Question:medium

A stretched sonometer wire is in unison with a tuning fork. When length of wire is increased by 1%, the number of beats heard per second is 5. Then the frequency of the fork is :

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When beats are formed after change, consider both cases: fork frequency slightly higher or lower.
Updated On: Apr 18, 2026
  • 500 Hz
  • 505 Hz
  • 255 Hz
  • 250 Hz
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
The frequency of a sonometer wire is inversely proportional to its length (\( f \propto 1/l \)). Beats are produced when there is a small difference between two frequencies.
: Key Formula or Approach:
1. \( f \cdot l = \text{constant} \).
2. Number of beats \( = |f_1 - f_2| \).
Step 2: Detailed Explanation:
Let the initial frequency of the wire be \( f_1 \), which is equal to the fork frequency because they are in unison.
Let the initial length be \( L \). So, \( f_1 \propto 1/L \).
The new length is \( L_2 = L + 0.01L = 1.01L \).
The new frequency \( f_2 \) will be:
\[ f_2 = \frac{f_1 \cdot L}{1.01L} = \frac{f_1}{1.01} \]
Since the length increased, the frequency decreased. The number of beats is:
\[ f_1 - f_2 = 5 \]
\[ f_1 - \frac{f_1}{1.01} = 5 \]
\[ f_1 \left( 1 - \frac{1}{1.01} \right) = 5 \]
\[ f_1 \left( \frac{0.01}{1.01} \right) = 5 \]
\[ f_1 = \frac{5 \times 1.01}{0.01} = 5 \times 101 = 505 \text{ Hz} \]
Step 3: Final Answer:
The frequency of the tuning fork is 505 Hz.
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