Question:
The fundamental frequency of sonometer wire is \(50\) Hz for some length and tension. If the length is increased by \(25\%\) by keeping the tension same, then the frequency change of second harmonic is
The fundamental frequency of sonometer wire is \(50\) Hz for some length and tension. If the length is increased by \(25\%\) by keeping the tension same, then the frequency change of second harmonic is
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Frequency is inversely proportional to length.
Updated On: Oct 1, 2026
- increase by \(20\%\)
- decrease by \(20\%\)
- increase by \(40\%\)
- decrease by \(40\%\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
For a wire fixed at both ends, the \(n\)th harmonic is \(f_n=\dfrac{n}{2l}\sqrt{\dfrac T\mu}\). At a fixed tension, \(f\propto\dfrac1l\).
Step 2: Second harmonic at the start:
\(f_2=2\times50=100\) Hz.
Step 3: After the change:
The length becomes \(1.25l\), so the frequency becomes \(\dfrac{100}{1.25}=80\) Hz.
Step 4: Percent change:
\[ \frac{80-100}{100}\times100=-20\% \]
The frequency decreases by \(20\%\). Option (B).
Final Answer:
The second harmonic falls from 100 Hz to 80 Hz, a 20 percent decrease. \[ \boxed{\text{Decrease by }20\%} \]
For a wire fixed at both ends, the \(n\)th harmonic is \(f_n=\dfrac{n}{2l}\sqrt{\dfrac T\mu}\). At a fixed tension, \(f\propto\dfrac1l\).
Step 2: Second harmonic at the start:
\(f_2=2\times50=100\) Hz.
Step 3: After the change:
The length becomes \(1.25l\), so the frequency becomes \(\dfrac{100}{1.25}=80\) Hz.
Step 4: Percent change:
\[ \frac{80-100}{100}\times100=-20\% \]
The frequency decreases by \(20\%\). Option (B).
Final Answer:
The second harmonic falls from 100 Hz to 80 Hz, a 20 percent decrease. \[ \boxed{\text{Decrease by }20\%} \]
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