NEET Waves PYQs & Practice Questions

10 free NEET Physics questions on Waves, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Doppler Effect · Organ Pipes - Fundamental Frequency · Types of Waves · Speed of Sound · Wave Equation · Superposition and Interference · Standing Waves · Beats · Organ Pipes · Transverse and Longitudinal Waves · Speed of a Transverse Wave on a String · Wave Speed, Frequency and Wavelength

1. Q1NEET 2026Harmonics in Open and Closed Pipes - Nodes

   For sound waves, if the number of nodes for the 5th harmonic of an open-ended pipe is n and that for the 9th harmonic of the same pipe with one of its ends closed is m, the ratio n/m is:

   • A. 9/5
   • B. 1
   • C. 3/5
   • D. 5/9
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   Answer: A — 9/5

   For an open pipe, the kth harmonic has k nodes, so the 5th harmonic has n=5 nodes. For a closed pipe (one end closed), only odd harmonics (1st, 3rd, 5th, 7th, 9th...) occur, and the kth odd harmonic (2k-1) has k nodes; for the 9th harmonic, 2k-1=9 so k=5, giving m=5 nodes. So n/m = 5/5 = 1.

2. Q2NEET 2025Organ Pipes - Fundamental Frequency

   A pipe open at both ends has a fundamental frequency f in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to :

   • A. f/2
   • B. f
   • C. 3f/2
   • D. 2f
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   Answer: A — f/2

   Open pipe of length l: fopen = v/(2l) = f. When dipped to half length, the air column becomes a closed pipe (closed at water surface, open at top) of length l/2: fclosed = v/(4×(l/2)) = v/(2l) = f. So the frequency remains f.

3. Q3NEET 2023Fundamental Frequency - Open vs Closed Pipe

   The ratio of frequencies of fundamental harmonic produced by an open pipe to that of closed pipe having the same length is:

   • A. 2 : 1
   • B. 1 : 3
   • C. 3 : 1
   • D. 1 : 2
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   Answer: —

   For an open pipe, fundamental frequency f0 = v/2l. For a closed pipe, fundamental frequency fc = v/4l. The ratio f0:fc = (v/2l):(v/4l) = 2:1.

4. Q4NEET 2022Wave Speed on a String

   If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is:

   • A. 1:1
   • B. sqrt(2):1
   • C. 1:sqrt(2)
   • D. 1:2
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   Answer: B — sqrt(2):1

   Wave speed v = sqrt(T/mu). If T doubles, v_final = sqrt(2)*v_initial. So the ratio of initial to final speed is v_i:v_f = 1:sqrt(2).

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5. Q5NEET 2022Transverse Wave on String

   If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is

   • A. 1 : 1
   • B. √2 : 1
   • C. 1 : √2
   • D. 1 : 2
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   Answer: B — √2 : 1

   v ∝ √T. If T doubles, v_final = √2 × v_initial, so v_initial : v_final = 1 : √2.

6. Q6NEET 2020Beats and determination of original frequency

   In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6 Hz. When 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 will be:

   • A. 523 Hz
   • B. 524 Hz
   • C. 536 Hz
   • D. 537 Hz
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   Answer: B — 524 Hz

   Initially, beat frequency = 6Hz, so fB = 530±6 = 536 or 524 Hz. Decreasing tension in B decreases its frequency. If fB=536Hz, decreasing it moves it further from 530 (beat increases), consistent with the given info; if fB=524, decreasing it would move it further from 530 too... per official key, original frequency of B = 536 Hz, since decreasing tension in B decreases fB away from 530 making beats increase to 7Hz when fB was originally higher at 536Hz.

7. Q7NEET 2018Resonance tube experiment: finding velocity of sound

   A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27°C two successive resonances are produced at 20 cm and 73 cm of column length. If the frequency of the tuning fork is 320 Hz, the velocity of sound in air at 27°C is

   • A. 330 m/s
   • B. 300 m/s
   • C. 350 m/s
   • D. 339 m/s
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   Answer: C — 350 m/s

   The distance between two successive resonances equals half the wavelength: λ/2 = 73-20 = 53 cm, so λ = 106 cm = 1.06 m. Velocity v = fλ = 320×1.06 ≈ 339 m/s.

8. Q8NEET 2018Organ pipes: open pipe length from closed pipe harmonic matching

   The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is

   • A. 13.2 cm
   • B. 16 cm
   • C. 12.5 cm
   • D. 8 cm
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   Answer: —

   Third harmonic of closed pipe = 3v/(4×20). Fundamental of open pipe = v/(2L_open). Equating: v/(2L_open) = 3v/80, giving L_open = 80/(2×3) = 13.33 ≈ 13.2 cm.

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9. Q9NEET 2017Doppler effect

   Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5 m/s respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is [velocity of sound 340 m/s]:

   • A. 361 Hz
   • B. 411 Hz
   • C. 448 Hz
   • D. 350 Hz
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   Answer: B — 411 Hz

   Apparent frequency f' = f×(v+v_observer)/(v-v_source) = 400×(340+16.5)/(340-22) = 400×356.5/318 ≈ 448 Hz.

10. Q10NEET 2017Organ pipes (closed pipe harmonics)

    The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz. What is the fundamental frequency of the system?

    • A. 20 Hz
    • B. 30 Hz
    • C. 40 Hz
    • D. 10 Hz
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    Answer: —

    In a closed pipe, only odd harmonics exist: (2n−1)f₀. Difference between consecutive allowed harmonics = 2f₀ = 260−220 = 40 Hz, so f₀ = 20 Hz.