Question:
A sonometer wire 49 cm long is in unison with a tuning fork of frequency '$n$'. If the length of the wire is decreased by 1 cm and it is vibrated with the same tuning fork, 6 beats are heard per second. The value of '$n$' is
A sonometer wire 49 cm long is in unison with a tuning fork of frequency '$n$'. If the length of the wire is decreased by 1 cm and it is vibrated with the same tuning fork, 6 beats are heard per second. The value of '$n$' is
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Use the conservation of the product $nL = \text{constant}$ for sonometer problems involving frequency shifts.
Updated On: Apr 28, 2026
- 256 Hz
- 288 Hz
- 320 Hz
- 384 Hz
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The Correct Option is A
Solution and Explanation
Step 1: The frequency of a sonometer wire is inversely proportional to its length $L$ for constant tension and linear density: $n \propto \frac{1}{L}$. This implies $n_1 L_1 = n_2 L_2$.
Step 2: Initially, $n_1 = n$ and $L_1 = 49$ cm. When the length is decreased by 1 cm, $L_2 = 48$ cm. The new frequency $n_2$ is \[ n_2 = \frac{n_1 L_1}{L_2} = \frac{n \times 49}{48} \]
Step 3: The beat frequency is given as $|n_2 - n_1| = 6$ beats per second. Since $L_2<L_1$, then $n_2>n_1$, so \[ \frac{49}{48}n - n = 6 \]
Step 4: Simplify and solve for $n$: \[ \frac{n}{48} = 6 \Rightarrow n = 6 \times 48 = 288 \text{ Hz} \]
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