Q.1 The fundamental frequency of a stretched string depends on which of the following?
Length, mass per unit length, and tension
Length only
Mass only
Tension only
Explanation - The fundamental frequency of a stretched string is given by f = (1/2L)√(T/μ), where L is the length, T is the tension, and μ is the linear mass density.
Correct answer is: Length, mass per unit length, and tension
Q.2 If the length of a vibrating string is doubled, the fundamental frequency becomes?
Doubled
Halved
Unchanged
Four times
Explanation - Since f = 1/(2L)√(T/μ), doubling L reduces f by half.
Correct answer is: Halved
Q.3 In a closed organ pipe, the fundamental frequency is produced by:
Antinode at closed end, node at open end
Node at closed end, antinode at open end
Antinode at both ends
Node at both ends
Explanation - In closed organ pipes, a node forms at the closed end and an antinode at the open end.
Correct answer is: Node at closed end, antinode at open end
Q.4 A string of length L fixed at both ends has fundamental frequency f. If tension is quadrupled, the new frequency is:
f
2f
4f
f/2
Explanation - Frequency is proportional to √T. Quadrupling tension makes frequency double.
Correct answer is: 2f
Q.5 In an open pipe, the distance between successive nodes is:
λ
λ/2
λ/4
2λ
Explanation - Successive nodes or antinodes in an open pipe are separated by half a wavelength.
Correct answer is: λ/2
Q.6 Which harmonic is missing in a closed organ pipe?
Odd harmonics
Even harmonics
All harmonics
No harmonics
Explanation - Closed organ pipes support only odd harmonics due to boundary conditions.
Correct answer is: Even harmonics
Q.7 If the fundamental frequency of a closed pipe is 100 Hz, its third harmonic frequency is:
200 Hz
300 Hz
400 Hz
500 Hz
Explanation - Closed pipes support odd harmonics: f, 3f, 5f... So, 3 × 100 = 300 Hz.
Correct answer is: 300 Hz
Q.8 A tuning fork produces resonance with a pipe closed at one end when the pipe length is 0.25 m. If the speed of sound is 340 m/s, the frequency of the tuning fork is:
340 Hz
680 Hz
850 Hz
1360 Hz
Explanation - For closed pipe, L = λ/4 → λ = 4L = 1 m → f = v/λ = 340/1 = 340 Hz.
Correct answer is: 340 Hz
Q.9 If a string vibrates in its second harmonic, the number of nodes excluding ends is:
0
1
2
3
Explanation - In second harmonic, one additional node forms in between the ends.
Correct answer is: 1
Q.10 In an open organ pipe, the fundamental frequency corresponds to:
L = λ/4
L = λ/2
L = λ
L = 2λ
Explanation - Open pipe has antinode at both ends, so fundamental wavelength is λ = 2L.
Correct answer is: L = λ/2
Q.11 If frequency of vibration doubles, the wavelength of sound in a pipe becomes:
Double
Half
Same
One-fourth
Explanation - Since v = fλ, doubling f makes λ half for constant v.
Correct answer is: Half
Q.12 Which of the following is correct for resonance tube experiment?
Resonance occurs when air column length matches an integer multiple of wavelength
Resonance occurs when air column length matches a fraction of wavelength
Resonance does not depend on wavelength
Resonance occurs only at half wavelength
Explanation - Air column resonates when its length fits quarter wavelength conditions, i.e., odd multiples of λ/4.
Correct answer is: Resonance occurs when air column length matches an integer multiple of wavelength
Q.13 If a closed pipe produces its first overtone, it corresponds to which harmonic?
Second harmonic
Third harmonic
Fourth harmonic
Fifth harmonic
Explanation - Closed pipes have only odd harmonics, so first overtone is the 3rd harmonic.
Correct answer is: Third harmonic
Q.14 The speed of sound in air is 340 m/s. If an open pipe of length 0.5 m resonates in fundamental mode, frequency is:
170 Hz
340 Hz
680 Hz
850 Hz
Explanation - For open pipe, L = λ/2 → λ = 2L = 1 m → f = v/λ = 340 Hz.
Correct answer is: 340 Hz
Q.15 Two organ pipes, one open and one closed, have same fundamental frequency. Their lengths ratio is:
1:1
1:2
2:1
4:1
Explanation - For open: L = λ/2, for closed: L = λ/4. To have same f, closed pipe must be half length of open pipe → ratio = 2:1.
Correct answer is: 2:1
Q.16 In a vibrating string, successive harmonics differ in frequency by:
Constant value
Increasing value
Decreasing value
Variable value
Explanation - Frequencies are f, 2f, 3f, etc. Successive difference = f, a constant.
Correct answer is: Constant value
Q.17 When both ends of a pipe are closed, the fundamental mode corresponds to:
L = λ/2
L = λ/4
L = λ
L = 2λ
Explanation - Closed at both ends behaves like string fixed at both ends. Fundamental wavelength = 2L.
Correct answer is: L = λ/2
Q.18 If string length is reduced to half keeping tension same, fundamental frequency becomes:
f/2
f
2f
4f
Explanation - f ∝ 1/L. Halving L doubles frequency.
Correct answer is: 2f
Q.19 The first overtone of an open pipe corresponds to:
2nd harmonic
3rd harmonic
4th harmonic
5th harmonic
Explanation - In open pipe, harmonics are f, 2f, 3f... The first overtone = 2nd harmonic.
Correct answer is: 2nd harmonic
Q.20 In a closed pipe, the fundamental frequency is 200 Hz. The frequency of second overtone is:
400 Hz
600 Hz
800 Hz
1000 Hz
Explanation - Closed pipe supports odd harmonics: f, 3f, 5f... 2nd overtone = 5f = 1000 Hz. Oops → Correction: Fundamental 200 Hz → 3rd harmonic = 600 Hz = 1st overtone. So 2nd overtone = 1000 Hz. Correct answer = 1000 Hz.
Correct answer is: 600 Hz
Q.21 Beats are heard when two organ pipes produce frequencies of 256 Hz and 260 Hz. Beat frequency is:
2 Hz
3 Hz
4 Hz
5 Hz
Explanation - Beat frequency = |f1 - f2| = 260 - 256 = 4 Hz.
Correct answer is: 4 Hz
Q.22 In a stretched string, tension is increased by 9 times. The frequency becomes:
3 times
9 times
√9 times
Unchanged
Explanation - f ∝ √T. Increasing T by 9 makes f × 3.
Correct answer is: 3 times
Q.23 The vibration of a string fixed at both ends is an example of:
Longitudinal waves
Transverse stationary waves
Progressive waves
Electromagnetic waves
Explanation - The string displacement is transverse and forms stationary wave patterns.
Correct answer is: Transverse stationary waves
Q.24 The third harmonic frequency of an open organ pipe of length L is given by:
v/2L
3v/2L
3v/4L
v/4L
Explanation - For open pipe: nth harmonic = nv/2L. For n=3 → 3v/2L.
Correct answer is: 3v/2L
Q.25 A resonance column experiment is performed with tuning fork of 512 Hz. If speed of sound is 320 m/s, the first resonance length is:
0.156 m
0.312 m
0.625 m
1.25 m
Explanation - λ = v/f = 320/512 = 0.625 m. First resonance = λ/4 = 0.156 m.
Correct answer is: 0.156 m