CBSE Class 11 Physics Wave Motion Objective Questions

CBSE CLASS 12 PHYSICS
WAVE MOTION
OBJECTIVE QUESTIONS

OBJECTIVES

1. The displacement of a wave is represented by
y = 0.25 × 10^-3 sin (500 t - 0.05 x)
where all the quantities are in their S.I. unit. The maximum particle velocity of the medium is .
(a) 0.5 ms^-1
(b) 0.75 ms^-1
(c) 0.125 ms^-1
(d) 0.150 ms^-1

Answer: (c) 0.125 ms^-1

1. Compare with y = A sin (ωt - kx): A = 0.25 × 10^-3 m and ω = 500 rad/s.
2. Particle velocity = dy/dt = Aω cos (ωt - kx), so the maximum particle velocity = Aω.
3. Aω = 0.25 × 10^-3 × 500 = 0.125 m/s.
4. Hence, the correct option is (c).

Teacher's Note:
a) Do not confuse the particle velocity (Aω) with the wave velocity (ω/k = 10⁴ m/s here).
b) The wave speed ω/k is a property of the medium, whereas particle velocity depends on amplitude.

2. If the ratio of the amplitudes of two interfering beams be 2 : 3, the ratio of the minimum and maximum intensities of sound would be
(a) 4 : 9
(b) √2 : √3
(c) 1 : 5
(d) 1 : 25

Answer: (d) 1 : 25

1. Let the amplitudes be 2a and 3a. I_min ∝ (3a - 2a)² = a² and I_max ∝ (3a + 2a)² = 25a².
2. I_min : I_max = 1 : 25.
3. Hence, the correct option is (d).

Teacher's Note:
a) Intensity is proportional to the square of the resultant amplitude.
b) The ratio of the amplitudes, not intensities, is given in the question.

3. In the interference of waves from two sources of intensities I₀ and 4I₀, the intensity at a point where the phase difference is π is
(a) 5I₀
(b) 3I₀
(c) 2I₀
(d) I₀

Answer: (d) I₀

1. I = I₁ + I₂ + 2√(I₁I₂) cos φ.
2. With φ = π: I = I₀ + 4I₀ - 2√(4I₀²) = 5I₀ - 4I₀ = I₀.
3. Hence, the correct option is (d).

Teacher's Note:
a) At a phase difference of π the intensity is the minimum possible, (√I₁ - √I₂)² = (2√I₀ - √I₀)².
b) The minimum is zero only when the two intensities are equal.

4. In stationary waves
(a) pressure nodes and antinodes coincide with displacement nodes and antinodes
(b) velocity nodes and antinodes coincide with displacement nodes and antinodes
(c) pressure nodes and antinodes coincide with velocity nodes and antinodes
(d) none of these

Answer: (b) velocity nodes and antinodes coincide with displacement nodes and antinodes

1. The particle velocity is the time derivative of the displacement, so it is zero wherever the displacement is always zero (nodes) and largest where the displacement is largest (antinodes).
2. Pressure variation is largest where the particles are closest and farthest, i.e. at displacement nodes, so pressure nodes coincide with displacement antinodes.
3. So only statement (b) is correct.

Teacher's Note:
a) Pressure and displacement are 90° out of phase in space, so their nodes and antinodes interchange.
b) In an open organ pipe the ends are displacement antinodes and pressure nodes.

5. A transverse wave is described by the equation
y = y₀ sin 2π(ft - x/λ)
The maximum particle velocity is equal to four times the wave velocity if
(a) λ = \( \frac{\pi y_0}{4} \)
(b) λ = \( \frac{\pi y_0}{2} \)
(c) λ = πy₀
(d) λ = 2πy₀

Answer: (b) λ = \( \frac{\pi y_0}{2} \)

1. Maximum particle velocity = Aω = 2πf y₀.
2. Wave velocity = fλ.
3. 2πf y₀ = 4 fλ, so λ = \( \frac{2\pi y_0}{4} = \frac{\pi y_0}{2} \).
4. Hence, the correct option is (b).

Teacher's Note:
a) The frequency f cancels from both sides.
b) Particle velocity depends on the amplitude, but wave velocity does not.

6. The fundamental frequency of a vibrating organ pipe is 200 Hz.
(a) The first overtone is 400 Hz
(b) The first overtone may be 400 Hz
(c) The first overtone may be 600 Hz
(d) 600 Hz is an overtone.

Answer: (b), (c) and (d)

1. The pipe may be open at both ends or closed at one end, and the question does not say which.
2. Open pipe: harmonics are 200, 400, 600 Hz, so the first overtone is 400 Hz. Closed pipe: harmonics are 200, 600, 1000 Hz, so the first overtone is 600 Hz.
3. So the first overtone may be 400 Hz or 600 Hz, and 600 Hz is an overtone in both cases.
4. Statement (a) is not always true, so the correct options are (b), (c) and (d).

Teacher's Note:
a) This question has more than one correct option, as the printed key (b, c, d) shows.
b) Statement (a) states the first overtone must be 400 Hz, which holds only for an open pipe.

7. In the equation y = a sin k(x - pt), where k = \( \frac{2\pi}{\lambda} \), the speed of the wave is
(a) p
(b) k p
(c) k/p
(d) p/k

Answer: (a) p

1. Write the equation as y = a sin (kx - kpt) and compare with y = a sin (kx - ωt): ω = kp.
2. Wave speed v = ω/k = kp/k = p.
3. Hence, the correct option is (a).

Teacher's Note:
a) A wave y = f(x - pt) travels along +x with speed p.
b) Here p has the unit of velocity, and kp is the angular frequency.

8. The speed of sound through a gaseous medium bears a constant ratio with the rms speed of its molecules That constant ratio is
(a) \( \sqrt{\frac{\gamma}{3}} \)
(b) γ - 1
(c) \( \sqrt{\frac{2\gamma}{3}} \)
(d) γ

Answer: (a) \( \sqrt{\frac{\gamma}{3}} \)

1. Speed of sound v = √(γP/ρ) and rms speed v_rms = √(3P/ρ).
2. v / v_rms = \( \sqrt{\frac{\gamma P/\rho}{3P/\rho}} = \sqrt{\frac{\gamma}{3}} \).
3. Hence, the correct option is (a).

Teacher's Note:
a) The ratio is less than 1 for all gases, so sound is slower than the molecules themselves.
b) Both speeds depend on √(P/ρ), which cancels in the ratio.

9. The temperature at which the velocity of sound is double its velocity at 0°C is
(a) 546° C
(b) 819°C
(c) 1092° C
(d) 273° C

Answer: (b) 819°C

1. The speed of sound v ∝ √T, with T in kelvin.
2. For v to double, T must become 4 times: T = 4 × 273 = 1092 K.
3. In celsius this is 1092 - 273 = 819 °C.
4. Hence, the correct option is (b).

Teacher's Note:
a) Always convert to kelvin before using v ∝ √T.
b) Option (c) 1092 is the kelvin value, which is the usual trap.

10. In the fourth harmonic of an open organ pipe, there are
(a) 2N, 3 A
(b) 3N, 4A
(c) 4N, 5A
(d) 5N, 4A

Answer: (c) 4N, 5A

1. An open pipe has an antinode at each end. In the n-th harmonic there are n nodes and (n + 1) antinodes.
2. For the fourth harmonic: 4 nodes and 5 antinodes.
3. Hence, the correct option is (c).

Teacher's Note:
a) Draw the standing wave: the first harmonic has 1 node and 2 antinodes.
b) In the n-th harmonic the pipe contains n half-wavelengths, L = nλ/2.

11. The length of an open pipe is 25 cm and the velocity of sound in air is 350 ms^-1. The frequency in Hz of the second harmonic of the pipe is
(a) 350
(b) 700
(c) 1050
(d) 1400

Answer: (d) 1400

1. For an open pipe the fundamental is f₁ = v/2L = 350 / (2 × 0.25) = 700 Hz.
2. The second harmonic is f₂ = 2f₁ = 1400 Hz.
3. Hence, the correct option is (d).

Teacher's Note:
a) Use L in metres (25 cm = 0.25 m).
b) For the open pipe, harmonic number n gives f = nv/2L; 700 Hz is the first harmonic, not the second.

12. A stretched string vibrates with a frequency of 30 Hz in its fundamental mode when the supports are 60 cm apart. The velocity of transverse waves in the string is
(a) 9 ms^-1
(b) 18 9 ms^-1
(c) 36 ms^-1
(d) 72 ms^-1

Answer: (c) 36 ms^-1

1. In the fundamental mode, L = λ/2, so λ = 2 × 0.60 = 1.2 m.
2. v = fλ = 30 × 1.2 = 36 m/s.
3. Hence, the correct option is (c).

Teacher's Note:
a) The string is fixed at both ends, so the fundamental has λ = 2L.
b) The speed does not depend on the mode; only the frequency does.

13. A string B has twice the length, twice the diameter, twice the tension and twice the density of another string A. Which of the following alternatives expresses the relation between the frequencies of A and B?
(a) n_B = 4n_A
(b) n_A = 4n_B
(c) n_A = 2n_B
(d) 2n_B = n_A

Answer: (b) n_A = 4n_B

1. Frequency n = \( \frac{1}{2L}\sqrt{\frac{T}{\mu}} \), where the mass per unit length μ = ρ × (π d²/4) ∝ ρd².
2. For B: L is 2 times, T is 2 times and μ is 2 × 2² = 8 times that of A.
3. n_B/n_A = (1/2) × √(2/8) = (1/2)(1/2) = 1/4.
4. So n_A = 4n_B. Hence, the correct option is (b).

Teacher's Note:
a) The printed key gives (a), which would need n_B to be larger; B is longer, thicker and heavier per unit length, so its frequency is lower.
b) The mass per unit length depends on the square of the diameter.

14. The equation y = A sin²(kx - ωt) represents a wave motion with
(a) amplitude A frequency ω/2π
(b) amplitude A/2, frequency ω/π
(c) amplitude 2A, frequency ω/4π
(d) does not represent a wave motion.

Answer: (b) amplitude A/2, frequency ω/π

1. Use sin²θ = (1 - cos 2θ)/2: y = A/2 - (A/2) cos (2kx - 2ωt).
2. The oscillating part has amplitude A/2 and angular frequency 2ω.
3. Frequency = 2ω/2π = ω/π.
4. Hence, the correct option is (b).

Teacher's Note:
a) The constant A/2 only shifts the mean position and does not affect the wave.
b) Squaring the sine doubles the frequency and halves the amplitude.

15. A wave pulse, travelling on a two-piece string, gets partially reflected and partially transmitted at the junction. The reflected wave is inverted in shape as compared to the incident one. If the incident wave has wavelength λ and the transmitted wave λ',
(a) λ' > λ
(b) λ' = λ
(c) λ' < λ
(d) nothing can be said about the relation of λ and λ'

Answer: (c) λ' < λ

1. A reflected pulse is inverted when the wave goes from a lighter string to a heavier string (a denser medium).
2. The speed on the heavier string is lower (v = √(T/μ)), while the frequency stays the same.
3. λ = v/f, so the wavelength on the heavier string is smaller: λ' < λ.
4. Hence, the correct option is (c).

Teacher's Note:
a) The frequency is fixed by the source and does not change at the junction.
b) If the reflection were not inverted, the second string would be lighter and λ' would be greater.

16. Two waves represented by y = a sin (ωt - kx) and y = a cos (ωt - kx) are superposed. The resultant wave will have an amplitude
(a) a
(b) √2 a
(c) 2 a
(d) 0

Answer: (b) √2 a

1. a cos θ = a sin (θ + π/2), so the two waves differ in phase by π/2.
2. Resultant amplitude = √(a² + a² + 2a² cos 90°) = √2 a.
3. Hence, the correct option is (b).

Teacher's Note:
a) Equivalent method: sin θ + cos θ = √2 sin (θ + π/4).
b) A phase difference of 90° adds the amplitudes like the sides of a right triangle.

17. Two wires A and B having identical geometrical construction, are stretched from their natural length by equal amount. The Young's modulus of the wires are Y_A and Y_B whereas the densities are ρ_A and ρ_B. It is given that Y_A > Y_B and ρ_A > ρ_B. A transverse signal started at one end takes a time t₁ to reach the other end for A and t₂ for B.
(a) t₁ < t₂
(b) t₁ = t₂
(c) t₁ > t₂
(d) The information is insufficient to find the relation between t₁ and t₂

Answer: (d) The information is insufficient to find the relation between t₁ and t₂

1. Tension T = Y A ε, where ε is the equal strain, and the mass per unit length is μ = ρ A.
2. Wave speed v = √(T/μ) = √(Yε/ρ), so v depends on the ratio Y/ρ.
3. Since Y_A > Y_B and ρ_A > ρ_B, the ratio Y/ρ can be larger for either wire.
4. So t₁ and t₂ cannot be compared, and the correct option is (d).

Teacher's Note:
a) A larger Y raises the speed and a larger ρ lowers it, so the effects compete.
b) More data (the actual values of Y/ρ) are needed.

18. Two sine waves travel in the same direction in a medium. The amplitude of each wave is A and the phase difference between the two waves is 120°. The resultant amplitude will be
(a) A
(b) 2A
(c) 4A
(d) √2A

Answer: (a) A

1. Resultant amplitude = √(A² + A² + 2A² cos 120°) = √(2A² - A²) = A.
2. Equivalent: 2A cos (φ/2) = 2A cos 60° = A.
3. Hence, the correct option is (a).

Teacher's Note:
a) For equal amplitudes, the resultant is 2A cos (φ/2).
b) At 120° the resultant amplitude equals that of each wave.

19. A wave is represented by the equation
y = (0.001 mm) sin [(50 s^-1)t + (2.0 m^-1)x].
(a) The wave velocity = 100 m/s
(b) The wavelength = 2.0 m.
(c) The frequency = 25/π Hz.
(d) The amplitude = 0.001 mm.

Answer: (c) and (d)

1. Compare with y = A sin (ωt + kx): A = 0.001 mm, ω = 50 s^-1, k = 2.0 m^-1.
2. Wave velocity = ω/k = 50/2 = 25 m/s, so (a) is false. Wavelength = 2π/k = π m, so (b) is false.
3. Frequency = ω/2π = 50/2π = 25/π Hz, so (c) is true, and the amplitude is 0.001 mm, so (d) is true.
4. Hence, the correct options are (c) and (d).

Teacher's Note:
a) More than one option is correct here, as the printed key (c, d) shows.
b) The plus sign shows that the wave travels in the negative x direction.

20. A standing wave is produced on a string clamped at one end and free at the other. The length of the string
(a) must be an integral multiple of λ/4
(b) must be an integral multiple of λ/2
(c) must be an integral multiple of λ
(d) may be an integral multiple of λ/2

Answer: (a) must be an integral multiple of λ/4

1. The clamped end is a node and the free end is an antinode.
2. The distance between a node and an adjacent antinode is λ/4, so L = (2n + 1) λ/4, an odd multiple of λ/4.
3. So L is an integral multiple of λ/4 (always an odd one).
4. Hence, the correct option is (a).

Teacher's Note:
a) The same condition applies to a closed organ pipe.
b) Option (b) holds for strings with both ends fixed.

21. Mark out the correct options.
(a) The energy of an small part of a string remains constant in a travelling wave.
(b) The energy of an small part of a string remains constant in a standing wave.
(c) The energies of all the small parts of equal length are equal in a travelling wave.
(d) energy trapped between two nodes is constant

Answer: (d) energy trapped between two nodes is constant

1. In a travelling wave the energy of a small part changes with time, as the wave carries energy past it, so (a) is false; at one instant the parts have different energies, so (c) is false.
2. In a standing wave the energy of a small part changes too (kinetic changes to potential), so (b) is false.
3. No energy crosses a node, so the energy trapped between two nodes is constant, and (d) is true.
4. Hence, the correct option is (d).

Teacher's Note:
a) Nodes never move, so no energy flows through them.
b) The energy of one loop keeps changing between kinetic and potential forms but its total stays constant.

22. In a stationary wave,
(a) all the particles of the medium vibrate in phase
(b) all the antinodes vibrate in phase
(c) the alternate antinodes vibrate in phase
(d) all the particles between consecutive nodes vibrate in phase.

Answer: (c) and (d)

1. In a stationary wave, all particles between two consecutive nodes move together in phase, so (d) is true.
2. Particles in adjacent loops are in opposite phase (a phase difference of π), so (a) and (b) are false.
3. So alternate antinodes (two loops apart) vibrate in phase, and (c) is true.
4. Hence, the correct options are (c) and (d).

Teacher's Note:
a) The question has more than one right answer, as the printed key (c, d) shows.
b) The phase changes by π each time you cross a node.

23. An open organ pipe of length L vibrates in its fundamental mode. The variation is maximum
(a) at the two ends
(b) at the middle of the pipe
(c) at distance L/4 inside the ends
(d) at distance L/8 inside the ends.

Answer: (b) at the middle of the pipe

1. The variation here means the pressure variation (and density variation).
2. In the fundamental mode of an open pipe there are displacement antinodes at the two ends and a displacement node in the middle.
3. Pressure variation is largest at the displacement node, which is the middle of the pipe.
4. Hence, the correct option is (b).

Teacher's Note:
a) At the open ends the pressure stays equal to the atmospheric pressure, so the pressure variation there is zero.
b) Displacement and pressure variations are out of step along the pipe.

24. A cylindrical tube, open at both ends, has a fundamental frequency v. The tube is dipped vertically in water so that half of its length is inside the water. The new fundamental frequency is .
(a) v/4
(b) v/2
(c) v
(d) 2v

Answer: (c) v

1. The open tube of length L has v = v_s/2L, where v_s is the speed of sound.
2. After dipping, the air column has length L/2 and is closed at the bottom, so its fundamental is v_s/(4 × L/2) = v_s/2L.
3. This equals the original v. Hence, the correct option is (c).

Teacher's Note:
a) An open pipe of length L and a closed pipe of length L/2 have the same fundamental frequency.
b) The new frequency does not change because the closed pipe has λ = 4 × (L/2) = 2L, the same as the open pipe.

25. A small source of sound moves on a circle as shown in figure and an observer is sitting at O. Let v₁, v₂, v₃ be the frequencies heard when the source is at A, B and C respectively.
(a) v₁ > v₂ > v₃
(b) v₁ = v₂ > v₃
(c) v₂ > v₃ > v₁
(d) v₁ > v₃ > v₂

[Figure: A circle with its centre marked by a dot. Point A is at the top, point B at the bottom and point C on the right end of the horizontal diameter. A dashed horizontal line runs from the centre through C to the observer O far to the right. An arrowhead on the left side of the circle points upward, showing the direction of motion of the source.]

Answer: (d) v₁ > v₃ > v₂

1. In the figure the arrow on the left of the circle points upwards, so the source moves clockwise: at A it moves towards the right, at B towards the left, and at C vertically downwards.
2. At A the source has a component of velocity towards O, so the observed frequency is raised: v₁ is greater than the true frequency.
3. At C the velocity is perpendicular to the line CO, so there is no Doppler shift: v₃ equals the true frequency. At B the source moves away from O, so v₂ is less than the true frequency.
4. So v₁ > v₃ > v₂. Hence, the correct option is (d).

Teacher's Note:
a) The printed key (c) holds only if the source moves anticlockwise; the arrow in the figure shows clockwise motion.
b) Only the component of the source velocity along the line to the observer matters.