Waves – Principle of Superposition of Waves and Organ Pipe: NEET MCQs with Answers

15 free questions on Waves – Principle of Superposition of Waves and Organ Pipe 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: Principle of Superposition of Waves and Organ Pipe: Principle of superposition of waves, Standing waves in strings and organ pipes, fundamental mode and harmonics.

Practise all 276 Waves – Principle of Superposition of Waves and Organ Pipe questions free Take a free timed test

Question 1

Previous-year question
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 [NEET 2018]
  1. 13.2 cm
  2. 8 cm
  3. 16 cm
  4. 12.5 cm
Show answer
Answer: (A) 13.2 cm
Explanation: The third harmonic of the closed pipe is 3v/4L_c and the fundamental of the open pipe is v/2L₀. Equating them gives L₀ = (2/3)L_c = (2/3) × 20 = 13.3 cm, which is the listed value of about 13.2 cm.

Question 2

Previous-year question
A tuning fork is used to produce resonance in a glass tube. The length of the air column in the tube can be adjusted by a variable piston. At room temperature 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 [NEET 2018]
  1. 330 m s⁻¹
  2. 339 m s⁻¹
  3. 300 m s⁻¹
  4. 350 m s⁻¹
Show answer
Answer: (B) 339 m s⁻¹
Explanation: Successive resonances differ by λ/2, so λ = 2(73 − 20) = 106 cm. Then v = fλ = 320 × 1.06 ≈ 339 m s⁻¹. Using the difference removes the end correction.

Question 3

Previous-year question
The two nearest harmonics of a tube closed at one end and open at the other end are 220 Hz and 260 Hz. What is the fundamental frequency of the system? [NEET 2017]
  1. 20 Hz
  2. 30 Hz
  3. 40 Hz
  4. 10 Hz
Show answer
Answer: (A) 20 Hz
Explanation: A closed pipe supports only odd harmonics, so consecutive ones differ by 2f₁. Hence 2f₁ = 260 − 220 = 40 Hz and f₁ = 20 Hz.

Question 4

Previous-year question
An air column, closed at one end and open at the other, resonates with a tuning fork when the smallest length of the column is 50 cm. The next larger length of the column resonating with the same tuning fork is [NEET 2016]
  1. 66.7 cm
  2. 100 cm
  3. 150 cm
  4. 200 cm
Show answer
Answer: (C) 150 cm
Explanation: A closed pipe resonates at L = λ/4, 3λ/4, 5λ/4 … With λ/4 = 50 cm, the next length is 3 × 50 = 150 cm.

Question 5

Previous-year question
A string is stretched between two fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is [AIPMT 2015]
  1. 205 Hz
  2. 10.5 Hz
  3. 105 Hz
  4. 155 Hz
Show answer
Answer: (C) 105 Hz
Explanation: Because there is no resonance in between, the two are consecutive harmonics and their difference is the fundamental: 420 − 315 = 105 Hz.

Question 6

Previous-year question
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? [NEET, 2017]
  1. 20 Hz
  2. 30 Hz
  3. 40 Hz
  4. 10 Hz
Show answer
Answer: (A) 20 Hz
Explanation: A closed tube supports only odd harmonics: f₀, 3f₀, 5f₀, … Consecutive harmonics therefore differ by 2f₀. So 2f₀ = 260 − 220 = 40 Hz ⇒ f₀ = 20 Hz (220 = 11f₀ and 260 = 13f₀).

Question 7

Previous-year question
An air column, closed at one end and open at the other, resonates with a tuning fork when the smallest length of the column is 50 cm. The next larger length of the column resonating with the same tuning fork is: [AIPMT, 2016]
  1. 66.7 cm
  2. 100 cm
  3. 150 cm
  4. 200 cm
Show answer
Answer: (C) 150 cm
Explanation: A closed air column resonates at odd multiples of λ/4: L = λ/4, 3λ/4, 5λ/4, … The smallest length 50 cm corresponds to λ/4, so the next resonating length is 3λ/4 = 3 × 50 = 150 cm.
Doing well? Create a free account to take timed tests on this topic and see your weak areas.

Question 8

Previous-year question
A string is stretched between two fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is: [Re-AIPMT, 2015]
  1. 105 Hz
  2. 155 Hz
  3. 205 Hz
  4. 10.5 Hz
Show answer
Answer: (A) 105 Hz
Explanation: Resonant frequencies of a string are integral multiples of the fundamental: fₙ = nf₀. Since 420 and 315 Hz are consecutive resonances, their difference equals the fundamental: f₀ = 420 − 315 = 105 Hz (420 = 4 × 105, 315 = 3 × 105).

Question 9

Previous-year question
The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is: [AIPMT, 2015]
  1. 120 cm
  2. 140 cm
  3. 80 cm
  4. 100 cm
Show answer
Answer: (A) 120 cm
Explanation: Fundamental of closed pipe = v/4L₁ = v/(4 × 20) = v/80. Second overtone (third harmonic) of open pipe = 3v/2L₂. Equating: v/80 = 3v/2L₂ ⇒ L₂ = 3 × 80/2 = 120 cm.

Question 10

Previous-year question
The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are: (velocity of sound = 340 m/s) [AIPMT, 2014]
  1. 4
  2. 5
  3. 7
  4. 6
Show answer
Answer: (D) 6
Explanation: For a closed pipe, fₘ = (2m + 1)v/4L = (2m + 1) × 340/(4 × 0.85) = (2m + 1) × 100 Hz. Allowed frequencies below 1250 Hz: 100, 300, 500, 700, 900 and 1100 Hz — a total of 6.

Question 11

Previous-year question
If n₁, n₂ and n₃ are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by: [AIPMT, 2014]
  1. 1/n = 1/n₁ + 1/n₂ + 1/n₃
  2. 1/√n = 1/√n₁ + 1/√n₂ + 1/√n₃
  3. √n = √n₁ + √n₂ + √n₃
  4. n = n₁ + n₂ + n₃
Show answer
Answer: (A) 1/n = 1/n₁ + 1/n₂ + 1/n₃
Explanation: Fundamental frequency of a stretched string, n = v/2l ⇒ l ∝ 1/n (same v for all segments of the same string). Adding the lengths: l = l₁ + l₂ + l₃ ⇒ v/2n = v/2n₁ + v/2n₂ + v/2n₃ ⇒ 1/n = 1/n₁ + 1/n₂ + 1/n₃.

Question 12

Previous-year question
If we study the vibration of a pipe open at both ends, then the following statement is not true: [NEET, 2013]
  1. Pressure change will be maximum at both ends
  2. Open end will be antinode
  3. Odd harmonics of the fundamental frequency will be generated
  4. All harmonics of the fundamental frequency will be generated
Show answer
Answer: (A) Pressure change will be maximum at both ends
Explanation: An open end is a displacement antinode and hence a pressure node, so the pressure change there is minimum, not maximum. The other statements hold: open ends are antinodes, and an open pipe produces all harmonics, including the odd ones.

Question 13

Previous-year question
When a string is divided into three segments of length l₁, l₂ and l₃, the fundamental frequencies of these three segments are ν₁, ν₂ and ν₃ respectively. The original fundamental frequency (ν) of the string is: [AIPMT Pre., 2012]
  1. 1/ν = 1/ν₁ + 1/ν₂ + 1/ν₃
  2. 1/√ν = 1/√ν₁ + 1/√ν₂ + 1/√ν₃
  3. √ν = √ν₁ + √ν₂ + √ν₃
  4. ν = ν₁ + ν₂ + ν₃
Show answer
Answer: (A) 1/ν = 1/ν₁ + 1/ν₂ + 1/ν₃
Explanation: For the same string (same tension and mass per unit length), the fundamental frequency ν = v/2l, so l ∝ 1/ν. Since the total length l = l₁ + l₂ + l₃, substituting l = v/2ν etc. gives 1/ν = 1/ν₁ + 1/ν₂ + 1/ν₃.

Question 14

A stretched string vibrates in a stationary wave with several loops. Which statement about the phases of its particles is correct?
  1. All particles lying between two consecutive nodes vibrate in the same phase.
  2. All particles of the string vibrate in the same phase.
  3. Any two adjacent antinodes vibrate with a phase difference of π/2.
  4. Any two adjacent antinodes vibrate in the same phase.
Show answer
Answer: (A) All particles lying between two consecutive nodes vibrate in the same phase.
Explanation: Within one loop every particle reaches its extreme position and passes through its mean position at the same moment, differing only in amplitude. The phase reverses on crossing a node, so adjacent loops, and hence adjacent antinodes, are in opposite phase. Only alternate antinodes swing together.

Question 15

An organ pipe has a fundamental frequency of 150 Hz. It is not stated whether the pipe is open at both ends or closed at one end. Which of these frequencies is certain to be an overtone of the pipe, whichever type it is?
  1. 450 Hz
  2. 600 Hz
  3. 900 Hz
  4. 300 Hz
Show answer
Answer: (A) 450 Hz
Explanation: An open pipe gives all integer multiples of 150 Hz, while a closed pipe gives only the odd multiples: 450, 750, 1050 Hz and so on. A frequency common to both must be an odd multiple of 150 Hz above the fundamental. Only 450 Hz, the third multiple, qualifies. The values 300, 600 and 900 Hz are even multiples and occur only if the pipe is open.

Want more Waves – Principle of Superposition of Waves and Organ Pipe questions?

E-GYAN NIDHI has 276 questions on this topic and 63,000+ in total, all free: timed CBT tests, NEET marking and instant analysis.

Register free Free Sunday Mock Test

More topics in OSCILLATIONS AND WAVES

↑