CBSE Class 11 Physics Wave Motion Solved Examples

Wave motion is the transfer of energy through a medium by a disturbance that travels in space. This Class 12 Physics solved examples page on wave motion works through standing waves on strings, beats, sonometer wires, vibrations of air columns in organ pipes, velocity of sound in gases, and the Doppler effect for moving sources, observers and winds, with step-by-step solutions for each problem.

SOLVED EXAMPLES

Ex.1 The vibration of a string of length 60 cm fixed at both ends are represented by the equation :
y = 4 sin \( \left( \frac{\pi x}{15} \right) \) cos (96 π t) where x and y are in cm and t is in seconds.
(a) What is the maximum displacement of a point at x = 5 cm ?
(b) Where are the nodes located along the string ?
(c) What is the velocity of a particle at x = 7.5 cm & at t = 0.25 sec.
(d) Write down the components waves which give the above wave on superposition.

Answer:
1. Standard standing wave form: y = 2a sin(\( \frac{2\pi x}{\lambda} \)) cos(\( \frac{2\pi v t}{\lambda} \)).
2. Comparing with the given equation: \( \frac{2\pi}{\lambda} \) = \( \frac{\pi}{15} \), so λ = 30 cm, and 2a = 4, so a = 2 cm.
3. \( \frac{2\pi v}{\lambda} \) = 96π, so v = 1440 cm/s.
4. (a) At x = 5 cm, the amplitude is y_max = 4 sin(\( \frac{\pi \times 5}{15} \)) = 4 sin(\( \frac{\pi}{3} \)) = 4 × \( \frac{\sqrt{3}}{2} \) = 2√3 cm, which is about 3.46 cm.
5. (b) Nodes are where sin(\( \frac{\pi x}{15} \)) = 0, so x = 0, 15, 30, 45 and 60 cm.
6. (c) ∂y/∂t = -4 sin(\( \frac{\pi x}{15} \)) × 96π sin(96πt).
7. At x = 7.5 cm, sin(\( \frac{\pi \times 7.5}{15} \)) = sin(\( \frac{\pi}{2} \)) = 1. At t = 0.25 s, sin(96π × 0.25) = sin(24π) = 0.
8. So the velocity of the particle is -4 × 1 × 96π × 0 = 0.
9. (d) Using 4 sin A cos B = 2 sin(A + B) + 2 sin(A - B), the component waves are:
y₁ = 2 sin(\( \frac{\pi x}{15} \) + 96πt) and y₂ = 2 sin(\( \frac{\pi x}{15} \) - 96πt).

Teacher's Note:
a) The amplitude of a standing wave depends on position, and it is 2a sin(kx) here.
b) Adjacent nodes are λ/2 = 15 cm apart.
c) A particle can have zero velocity at an instant even when it is not at a node.

Ex.2 A man standing in front of a mountain at a certain distance beats a drum at regular intervals. The drumming rate is gradually increased and he finds that the echo is not heard distinctly when the rate becomes 40 per minute. He then moves nearer to the mountain by 90 metres and finds that the echo is again not heard when the drumming rate becomes 60 per minute. Calculate :
(a) the distance between the mountain and the initial position of the man
(b) the velocity of sound.

Answer:
1. Let the initial distance be x metres and the speed of sound be v m/s. The echo travels 2x.
2. Time taken by the echo to return: \( \frac{2x}{v} \).
3. Interval between beats at 40 per minute = \( \frac{60}{40} \) = 1.5 s. The echo is not heard distinctly when it overlaps the next beat, so \( \frac{2x}{v} \) = 1.5 ..... (i)
4. After moving 90 m closer, the distance is (x - 90) and the echo path is 2(x - 90).
5. Interval between beats at 60 per minute = 1 s, so \( \frac{2(x - 90)}{v} \) = 1 ..... (ii)
6. Subtracting (ii) from (i): \( \frac{180}{v} \) = 0.5, so v = 360 m/s.
7. From (i): x = \( \frac{1.5 \times 360}{2} \) = 270 m.
8. Hence, (a) the initial distance is 270 m and (b) the velocity of sound is 360 m/s.

Teacher's Note:
a) The echo is not heard distinctly when its delay equals the interval between beats.
b) Moving 90 m closer shortens the echo path by 180 m.
c) Check: 2 × 270 / 360 = 1.5 s, which matches the first beat interval.

Ex.3 A simple harmonic wave of amplitude 8 units traverses a line of particles in the direction of positive x-axis. At any given instant of time, the displacement of a particle at a distance of 10 cm from the origin is +6 units and the displacement of this particle at a distance of 25 cm from the origin is +4 units. Find the wavelength of wave (given sin 48.6º = 0.75)

Answer:
1. Wave equation: y = a sin 2π(\( \frac{t}{T} \) - \( \frac{x}{\lambda} \)), with a = 8.
2. At x = 10 cm, y = 6: sin[2π(\( \frac{t}{T} \) - \( \frac{10}{\lambda} \))] = \( \frac{6}{8} \) = 0.75 = sin 48.6°.
3. So \( \frac{t}{T} \) - \( \frac{10}{\lambda} \) = \( \frac{48.6}{360} \) ..... (iii)
4. At x = 25 cm, y = 4: sin[2π(\( \frac{t}{T} \) - \( \frac{25}{\lambda} \))] = \( \frac{4}{8} \) = 0.5 = sin 30°.
5. So \( \frac{t}{T} \) - \( \frac{25}{\lambda} \) = \( \frac{30}{360} \) = \( \frac{1}{12} \) ..... (iv)
6. Subtracting (iv) from (iii): \( \frac{15}{\lambda} \) = \( \frac{48.6 - 30}{360} \) = \( \frac{18.6}{360} \).
7. λ = \( \frac{15 \times 360}{18.6} \) = 290.3 cm.

Teacher's Note:
a) The phase angles 48.6° and 30° are the smallest positive choices, and this gives λ = 290.3 cm.
b) Other phase choices for sin 48.6° and sin 30° give other wavelengths, so the answer uses this one.
c) Keep the phase units consistent when subtracting the two equations.

Ex.4 Two sources S₁ and S₂, separated by 2.0 m, vibrate according to equation y₁ = 0.03 sin (π t) and y₂ = 0.02 sin (π t) where y₁, y₂ and t are in M.K.S. units. They send out waves of velocity 1.5 m/s. Calculate the amplitude of the resultant motion of the particle co-linear with S₁ and S₂ and located at a point
(a) to the right of S₂
(b) to the left of S₂ and
(c) in the middle of S₁ and S₂.

Answer:
1. Amplitudes: a₁ = 0.03 m and a₂ = 0.02 m. Angular frequency ω = π, so T = 2 s and n = 0.5 Hz.
2. Wavelength λ = \( \frac{v}{n} \) = \( \frac{1.5}{0.5} \) = 3.0 m.
3. (a) For a point to the right of S₂, the path difference is S₁R - S₂R = 2 m.
4. Phase difference φ = \( \frac{2\pi}{\lambda} \) × 2 = \( \frac{4\pi}{3} \).
5. Resultant amplitude a = √(a₁² + a₂² + 2a₁a₂ cos φ) = √(0.0009 + 0.0004 - 0.0006) = 0.0265 m.
6. (b) For points on the left of S₁, the path difference is S₂P - S₁P = 2 m, so the amplitude is again 0.0265 m.
7. (c) For the midpoint Q, the path difference is zero, so φ = 0 and a = a₁ + a₂ = 0.03 + 0.02 = 0.05 m.

Teacher's Note:
a) The solution for part (b) takes the point to the left of S₁, where the path difference is 2 m.
b) A phase difference of 4π/3 reduces the amplitude below the sum of the two amplitudes.
c) At the midpoint the waves arrive in phase, so the amplitudes add.

Ex.5 A progressive and a stationary simple harmonic wave each has the same frequency of 250 Hz, and the same velocity of 30 m/s. Calculate
(a) the phase difference between two vibrating points on the progressive wave which are 10 cm apart,
(b) the equation of motion of the progressive wave if its amplitude is 0.03 m,
(c) the distance between nodes in the stationary wave,
(d) the equation of motion of the stationary wave if its amplitude is 0.01 m.

Answer:
1. Wavelength λ = \( \frac{v}{n} \) = \( \frac{30}{250} \) = 0.12 m = 12 cm.
2. (a) Phase difference for a path of 10 cm = \( \frac{2\pi}{\lambda} \) × 10 = \( \frac{2\pi \times 10}{12} \) = \( \frac{5\pi}{3} \).
3. (b) For the progressive wave, y = a sin 2π(\( \frac{t}{T} \) - \( \frac{x}{\lambda} \) + φ) with a = 0.03 m, 1/T = 250 Hz and 1/λ = \( \frac{25}{3} \) per m.
4. So y = 0.03 sin 2π(250t - \( \frac{25x}{3} \) + φ) m, where φ is the initial phase.
5. (c) Distance between nodes in a stationary wave = \( \frac{\lambda}{2} \) = \( \frac{12}{2} \) = 6 cm.
6. (d) With an antinode at x = 0, y = 2a cos(\( \frac{2\pi x}{\lambda} \)) sin(\( \frac{2\pi t}{T} \)).
7. Here a = 0.01 m, so 2a = 0.02, \( \frac{2\pi}{\lambda} \) = \( \frac{50\pi}{3} \), and \( \frac{2\pi}{T} \) = 500π.
8. So y = 0.02 cos(\( \frac{50\pi x}{3} \)) sin(500πt) m, where x and y are in m and t is in s.

Teacher's Note:
a) Phase difference depends on the path difference divided by the wavelength.
b) In a stationary wave, nodes are separated by half a wavelength.
c) The stationary wave amplitude is 2a, which is why 0.01 m gives 0.02 in the equation.

Ex.6 The length of the sonometer wire between two fixed ends is 100 cm. Where should the two bridges be placed so as to divide the wire into three segments whose fundamental frequencies are in the ratio of 1:2:3?

Answer:
1. Let the bridges be at C and D, so that the segments AC, CD and DB have lengths l₁, l₂ and l₃.
2. Total length: l₁ + l₂ + l₃ = 100 cm ..... (i)
3. Frequencies n₁ : n₂ : n₃ = 1 : 2 : 3.
4. For the same tension and wire, n ∝ \( \frac{1}{l} \), so n₁l₁ = n₂l₂ = n₃l₃.
5. Therefore l₁ = \( \frac{n_2}{n_1} \) l₂ = 2l₂ and l₃ = \( \frac{n_2}{n_3} \) l₂ = \( \frac{2}{3} \) l₂.
6. Substituting in (i): 2l₂ + l₂ + \( \frac{2}{3} \) l₂ = 100, so \( \frac{11}{3} \) l₂ = 100 and l₂ = 27.27 cm.
7. l₁ = 2 × 27.27 = 54.54 cm, and l₃ = \( \frac{2}{3} \) × 27.27 = 18.18 cm.

Teacher's Note:
a) Shorter segments have higher fundamental frequencies, so the longest segment has the lowest frequency.
b) The lengths are in the ratio 6 : 3 : 2, which adds up to 11 parts.
c) Check: 54.54 + 27.27 + 18.18 = 100 cm.

Ex.7 The fundamental frequency of a 1.5 m long, stretched steel wire is 175 Hertz. The density of steel is 7.8 × 10³ kg/m³.
(a) Find the speed of transverse waves in the wire.
(b) Calculate the longitudinal stress of the wire.
(c) If the tension in the wire is increased by 3%, calculate the percentage change in the frequency of the wire.

Answer:
1. (a) For the fundamental mode of a stretched wire fixed at both ends, the speed is v = n × 2l.
2. v = 175 × (2 × 1.5) = 525 m/s.
3. (b) The speed of a transverse wave is v = √(T/μ), where μ = Ad is the mass per unit length.
4. So \( \frac{T}{A} \) = v²d = (525)² × 7.8 × 10³ = 2.15 × 10⁹ N/m².
5. (c) Frequency n = \( \frac{1}{2l} \)√(T/μ), so n ∝ √T.
6. \( \frac{n'}{n} \) = √1.03 = 1.0149.
7. Percentage change = (1.0149 - 1) × 100 = 1.49%, which is about 1.5%.

Teacher's Note:
a) A 3% increase in tension gives only about 1.5% increase in frequency.
b) The longitudinal stress equals the tension per unit area of the wire.
c) Take the square root of the tension ratio, not the ratio itself.

Ex.8 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.

Answer:
1. Let p and q be the numbers of loops in the wires of radii r and 2r, with masses per unit length m₁ and m₂.
2. Frequencies: n₁ = \( \frac{p}{2l} \)√(T/m₁) and n₂ = \( \frac{q}{2l} \)√(T/m₂).
3. Since the wire is continuous, n₁ = n₂, so \( \frac{p}{q} \) = √(\( \frac{m_1}{m_2} \)).
4. m₁ = πr²d and m₂ = π(2r)²d, so \( \frac{m_1}{m_2} \) = \( \frac{1}{4} \).
5. \( \frac{p}{q} \) = √\( \frac{1}{4} \) = \( \frac{1}{2} \).
6. Hence the ratio of the number of loops is 1 : 2.

Teacher's Note:
a) The thinner wire has a smaller mass per unit length, so it vibrates with more loops at the same frequency.
b) The ratio of mass per unit length is the square of the ratio of radii.
c) The welded joint being a node is consistent with this 1 : 2 ratio.

Ex.9 A copper wire is held at the two ends by rigid supports. At 30ºC the wire is just taut with zero tension. Find the speed of transverse waves in this wire at 10ºC (α = 1.7 × 10^-5/ºC, Y = 1.3 × 10¹¹ N/m², d = 9 × 10³ kg/m³).

Answer:
1. Contraction on cooling from 30°C to 10°C: \( \frac{dl}{l} \) = α × Δt = 1.7 × 10^-5 × 20 = 3.4 × 10^-4.
2. Stress = Y × strain, so \( \frac{T}{A} \) = Y × \( \frac{dl}{l} \) = 1.3 × 10¹¹ × 3.4 × 10^-4 = 4.42 × 10⁷ N/m².
3. Speed v = √(T/m) = √(\( \frac{T}{Ad} \)) = √(\( \frac{4.42 \times 10^7}{9 \times 10^3} \)).
4. v ≈ 70 m/s.

Teacher's Note:
a) The wire is taut only because it cannot contract freely between the supports.
b) The temperature change must be in degrees, which is the same size as kelvin.
c) Use the stress, not the strain, when substituting into the wave speed formula.

Ex.10 The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz. The fundamental frequency of the closed organ pipe is 110 Hz. Find the lengths of the pipes.

Answer:
1. Let the lengths of the open and closed pipes be l₁ and l₂. Take v = 330 m/s.
2. Fundamental of the closed pipe: \( \frac{v}{4l_2} \) = 110, so l₂ = \( \frac{330}{440} \) = 0.75 m.
3. First overtone of the closed pipe: \( \frac{3v}{4l_2} \) = 3 × 110 = 330 Hz.
4. First overtone of the open pipe: n₁ = \( \frac{2v}{2l_1} \) = \( \frac{v}{l_1} \).
5. Beat condition: |\( \frac{v}{l_1} \) - 330| = 2.2, so \( \frac{330}{l_1} \) = 332.2 or 327.8.
6. l₁ = \( \frac{330}{332.2} \) = 0.993 m, or l₁ = \( \frac{330}{327.8} \) = 1.007 m.
7. Hence the open pipe is 0.993 m or 1.007 m long, and the closed pipe is 0.75 m long.

Teacher's Note:
a) The first overtone of an open pipe is twice its fundamental, while for a closed pipe it is three times.
b) The beat frequency can come from the open pipe being either higher or lower than 330 Hz, so both lengths are valid.
c) A closed pipe has only odd harmonics, so its first overtone is 3 times the fundamental.

Ex.11 The speed of sound in hydrogen is 1270 m/sec. Calculate the speed of sound in the mixture of oxygen and hydrogen in which they are mixed in 1 : 4 ratio.

Answer:
1. Let the volumes of oxygen and hydrogen be V₁ and V₂, with V₁ : V₂ = 1 : 4.
2. Density of oxygen to density of hydrogen: \( \frac{d_1}{d_2} \) = \( \frac{\text{Molecular weight of oxygen}}{\text{Molecular weight of hydrogen}} \) = \( \frac{32}{2} \) = 16.
3. Density of the mixture: d = \( \frac{V_1 d_1 + V_2 d_2}{V_1 + V_2} \) = \( \frac{1 \times 16 d_2 + 4 d_2}{5} \) = 4d₂.
4. Both gases are diatomic, so γ = 1.4 for each and for the mixture, and the pressure is the same.
5. v = √(\( \frac{\gamma P}{d} \)), so \( \frac{v_{mix}}{v_{H_2}} \) = √(\( \frac{d_2}{d} \)) = √\( \frac{1}{4} \) = \( \frac{1}{2} \).
6. v_mix = \( \frac{1270}{2} \) = 635 m/s.

Teacher's Note:
a) The mixture is denser than hydrogen, so sound is slower in it.
b) Use the mixture's average density by volume, not a simple average of the two densities.
c) Hydrogen and oxygen are both diatomic, which is why the γ values match.

Ex.12 The wavelength of the note emitted by a tuning fork of frequency 512 Hz in air at 17ºC is 66.6 cm. If the density of air at S.T.P. is 1.293 gm per litre, calculate γ for air.

Answer:
1. Speed of sound at 17°C: v = nλ = 512 × 0.666 = 340.99 m/s.
2. Speed at 0°C: v₀ = \( \frac{v}{1 + t/546} \) = \( \frac{340.99}{1 + 17/546} \) = 330.7 m/s.
3. v = √(\( \frac{\gamma P}{\rho} \)), so γ = \( \frac{v_0^2 \rho}{P} \).
4. Substituting P = 1.013 × 10⁵ N/m² and ρ = 1.293 kg/m³: γ = \( \frac{(330.7)^2 \times 1.293}{1.013 \times 10^5} \) = 1.40.

Teacher's Note:
a) The printed working gives v = 340.48 m/s, v₀ = 330 m/s and γ = 1.39, but 512 × 0.666 = 340.99 m/s, so the corrected values are v₀ ≈ 330.7 m/s and γ ≈ 1.40.
b) The value 1.40 matches the known γ for diatomic air.
c) Use the density at S.T.P. and the standard pressure together with the speed at 0°C.

Ex.13 The difference between the apparent frequency of a source of sound as perceived by an observer during its approach and recession is 2% of the natural frequency of the source. Find the velocity of the source. Take the velocity of sound as 350 m/s.

Answer:
1. Approaching source: n' = n \( \frac{v}{v - v_s} \) ≈ n(1 + \( \frac{v_s}{v} \)).
2. Receding source: n'' = n \( \frac{v}{v + v_s} \) ≈ n(1 - \( \frac{v_s}{v} \)).
3. Difference: n' - n'' = \( \frac{2nv_s}{v} \).
4. Given \( \frac{n' - n''}{n} \) = 0.02, so \( \frac{2v_s}{v} \) = 0.02.
5. v_s = 0.01 × 350 = 3.5 m/s.

Teacher's Note:
a) This approximation works because the source speed is much less than the speed of sound.
b) The percentage is of the natural frequency, not of the received frequency.
c) A 2% frequency change needs a source speed of only 1% of the speed of sound.

Ex.14 A whistle emitting a sound of frequency 440 Hz is tied to a string of 1.5 m length and rotated with an angular velocity of 20 rad/s in the horizontal plane. Calculate the range of frequencies heard by an observer stationed at a large distance from the whistle.

Answer:
1. Speed of the whistle: v_s = rω = 1.5 × 20 = 30 m/s.
2. Speed of sound v = 330 m/s and n = 440 Hz.
3. Maximum frequency, when the whistle approaches the observer: n_max = n \( \frac{v}{v - v_s} \) = 440 × \( \frac{330}{300} \) = 484 Hz.
4. Minimum frequency, when the whistle recedes from the observer: n_min = n \( \frac{v}{v + v_s} \) = 440 × \( \frac{330}{360} \) = 403 Hz.
5. Range of frequencies heard: 403 Hz to 484 Hz.

Teacher's Note:
a) The whistle moves in a circle, so its speed is constant but its direction changes.
b) Use the approach formula for the highest pitch and the recession formula for the lowest pitch.
c) The observer is far away, so the effect of the circular path on direction is ignored.

Ex.15 A train approaching a hill at a speed of 40 km/hr sounds a whistle of frequency 580 Hz when it is at a distance of 1 km from a hill. A wind with a speed of 40 km/hr is blowing in the direction of motion of the train. Find
(a) the frequency of the whistle as heard by the observer on the hill,
(b) the distance from the hill at which the echo from the hill is heard by the driver and its frequency
(Velocity of sound in air 1,200 km/hr).

Answer:
1. (a) Sound travels with the wind toward the hill, so the effective speed is v + w = 1200 + 40 = 1240 km/hr.
2. Apparent frequency: n' = n \( \frac{v + w}{(v + w) - v_s} \) = 580 × \( \frac{1240}{1200} \) = 599.3 Hz.
3. (b) Let the driver hear the echo when he is at a distance x km from the hill. The train covers (1 - x) km in \( \frac{1 - x}{40} \) hr.
4. The sound from the train's position at 1 km reaches the hill in \( \frac{1}{1240} \) hr, travelling with the wind.
5. The echo returns to the driver against the wind, at speed 1200 - 40 = 1160 km/hr, so the time taken by the echo to reach the point B' is t = \( \frac{x}{1200 - 40} \) hr, i.e. it covers x km in \( \frac{x}{1160} \) hr.
6. Equating the two times: \( \frac{1 - x}{40} \) = \( \frac{1}{1240} \) + \( \frac{x}{1160} \), which gives x ≈ 0.935 km.
7. So the driver hears the echo at about 935 m from the hill.
8. The hill hears 599.3 Hz and re-emits it. The driver moves toward the hill at 40 km/hr against the echo, so the frequency heard is n'' = 599.3 × \( \frac{1160 + 40}{1160} \) = 620 Hz.

Teacher's Note:
a) The sound's forward journey to the hill must be included in the echo time.
b) Wind adds to the sound speed toward the hill and subtracts from it on the return.
c) The echo is heard at a higher pitch because the driver is moving toward the hill.

Ex.16 A band playing music at a frequency f is moving towards a wall at a speed v_b. A motorist is following the band with a speed v_m. If v is the speed of sound, obtain an expression for the beat frequency heard by the motorist.

Answer:
1. Frequency of the band heard directly by the motorist: f' = f \( \frac{v + v_m}{v + v_b} \).
2. Frequency reaching the wall: f_w = f \( \frac{v}{v - v_b} \).
3. Frequency reaching the motorist after reflection from the wall: f'' = f_w \( \frac{v + v_m}{v} \) = f \( \frac{v + v_m}{v - v_b} \).
4. Beat frequency: Δf = f'' - f' = f(v + v_m) [\( \frac{1}{v - v_b} \) - \( \frac{1}{v + v_b} \)].
5. Δf = f(v + v_m) × \( \frac{2v_b}{v^2 - v_b^2} \) = \( \frac{2 f v_b (v + v_m)}{v^2 - v_b^2} \).

Teacher's Note:
a) The motorist hears the direct sound and the reflected sound from the wall.
b) The wall is a stationary observer for the band and a stationary source for the reflected sound.
c) The beat frequency is the difference between the two frequencies heard.

Ex.17 A source of sound is moving along circular orbit of radius 3 meters with an angular velocity of 10 rad/s. A sound detector located far away from the source is executing linear simple harmonic motion along the line BD as shown in the figure with an amplitude BC = CD = 6 meters. The frequency of oscillation of the detector is 5/π per second. The source is at the point A when the detector is at the point B. If the source emits a continuous sound wave of frequency 340 Hz, find the maximum and the minimum frequencies recorded by the detector.

[Figure: Circle of radius 3 m with centre marked, a point A on the circle at the right of the centre, and a straight line BD on the right, with points B, C and D marked. Between B and C the label reads 6 m and between C and D the label reads 6 m.]

Answer:
1. Speed of source: v_s = rω = 3 × 10 = 30 m/s.
2. Maximum speed of the detector: v₀ = Aω' = 6 × 2π × \( \frac{5}{\pi} \) = 60 m/s.
3. Maximum frequency, when the source and detector move toward each other: n_max = n \( \frac{v + v_0}{v - v_s} \) = 340 × \( \frac{330 + 60}{330 - 30} \) = 442 Hz.
4. Minimum frequency, when the source and detector move away from each other: n_min = n \( \frac{v - v_0}{v + v_s} \) = 340 × \( \frac{330 - 60}{330 + 30} \) = 255 Hz.
5. Maximum frequency is 442 Hz and minimum frequency is 255 Hz.

Teacher's Note:
a) The detector's speed is taken as its maximum speed, which is reached at the centre of its motion.
b) For a receding detector the numerator is v - v₀, which is why the printed minimum formula needs this sign.
c) Both the source and detector speeds are taken as positive when they move toward each other.

Sol.: Comparing given equation with equation of standing wave,