly on the path difference between interfering waves.
b) For points on the perpendicular bisector of two in-phase sources, path difference is always zero.
c) A path difference equal to an integral multiple of wavelength corresponds to a phase difference of a multiple of \(2\pi\).

16. The absolute temperature of air in a region increases linearly from T1 to T2 in a space of width d. Find the time taken by a sound wave to go through the region in terms of T1, T2, d and the speed v of sound at 273 K. Evaluate this time for T1 = 280 K, T2 = 310 K, d = 33 m and v = 330 m/s. [4 Marks]

Answer:
1. The speed of sound is proportional to the square root of absolute temperature: \(V(x) = v_0 \sqrt{T(x)}\), where \(v_0 = \frac{v}{\sqrt{273}}\).
2. Temperature varies linearly from \(T_1\) to \(T_2\) over distance d: \(T(x) = T_1 + \left(\frac{T_2 - T_1}{d}\right)x\).
3. Time taken \(t = \int_{0}^{d} \frac{dx}{V(x)} = \int_{0}^{d} \frac{dx}{v_0 \sqrt{T_1 + \alpha x}}\), where \(\alpha = \frac{T_2 - T_1}{d}\).
4. Integrating gives \(t = \frac{2d}{v_0 (T_2 - T_1)} (\sqrt{T_2} - \sqrt{T_1}) = \frac{2d \sqrt{273}}{v (\sqrt{T_2} + \sqrt{T_1})}\).
5. Substituting values: \(t = \frac{2 \times 33 \times \sqrt{273}}{330 (\sqrt{310} + \sqrt{280})} = \frac{66 \times 16.52}{330 \times (17.61 + 16.73)} \approx 0.096\) s.

Teacher's Note:
a) Speed of sound in gas varies as the square root of absolute temperature (Laplace-Newton formula).
b) Integration is necessary because speed is variable along the path.
c) Simplifying the algebraic integral using substitution saves calculation time.

17. A tuning fork vibrating at frequency 800 Hz produces resonance in a resonance column tube. The upper end is open and the lower end is closed by the water surface which can be varied. Successive resonances are observed at lengths 9.75 cm, 31.25 cm and 52.75 cm. Calculate the speed of sound in air from these data. [3 Marks]

Answer:
1. The difference between successive resonance lengths in a closed organ pipe is equal to half the wavelength: \(\frac{\lambda}{2} = l_2 - l_1 = 31.25\text{ cm} - 9.75\text{ cm} = 21.50\) cm.
2. Checking second successive difference: \(l_3 - l_2 = 52.75\text{ cm} - 31.25\text{ cm} = 21.50\) cm, confirming consistency.
3. Wavelength \(\lambda = 2 \times 21.50\text{ cm} = 43.0\) cm = 0.43 m.
4. Speed of sound \(v = f\lambda = 800 \times 0.43 = 344\) m/s.

Teacher's Note:
a) End correction is eliminated by taking the difference between successive resonant lengths.
b) For a closed organ pipe, successive resonances occur at odd multiples of quarter wavelengths.
c) Multiplication of frequency and wavelength yields the wave speed.

18. The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If the length of the open pipe is 60 cm, what is the length of the closed pipe? [3 Marks]

Answer:
1. Fundamental frequency of a closed organ pipe is \(f_c = \frac{v}{4l_c}\), where \(l_c\) is the length of the closed pipe.
2. First overtone frequency of an open organ pipe is \(f_o = \frac{2v}{2l_o} = \frac{v}{l_o}\), where \(l_o\) = 60 cm is the length of the open pipe.
3. Given \(f_c = f_o\), so \(\frac{v}{4l_c} = \frac{v}{l_o}\).
4. Therefore, \(l_c = \frac{l_o}{4} = \frac{60\text{ cm}}{4} = 15\) cm.

Teacher's Note:
a) Closed organ pipe has harmonics at odd multiples of fundamental frequency.
b) Open organ pipe has harmonics at all integer multiples of fundamental frequency.
c) Equating the respective frequencies allows direct cancellation of wave speed v.

19. The sound level at a point is increased by 30 dB. By what factor is the pressure amplitude increased? [3 Marks]

Answer:
1. Sound level in decibels is given by \(\beta = 20 \log_{10}\left(\frac{p_1}{p_0}\right)\), where \(p_1\) is the pressure amplitude.
2. The change in sound level is \(\Delta \beta = \beta_2 - \beta_1 = 20 \log_{10}\left(\frac{p_2}{p_1}\right)\).
3. Given \(\Delta \beta = 30\) dB, so \(30 = 20 \log_{10}\left(\frac{p_2}{p_1}\right) \implies \log_{10}\left(\frac{p_2}{p_1}\right) = 1.5\).
4. The factor of increase is \(\frac{p_2}{p_1} = 10^{1.5} = 10 \sqrt{10} \approx 31.62\).

Teacher's Note:
a) Sound intensity level is proportional to the square of pressure amplitude, giving a factor of 20 in decibel formula.
b) An increase of 20 dB corresponds to a factor of 10 in pressure amplitude.
c) Logarithmic properties simplify the ratio calculation.

1. (a) \(\frac{3}{25}\) m (b) 30 m/s (c) 15.7 m/s (d) \(2.47 \times 10^{4}\) m/s² [1 Mark]

Answer:
1. The given options represent the answers to various parts of Question 1 relating to wave parameters.
2. Part (a) value is \(\frac{3}{25}\) m, part (b) is 30 m/s, part (c) is 15.7 m/s, and part (d) is \(2.47 \times 10^{4}\) m/s².
3. These correspond to wavelength, wave speed, particle velocity, and maximum particle acceleration respectively.

Teacher's Note:
a) Basic wave kinematic formulas are applied directly here.
b) Ensure proper unit conversion before final calculation.
c) Maximum acceleration is given by \(\omega^{2}A\).

2. (a) 2.5 cm, 1.2 m/s (b) 0.03 m (c) zero [1 Mark]

Answer:
1. The given options represent the answers to various parts of Question 2.
2. Part (a) gives amplitude and wave speed as 2.5 cm and 1.2 m/s.
3. Part (b) gives wavelength or displacement as 0.03 m.
4. Part (c) gives the value as zero for displacement at nodes.

Teacher's Note:
a) Amplitude is the maximum displacement from the mean position.
b) Nodes are points of zero amplitude in standing waves.

3. 0.91 watt/m² [1 Mark]

Answer:
1. Intensity of the wave is calculated as 0.91 watt/m².
2. This represents the energy transferred per unit area per unit time.

Teacher's Note:
a) Intensity formula uses density, wave speed, angular frequency, and amplitude squared.
b) SI unit of sound intensity is watt per square meter.

4. 0.5 m/s [1 Mark]

Answer:
1. The calculated value for the required parameter is 0.5 m/s.
2. This corresponds to the wave propagation speed.

Teacher's Note:
a) Speed of wave is given by the ratio of coefficient of time to coefficient of position.

5. 27.04 N [1 Mark]

Answer:
1. The tension calculated in the string or medium is 27.04 N.
2. This is derived using wave velocity and linear mass density.

Teacher's Note:
a) Formula used is \(v = \sqrt{\frac{T}{\mu}}\).

6. 64.4 m/s [1 Mark]

Answer:
1. The speed of sound or wave motion is evaluated as 64.4 m/s.
2. Standard formulas for wave propagation in elastic media are applied.

Teacher's Note:
a) Accuracy in decimal calculations is essential here.

7. (a) 320 m/s (b) 10 cm (c) 40 cm (d) 400 Hz [1 Mark]

Answer:
1. Part (a) speed is 320 m/s.
2. Part (b) length is 10 cm.
3. Part (c) length is 40 cm.
4. Part (d) frequency is 400 Hz.

Teacher's Note:
a) Multiple sub-parts of a numerical problem tested together.

8. (a) 30 m/s, 30 cm (b) y = (1.0 cm) cos 2\(\pi\) [ \(\frac{x}{30\text{ cm}} - \frac{t}{0.01\text{ s}}\) ] (c) -5.4 m/s, 2.0 km/s² [1 Mark]

Answer:
1. Part (a) gives wave speed and wavelength.
2. Part (b) gives the wave equation.
3. Part (c) gives particle velocity and acceleration.

Teacher's Note:
a) Wave equation parameters must include proper signs for direction of propagation.

9. 10 cm [1 Mark]

Answer:
1. The required distance or amplitude value is 10 cm.
2. Derived from boundary conditions of the wave.

Teacher's Note:
a) Convert all units to standard SI units before reporting if required.

10. 180 Hz [1 Mark]

Answer:
1. The frequency of the wave is 180 Hz.
2. Calculated using fundamental frequency relations.

Teacher's Note:
a) Frequency is measured in Hertz or s^-1.

11. (a) 506 Hz (b) 490 Hz & 511 Hz [1 Mark]

Answer:
1. Part (a) frequency is 506 Hz.
2. Part (b) frequencies are 490 Hz and 511 Hz.

Teacher's Note:
a) Beat frequency concepts are used to find unknown frequencies.

12. \(\frac{2vv^{2}}{2vv - a}\) [1 Mark]

Answer:
1. The expression given in the answer key is \(\frac{2vv^{2}}{2vv - a}\).

Teacher's Note:
a) Doppler effect formula modification for moving reflector or source.

13. (a) \(\pi\)/2 (b) 2\(\pi\)/35 [1 Mark]

Answer:
1. Part (a) phase difference is \(\pi\)/2 radians.
2. Part (b) phase difference is 2\(\pi\)/35 radians.

Teacher's Note:
a) Phase difference is calculated using path difference formula \(\Delta \phi = \frac{2\pi}{\lambda} \Delta x\).

14. (a) \(1.7 \times 10^{-5}\) (b) \(1.1 \times 10^{-4}\) [1 Mark]

Answer:
1. Part (a) evaluated value is \(1.7 \times 10^{-5}\).
2. Part (b) evaluated value is \(1.1 \times 10^{-4}\).

Teacher's Note:
a) Express answers in proper scientific notation.

15. (a) zero (b) zero [1 Mark]

Answer:
1. Both parts evaluate to zero.

Teacher's Note:
a) Net work done or average energy flow over a full cycle in certain standing wave conditions can be zero.

16. \(\frac{2d}{v} \cdot \frac{\sqrt{273}}{\sqrt{T_1} + \sqrt{T_2}}\), 96 ms [1 Mark]

Answer:
1. The derived formula for the time taken is \(\frac{2d}{v} \cdot \frac{\sqrt{273}}{\sqrt{T_1} + \sqrt{T_2}}\).
2. The numerical value is 96 ms.

Teacher's Note:
a) Temperature dependence of velocity of sound is utilized here.

17. 344 m/s [1 Mark]

Answer:
1. Speed of sound in air at room temperature is 344 m/s.

Teacher's Note:
a) Standard value for speed of sound at 20°C.

18. 15 cm [1 Mark]

Answer:
1. The distance or length calculated is 15 cm.

Teacher's Note:
a) Fundamental node-antinode distance is quarter wavelength.

19. 32 [1 Mark]

Answer:
1. The count or ratio value is 32.

Teacher's Note:
a) Dimensionless number resulting from harmonic ratios.