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(B) Let the initial frequency of both wires be $n = 600 	ext{ Hz}$.The frequency of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}

Vedclass · Accepted Answer

$0.02$

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(B) Let the initial frequency of both wires be $n = 600 	ext{ Hz}$.The frequency of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.When the tension in one wire is increased to $T'$,its new frequency becomes $n' = \frac{1}{2l} \sqrt{\frac{T'}{m}}$.Given that the beat frequency is $6 	ext{ Hz}$,we have $n' - n = 6$.So,$n' = 600 + 6 = 606 	ext{ Hz}$.Taking the ratio of the two frequencies:$\frac{n'}{n} = \frac{\frac{1}{2l} \sqrt{\frac{T'}{m}}}{\frac{1}{2l} \sqrt{\frac{T}{m}}} = \sqrt{\frac{T'}{T}}$$\frac{606}{600} = \sqrt{\frac{T'}{T}}$$1.01 = \sqrt{\frac{T'}{T}}$Squaring both sides:$\frac{T'}{T} = (1.01)^2 = 1.0201 \approx 1.02$.The fractional increase in tension is $\frac{\Delta T}{T} = \frac{T' - T}{T} = \frac{T'}{T} - 1$.$\frac{\Delta T}{T} = 1.02 - 1 = 0.02$.

Two identical piano wires have a fundamental frequency of $600 \text{ Hz}$ 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?

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