Standing Waves in Strings and Organ Pipes, Fundamental Mode and Harmonics, Beats NEET Questions

Physics - Oscillations and Waves

Standing Waves in Strings and Organ Pipes, Fundamental Mode and Harmonics, Beats MCQ Questions


13.
The SECOND HARMONIC (first overtone) of a string fixed at both ends has:
A.
Two antinodes, three nodes (both ends + centre); wavelength λ₂ = L; frequency ν₂ = 2ν₁
B.
One antinode, two nodes; wavelength λ₂ = 2L; frequency ν₂ = ν₁
C.
Three antinodes, two nodes; wavelength λ₂ = 2L/3
D.
Two antinodes, two nodes; wavelength λ₂ = L/2
😑 View Answer Rough Work Error
ANSWER :
A. Two antinodes, three nodes (both ends + centre); wavelength λ₂ = L; frequency ν₂ = 2ν₁
14.
The nth harmonic of a string of length L fixed at both ends has n antinodes and (n + 1) nodes. For the 5th harmonic:
A.
4 antinodes, 5 nodes; wavelength λ₅ = 2L/5
B.
5 antinodes, 6 nodes; wavelength λ₅ = 2L/5; frequency ν₅ = 5v/(2L) = 5ν₁
C.
5 antinodes, 5 nodes; wavelength λ₅ = 2L/5
D.
5 antinodes, 6 nodes; wavelength λ₅ = 2L/4
😑 View Answer Rough Work Error
ANSWER :
B. 5 antinodes, 6 nodes; wavelength λ₅ = 2L/5; frequency ν₅ = 5v/(2L) = 5ν₁
15.
A string of length L fixed at both ends vibrates in its 3rd harmonic. The number of nodes, antinodes, and the ratio ν₃/ν₁ are:
A.
2 nodes, 3 antinodes; ν₃/ν₁ = 2
B.
3 nodes, 4 antinodes; ν₃/ν₁ = 3
C.
4 nodes (including both ends), 3 antinodes; ν₃/ν₁ = 3
D.
4 nodes, 3 antinodes; ν₃/ν₁ = 9
😑 View Answer Rough Work Error
ANSWER :
C. 4 nodes (including both ends), 3 antinodes; ν₃/ν₁ = 3
16.
If the fundamental frequency of a string is 200 Hz, the frequencies of its harmonics are:
A.
200, 600, 1000, 1400,… Hz — only odd multiples
B.
200, 400, 600, 800, 1000,… Hz — all integer multiples of 200 Hz
C.
Only 200 Hz — no higher harmonics possible
D.
200, 800, 1800,… Hz — n² × fundamental
😑 View Answer Rough Work Error
ANSWER :
B. 200, 400, 600, 800, 1000,… Hz — all integer multiples of 200 Hz
17.
The fundamental frequency of a string fixed at both ends is ν₁ = v/(2L). If the tension is increased to 4 times its original value (length unchanged), the new fundamental frequency is:
A.
ν₁ (unchanged — frequency doesn't depend on tension)
B.
ν₁/2
C.
2ν₁ — doubling, since v = √(T/μ) → v' = 2v, so ν₁' = v'/(2L) = 2ν₁
D.
4ν₁
😑 View Answer Rough Work Error
ANSWER :
C. 2ν₁ — doubling, since v = √(T/μ) → v' = 2v, so ν₁' = v'/(2L) = 2ν₁
18.
A string of linear mass density μ = 0.01 kg/m and length L = 0.5 m is under tension T = 25 N. Its fundamental frequency is:
A.
100 Hz
B.
ν₁ = v/(2L) = √(T/μ)/(2L) = √(25/0.01)/(2×0.5) = 50/1 = 50 Hz
C.
25 Hz
D.
200 Hz
😑 View Answer Rough Work Error
ANSWER :
B. ν₁ = v/(2L) = √(T/μ)/(2L) = √(25/0.01)/(2×0.5) = 50/1 = 50 Hz