Vibrations of Strings — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Vibrations of Strings MCQs with step-by-step solutions (21 questions). Part of Superposition of Waves. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Vibrations of Strings · easy · theory
The fundamental frequency of a stretched string of length $L$, tension $T$ and linear density $\mu$ is:
A. $2L\sqrt{\dfrac{T}{\mu}}$
B. $\dfrac{1}{2L}\sqrt{\dfrac{\mu}{T}}$
C. $\dfrac{1}{L}\sqrt{\dfrac{T}{\mu}}$
D. $\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$ ✓ Correct
Solution: In the fundamental mode $\lambda = 2L$, and the wave speed on the string is $\sqrt{\dfrac{T}{\mu}}$, so $n = \dfrac{v}{2L}$.
Q2 — Vibrations of Strings · medium · theory
A string fixed at both ends can support:
A. All integral harmonics of the fundamental ✓ Correct
B. Only even harmonics
C. Only odd harmonics
D. Only the fundamental frequency
Solution: With a node forced at each end, the allowed wavelengths are $\dfrac{2L}{p}$ for every integer $p$, giving frequencies $n, 2n, 3n, \ldots$
Q3 — Vibrations of Strings · easy · theory
The law of length for a vibrating string states that, at constant tension and linear density, the fundamental frequency is:
A. Independent of the length
B. Proportional to the square of the length
C. Directly proportional to the vibrating length
D. Inversely proportional to the vibrating length ✓ Correct
Solution: From $n = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$, shortening the string raises the pitch — which is how a guitarist frets a note.
Q4 — Vibrations of Strings · easy · theory
The law of tension for a vibrating string states that the fundamental frequency is proportional to:
A. $T^2$
B. $T$
C. $\dfrac{1}{T}$
D. $\sqrt{T}$ ✓ Correct
Solution: Only the square root of the tension enters $n = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$, so the tension must be quadrupled to double the pitch.
Q5 — Vibrations of Strings · medium · theory
The fundamental frequency of a vibrating string varies with its linear density $\mu$ as:
A. $\mu$
B. $\sqrt{\mu}$
C. $\dfrac{1}{\mu}$
D. $\dfrac{1}{\sqrt{\mu}}$ ✓ Correct
Solution: A heavier string of the same length and tension vibrates more slowly, since $n \propto \dfrac{1}{\sqrt{\mu}}$.
Q6 — Vibrations of Strings · medium · theory
The first overtone of a string fixed at both ends corresponds to the:
A. Fourth harmonic
B. Fundamental
C. Third harmonic
D. Second harmonic ✓ Correct
Solution: Since all harmonics are allowed, the first tone above the fundamental is $2n$, the second harmonic.
Q7 — Vibrations of Strings · easy · theory
The instrument used to verify the laws of a vibrating string is the:
A. Barometer
B. Potentiometer
C. Sonometer ✓ Correct
D. Resonance tube
Solution: A sonometer lets the length, tension and material of a stretched wire be varied independently while the frequency is compared against a tuning fork.
Q8 — Vibrations of Strings · easy · theory
In the fundamental mode of a string fixed at both ends, the ends are:
A. Antinodes, with a node at the midpoint
B. Nodes, with an antinode at the midpoint ✓ Correct
C. Both antinodes
D. Points of maximum velocity
Solution: The fixed ends cannot move, so they must be nodes; the single loop between them has its antinode at the centre.
Q9 — Vibrations of Strings · medium · numerical
A stretched wire of length $1\text{ m}$ vibrates in its fundamental mode at $250\text{ Hz}$. The speed of transverse waves along it is:
A. $1000\text{ m/s}$
B. $250\text{ m/s}$
C. $500\text{ m/s}$ ✓ Correct
D. $125\text{ m/s}$
Solution: In the fundamental mode $\lambda = 2L = 2\text{ m}$, so $v = n\lambda = 250 \times 2 = 500\text{ m/s}$.
Q10 — Vibrations of Strings · hard · numerical
The tension in a stretched string is increased by $69\%$. The percentage increase in its fundamental frequency is:
A. $69\%$
B. $30\%$ ✓ Correct
C. $25\%$
D. $13\%$
Solution: $n \propto \sqrt{T}$, so $n' = \sqrt{1.69}\,n = 1.30n$ — a rise of $30\%$.
Q11 — Vibrations of Strings · easy · numerical
If the vibrating length of a string is halved at constant tension, its fundamental frequency:
A. Doubles ✓ Correct
B. Halves
C. Becomes four times
D. Remains unchanged
Solution: $n \propto \dfrac{1}{L}$, so halving the length doubles the frequency.
Q12 — Vibrations of Strings · easy · numerical
The tension in a string is made four times as large. Its fundamental frequency becomes:
A. Sixteen times as large
B. Twice as large ✓ Correct
C. Half as large
D. Four times as large
Solution: $n \propto \sqrt{T}$, so a four-fold tension doubles the frequency.
Q13 — Vibrations of Strings · medium · numerical
The linear density of a string is made four times as large, keeping length and tension the same. Its fundamental frequency becomes:
A. One-fourth as large
B. Half as large ✓ Correct
C. Twice as large
D. Four times as large
Solution: $n \propto \dfrac{1}{\sqrt{\mu}}$, so a four-fold linear density halves the frequency.
Q14 — Vibrations of Strings · hard · numerical
A string of length $0.5\text{ m}$ and linear density $0.01\text{ kg/m}$ is stretched with a tension of $100\text{ N}$. Its fundamental frequency is:
A. $200\text{ Hz}$
B. $10\text{ Hz}$
C. $50\text{ Hz}$
D. $100\text{ Hz}$ ✓ Correct
Solution: $\sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{100}{0.01}} = 100\text{ m/s}$, so $n = \dfrac{100}{2 \times 0.5} = 100\text{ Hz}$.
Q15 — Vibrations of Strings · easy · numerical
A string has a fundamental frequency of $100\text{ Hz}$. Its third harmonic has a frequency of:
A. $300\text{ Hz}$ ✓ Correct
B. $150\text{ Hz}$
C. $400\text{ Hz}$
D. $200\text{ Hz}$
Solution: For a string fixed at both ends the harmonics are integral multiples: the third harmonic is $3 \times 100 = 300\text{ Hz}$.
Q16 — Vibrations of Strings · medium · numerical
Transverse waves travel at $400\text{ m/s}$ along a string of length $2\text{ m}$ fixed at both ends. Its fundamental frequency is:
A. $200\text{ Hz}$
B. $100\text{ Hz}$ ✓ Correct
C. $800\text{ Hz}$
D. $50\text{ Hz}$
Solution: $n = \dfrac{v}{2L} = \dfrac{400}{4} = 100\text{ Hz}$.
Q17 — Vibrations of Strings · medium · numerical
A sonometer wire of vibrating length $0.6\text{ m}$ gives $200\text{ Hz}$. When the length is reduced to $0.4\text{ m}$ at the same tension, the frequency becomes:
A. $300\text{ Hz}$ ✓ Correct
B. $400\text{ Hz}$
C. $133\text{ Hz}$
D. $150\text{ Hz}$
Solution: $n \propto \dfrac{1}{L}$, so $n_2 = 200 \times \dfrac{0.6}{0.4} = 300\text{ Hz}$.
Q18 — Vibrations of Strings · medium · numerical
Two wires of the same material and tension have lengths in the ratio $1 : 2$. The ratio of their fundamental frequencies is:
A. $1 : 4$
B. $2 : 1$ ✓ Correct
C. $1 : 2$
D. $4 : 1$
Solution: $n \propto \dfrac{1}{L}$, so the shorter wire has the higher frequency: $n_1 : n_2 = 2 : 1$.
Q19 — Vibrations of Strings · hard · numerical
A string of length $1\text{ m}$ and linear density $0.04\text{ kg/m}$ carries a tension of $400\text{ N}$. Its fundamental frequency is:
A. $200\text{ Hz}$
B. $50\text{ Hz}$ ✓ Correct
C. $25\text{ Hz}$
D. $100\text{ Hz}$
Solution: $\sqrt{\dfrac{400}{0.04}} = \sqrt{10000} = 100\text{ m/s}$, so $n = \dfrac{100}{2 \times 1} = 50\text{ Hz}$.
Q20 — Vibrations of Strings · medium · numerical
By what factor must the tension in a string be increased to double its fundamental frequency?
A. $4$ ✓ Correct
B. $\sqrt{2}$
C. $2$
D. $8$
Solution: $n \propto \sqrt{T}$, so doubling $n$ requires $T$ to be multiplied by $2^2 = 4$.
Q21 — Vibrations of Strings · medium · numerical
A wire vibrates at $256\text{ Hz}$. If the tension is made four times as large, the new frequency is:
A. $64\text{ Hz}$
B. $512\text{ Hz}$ ✓ Correct
C. $1024\text{ Hz}$
D. $128\text{ Hz}$
Solution: $n \propto \sqrt{T}$, so $n' = 2 \times 256 = 512\text{ Hz}$.