Two identical piano wires kept under the same tension $T$ have a fundamental frequency of $600 ext{ Hz}$. Th

Question

Two identical piano wires kept under the same tension $T$ have a fundamental frequency of $600 	ext{ Hz}$. The fractional increase in the tension of one of the wires which will lead to the occurrence of $6 	ext{ beats/s}$ when both the wires oscillate together would be:

Accepted Answer

1. **Identify the relation between frequency and tension:** The fundamental frequency $f$ of a stretched wire is given by $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$, which means $f \propto \sqrt{T}$ .
2. **Determine the fractional change relation:** By taking the natural logarithm on both sides and differentiating, we get the relationship for small changes: $\frac{\Delta f}{f} = \frac{1}{2} \frac{\Delta T}{T}$.
3. **Calculate the fractional change in frequency:** The beat frequency is the difference between the new and old frequencies, so $\Delta f = 6 	ext{ Hz}$ . The original frequency is $f = 600 	ext{ Hz}$. 
   Therefore, the fractional change in frequency is $\frac{\Delta f}{f} = \frac{6}{600} = 0.01$.
4. **Calculate the fractional increase in tension:** Rearranging the relation from step 2, the fractional change in tension is $\frac{\Delta T}{T} = 2 	imes \frac{\Delta f}{f}$.
   Substitute the value: $\frac{\Delta T}{T} = 2 	imes 0.01 = 0.02$.
The fractional increase in tension is $0.02$.

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