Waves – Beats: NEET MCQs with Answers

9 free questions on Waves – Beats for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Beats

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Question 1

Previous-year question
In a guitar, two strings A and B made of the same material are slightly out of tune and produce beats of frequency 6 Hz. When the tension in B is slightly decreased, the beat frequency increases to 7 Hz. If the frequency of A is 530 Hz, the original frequency of B is [NEET 2020]
  1. 524 Hz
  2. 536 Hz
  3. 537 Hz
  4. 523 Hz
Show answer
Answer: (A) 524 Hz
Explanation: Reducing the tension lowers the frequency of B. Since this increases the beat frequency, B must already lie below A, so f_B = 530 − 6 = 524 Hz.

Question 2

Previous-year question
A source of unknown frequency gives 4 beats/s when sounded with a source of known frequency 250 Hz. The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency 513 Hz. The unknown frequency is: [NEET, 2013]
  1. 260 Hz
  2. 254 Hz
  3. 246 Hz
  4. 240 Hz
Show answer
Answer: (B) 254 Hz
Explanation: First condition: |f − 250| = 4 ⇒ f = 246 or 254 Hz. Second condition: |2f − 513| = 5 ⇒ 2f = 508 or 518 ⇒ f = 254 or 259 Hz. The common value is f = 254 Hz.

Question 3

Previous-year question
Two sources of sound placed close to each other are emitting progressive waves given by: y₁ = 4 sin 600πt and y₂ = 5 sin 608πt. An observer located near these two sources of sound will hear: [AIPMT Pre., 2012]
  1. 8 beats per second with intensity ratio 81:1 between waxing and waning
  2. 4 beats per second with intensity ratio 81:1 between waxing and waning
  3. 4 beats per second with intensity ratio 25:16 between waxing and waning
  4. 8 beats per second with intensity ratio 25:16 between waxing and waning
Show answer
Answer: (B) 4 beats per second with intensity ratio 81:1 between waxing and waning
Explanation: f₁ = 600π/2π = 300 Hz and f₂ = 608π/2π = 304 Hz, so beat frequency = 4 per second. Intensity ratio between maxima (waxing) and minima (waning) = (a₁ + a₂)²/(a₁ − a₂)² = (4 + 5)²/(4 − 5)² = 81 : 1.

Question 4

Previous-year question
Two identical piano wires, kept under the same tension T, have a fundamental frequency of 600 Hz. The fractional increase in the tension of one of the wires which will lead to occurrence of 6 beats/s when both the wires oscillate together would be: [AIPMT Mains, 2011]
  1. 0.01
  2. 0.02
  3. 0.03
  4. 0.04
Show answer
Answer: (B) 0.02
Explanation: f ∝ √T, so Δf/f = (1/2)(ΔT/T). For 6 beats, Δf = 6 Hz: ΔT/T = 2 × Δf/f = 2 × 6/600 = 0.02.

Question 5

Previous-year question
A tuning fork of frequency 512 Hz makes 4 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was: [AIPMT Pre., 2010]
  1. 510 Hz
  2. 514 Hz
  3. 516 Hz
  4. 508 Hz
Show answer
Answer: (D) 508 Hz
Explanation: Initially |f − 512| = 4, so f = 508 or 516 Hz. Increasing the tension raises the string's frequency (f ∝ √T). Since the beats decreased, the string frequency must have moved towards 512 Hz — possible only if it was 508 Hz (rising towards 510 Hz gives 2 beats). If it were 516 Hz, the beats would have increased.

Question 6

Previous-year question
Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneously, the number of beats is: [AIPMT, 2009]
  1. 3
  2. 5
  3. 7
  4. 8
Show answer
Answer: (C) 7
Explanation: v = √(T/μ) = √(20/0.001) ≈ 141.4 m/s. f₁ = v/2l₁ = 141.4/(2 × 0.516) ≈ 137 Hz; f₂ = v/2l₂ = 141.4/(2 × 0.491) ≈ 144 Hz. Number of beats = f₂ − f₁ ≈ 7 per second.

Question 7

Two sound waves of equal amplitude and frequencies 400 Hz and 410 Hz arrive at a point in phase at t = 0. At t = 1/60 s, the resultant intensity at that point, as a fraction of the maximum intensity, is:
  1. √3/2
  2. 1/2
  3. 3/4
  4. 1/4
Show answer
Answer: (C) 3/4
Explanation: The phase difference grows at the rate 2π(f₂ − f₁), so at t = 1/60 s it is 2π × 10 × 1/60 = π/3. For equal amplitudes the resultant intensity is I = I_max cos²(Δφ/2). This gives I = I_max cos²(π/6) = 3I_max/4.
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Question 8

Two tuning forks of frequencies 340 Hz and 346 Hz are sounded together, and the two waves of equal amplitude reach a listener in phase at t = 0, so the sound is loudest at that instant. After how long is the sound at the listener first at its minimum?
  1. 1/12 s
  2. 1/6 s
  3. 1/24 s
  4. 1/3 s
Show answer
Answer: (A) 1/12 s
Explanation: The beat frequency is 346 − 340 = 6 Hz, so successive maxima are separated by 1/6 s. A minimum occurs midway between two maxima, when the phase difference has grown to π. The first minimum therefore comes after half a beat period, which is 1/12 s.

Question 9

Two sound sources of equal amplitude have frequencies 440 Hz and 446 Hz. At a fixed point their waves arrive in phase at t = 0. After how long does the resultant intensity at that point first fall to zero?
  1. 1/24 s
  2. 1/12 s
  3. 1/6 s
  4. 1/3 s
Show answer
Answer: (B) 1/12 s
Explanation: The phase difference between the two waves grows with time as 2π(f₂ − f₁)t = 2π(6)t. The intensity becomes zero when this phase difference first equals π. So 12πt = π, which gives t = 1/12 s.

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