Question

Two identical piano wires have a fundamental frequency of \(600\) cycles per second when kept under the same tension. What fractional increase in the tension of one wire will lead to the occurrence of \(6\) beats per second when both wires vibrate simultaneously?

• A. \(0.01\)
• B. \(0.02\)
• C. \(0.03\)
• D. \(0.04\)

Hint:

For string vibrations: \(f\propto\sqrt{T}\Rightarrow \frac{\Delta f}{f}=\frac{1}{2}\frac{\Delta T}{T}\).

Answer 1

The Correct Option is B

Step 1: Use frequency relation with tension.
For a stretched string:
\[ f \propto \sqrt{T} \]
Step 2: Beat frequency condition.
One wire remains at \(600\,Hz\).
Other wire is adjusted so that beat frequency is \(6\,Hz\):
\[ |f_2 - f_1| = 6 \Rightarrow f_2 = 606\,Hz \]
Step 3: Use fractional change approximation.
\[ \frac{\Delta f}{f} = \frac{1}{2}\frac{\Delta T}{T} \]
Here:
\[ \Delta f = 6,\quad f = 600 \]
So:
\[ \frac{6}{600} = \frac{1}{2}\frac{\Delta T}{T} \]
\[ 0.01 = \frac{1}{2}\frac{\Delta T}{T} \Rightarrow \frac{\Delta T}{T} = 0.02 \]
Final Answer: \[ \boxed{0.02} \]