Question:

A tuning fork gives \(5\) beats per second with a \(33\) cm length of sonometer wire. If the length of the wire is shortened by \(1\) cm, the number of beats is still the same. The frequency of the fork is

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Wire frequency is inversely proportional to length; equal beats on both sides put the fork frequency between the two wire frequencies.
Updated On: Oct 1, 2026
  • \(320\) Hz
  • \(325\) Hz
  • \(330\) Hz
  • \(340\) Hz
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The Correct Option is B

Solution and Explanation

Step 1: Set Up:
For a sonometer wire, \(f\propto\dfrac1L\). Shortening the wire from 33 cm to 32 cm raises its frequency. Let the fork frequency be \(n\).

Step 2: Use the Beat Condition:
With 33 cm the wire frequency is \(f_1\) and with 32 cm it is \(f_2>f_1\). Both give 5 beats, so the fork lies between them: \(n-f_1=5\) and \(f_2-n=5\). So \(f_1=n-5\) and \(f_2=n+5\).

Step 3: Ratio of Frequencies:
\[ \frac{f_2}{f_1}=\frac{33}{32}\Rightarrow\frac{n+5}{n-5}=\frac{33}{32} \]
\[ 32n+160=33n-165\Rightarrow n=325\ \text{Hz} \]

Step 4: Check:
\(f_1=320\) Hz and \(f_2=330\) Hz, and \(330/320=33/32\). Both give 5 beats with the 325 Hz fork. So (B) is correct.

Final Answer:
The frequency of the fork is 325 Hz, option (B). \[ \boxed{\text{(B) } 325\ \text{Hz}} \]
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