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Superposition, reflection and standing waves in strings and organ pipes: NEET MCQs with answers

57 questions on this topic of Oscillations and Waves (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A string of length L clamped at both ends vibrates in standing wave mode such that its third overtone has frequency f₄. How many nodes (including the clamped ends) are formed along the string?

  1. 3
  2. 5
  3. 4
  4. 6
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Answer: (B) 5

The fundamental mode is the 1st harmonic (2 nodes). The third overtone corresponds to the 4th harmonic (n = 4). For the nth harmonic of a string fixed at both ends, the number of displacement nodes is n + 1. For n = 4, the number of nodes is 4 + 1 = 5.

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2. The fundamental frequency of a sonometer wire under a certain tension is 200 Hz. If the vibrating length is decreased by 20% while keeping the tension constant, the new fundamental frequency will be:

  1. 160 Hz
  2. 250 Hz
  3. 220 Hz
  4. 280 Hz
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Answer: (B) 250 Hz

The fundamental frequency is inversely proportional to length (f ∝ 1/L). If the new length is L' = 0.8 L, the new frequency is f' = f / 0.8 = 200 / 0.8 = 250 Hz.

hard

3. A resonance air column with a tuning fork of frequency 500 Hz gives first resonance at length l₁ = 15.5 cm and second resonance at length l₂ = 49.5 cm. The speed of sound in air calculated from this experiment is:

  1. 340 m/s
  2. 330 m/s
  3. 350 m/s
  4. 320 m/s
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Answer: (A) 340 m/s

In a resonance tube apparatus, the difference between successive resonance lengths is half of the wavelength: l₂ - l₁ = λ/2. Therefore, λ = 2(49.5 - 15.5) = 2 × 34.0 cm = 68.0 cm = 0.68 m. The speed of sound is v = f × λ = 500 × 0.68 = 340 m/s.

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4. Two open organ pipes have lengths of 0.75 m and 0.77 m respectively. When both sound together in their fundamental modes, they produce 6 beats per second. The speed of sound in air is:

  1. 330 m/s
  2. 340 m/s
  3. 352 m/s
  4. 346.5 m/s
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Answer: (D) 346.5 m/s

The fundamental frequency of an open organ pipe of length L is f = v / (2L). The beat frequency is Δf = (v / 2) [1/L₁ - 1/L₂]. Substituting the given values: 6 = (v / 2) [1/0.75 - 1/0.77] = (v / 2) [(0.77 - 0.75) / (0.75 × 0.77)] = (v / 2) [0.02 / 0.5775]. Solving for v gives v = (12 × 0.5775) / 0.02 = 346.5 m/s.

easy

5. When a transverse wave pulse traveling along a light string reaches a junction with a much heavier string attached under the same tension, the reflected pulse experiences:

  1. A phase change of π radians and a decrease in speed
  2. Zero phase change with no change in speed
  3. Zero phase change and an increase in speed
  4. A phase change of π radians with no change in speed
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Answer: (D) A phase change of π radians with no change in speed

The junction with a denser (heavier) medium behaves as a rigid boundary. Hence, reflection at this interface introduces a phase reversal of π radians. Since the reflected wave remains in the original medium (the lighter string), its wave speed remains unchanged.

hard

6. A sonometer wire of length 110 cm is divided into three segments by two movable bridges. If the fundamental frequencies of the three segments are in the ratio 1 : 2 : 3, what are the lengths of the three segments?

  1. 66 cm, 33 cm, 11 cm
  2. 50 cm, 40 cm, 20 cm
  3. 60 cm, 30 cm, 20 cm
  4. 55 cm, 35 cm, 20 cm
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Answer: (C) 60 cm, 30 cm, 20 cm

Since tension and linear density are uniform, frequency is inversely proportional to length (f ∝ 1/l). Given f₁ : f₂ : f₃ = 1 : 2 : 3, we have l₁ : l₂ : l₃ = 1 : 1/2 : 1/3 = 6 : 3 : 2. Let the lengths be 6x, 3x, and 2x. Then 6x + 3x + 2x = 11x = 110 cm, giving x = 10 cm. The lengths are 60 cm, 30 cm, and 20 cm.

hard

7. A pipe of length 30 cm open at both ends is vibrating in its second harmonic. If the speed of sound in air is 330 m/s, what is the distance between two successive pressure nodes?

  1. 7.5 cm
  2. 15 cm
  3. 30 cm
  4. 60 cm
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Answer: (B) 15 cm

For an open pipe in its second harmonic (n = 2), L = 2(λ/2) = λ, so λ = 30 cm. Pressure nodes are spaced by half a wavelength (λ/2). Therefore, the distance between two successive pressure nodes is λ/2 = 30 / 2 = 15 cm.

easy

8. A stationary wave is represented by y = 4 sin(πx/15) cos(96πt), where x and y are in cm and t is in seconds. The distance between two successive nodes is:

  1. 15 cm
  2. 7.5 cm
  3. 30 cm
  4. 45 cm
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Answer: (A) 15 cm

Comparing with the standard equation y = 2A sin(kx) cos(ωt), we have k = π/15 cm⁻¹. Since k = 2π/λ, the wavelength λ = 30 cm. The distance between two consecutive nodes is λ/2 = 30/2 = 15 cm.

easy

9. Consider the following two statements regarding standing waves in an air column: Statement I: At the open end of an organ pipe, a displacement antinode and a pressure node are formed. Statement II: In a closed organ pipe, only odd harmonics are produced, whereas in an open organ pipe, all harmonics are present. Choose the correct option:

  1. Both Statement I and Statement II are correct.
  2. Statement I is correct but Statement II is incorrect.
  3. Statement I is incorrect but Statement II is correct.
  4. Both Statement I and Statement II are incorrect.
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Answer: (A) Both Statement I and Statement II are correct.

At an open end, particles can oscillate freely with maximum amplitude (displacement antinode), meaning the pressure remains equal to atmospheric pressure (pressure node). Also, closed pipes support only odd harmonics (1, 3, 5...), whereas open pipes support all integral harmonics (1, 2, 3...). Thus, both statements are correct.

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10. Taking the speed of sound as 340 m/s, the fundamental frequency of a pipe 34 cm long that is open at both ends is:

  1. 250 Hz
  2. 1000 Hz
  3. 500 Hz
  4. 750 Hz
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Answer: (C) 500 Hz

Open pipe: f = v/2L = 340/(2 × 0.34) = 500 Hz.

easy

11. An organ pipe closed at one end has a fundamental frequency of 180 Hz. Which of the following frequencies CANNOT be produced as an acoustic resonance in this pipe?

  1. 540 Hz
  2. 1260 Hz
  3. 900 Hz
  4. 720 Hz
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Answer: (D) 720 Hz

A pipe closed at one end only supports odd harmonics: f_n = (2n - 1)f₁ for n = 1, 2, 3... The allowed frequencies are 180 Hz, 540 Hz (3×), 900 Hz (5×), 1260 Hz (7×), etc. The frequency 720 Hz is the 4th harmonic (an even multiple), which is not permitted.

easy

12. In a stationary wave formed on a stretched string, the phase difference between the oscillations of any two particles situated in adjacent loops is:

  1. π rad
  2. π/2 rad
  3. 0 rad
  4. 2π rad
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Answer: (A) π rad

All particles within a single loop (between two adjacent nodes) vibrate in phase with each other. Particles in two consecutive (adjacent) loops vibrate in opposite phases, so the phase difference between them is π radians.

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