Question:
There are two wires vibrating with same frequency of \(500\) Hz. Tension in one wire is increased by 'x' % so that \(5\) beats per second are heard. The value of 'x' in % is
There are two wires vibrating with same frequency of \(500\) Hz. Tension in one wire is increased by 'x' % so that \(5\) beats per second are heard. The value of 'x' in % is
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Frequency varies as the square root of tension, so a small change in tension gives half that change in frequency.
Updated On: Nov 12, 2026
- \(5\)
- \(4\)
- \(3\)
- \(2\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Frequency of a wire is \(n\propto\sqrt{T}\), so \(\dfrac{\Delta n}{n}=\dfrac{1}{2}\dfrac{\Delta T}{T}\).
Step 2: Beats
Beat frequency equals \(\Delta n=5\) Hz, so \(\dfrac{\Delta n}{n}=\dfrac{5}{500}=1\%\).
Step 3: Calculation
\(\dfrac{\Delta T}{T}=2\times1\%=2\%\).
Final Answer:
The tension must be increased by 2 percent. \[ \boxed{\text{(D) }2} \]
Frequency of a wire is \(n\propto\sqrt{T}\), so \(\dfrac{\Delta n}{n}=\dfrac{1}{2}\dfrac{\Delta T}{T}\).
Step 2: Beats
Beat frequency equals \(\Delta n=5\) Hz, so \(\dfrac{\Delta n}{n}=\dfrac{5}{500}=1\%\).
Step 3: Calculation
\(\dfrac{\Delta T}{T}=2\times1\%=2\%\).
Final Answer:
The tension must be increased by 2 percent. \[ \boxed{\text{(D) }2} \]
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