Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

$0.02$

$0.03$

$0.04$

$0.01$

Step-by-Step Solution

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 $f \propto \sqrt{T}$ .
Determine the fractional change relation: By taking the natural logarithm on both sides and differentiating, we get the relationship for small changes: $\frac{\Delta f}{f} = \frac{1}{2} \frac{\Delta T}{T}$ .
Calculate the fractional change in frequency: The beat frequency is the difference between the new and old frequencies, so $\Delta f = 6 \text{ Hz}$ . The original frequency is $f = 600 \text{ Hz}$ . Therefore, the fractional change in frequency is $\frac{\Delta f}{f} = \frac{6}{600} = 0.01$ .
Calculate the fractional increase in tension: Rearranging the relation from step 2, the fractional change in tension is $\frac{\Delta T}{T} = 2 \times \frac{\Delta f}{f}$ . Substitute the value: $\frac{\Delta T}{T} = 2 \times 0.01 = 0.02$ . The fractional increase in tension is $0.02$ .

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