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(D) The frequency of the fundamental tone of a stretched wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $T$ is the tension,$l$ is the le

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$-3.2 \%$

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(D) The frequency of the fundamental tone of a stretched wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $T$ is the tension,$l$ is the length,and $m$ is the mass per unit length.Since $m = 	ext{Area} 	imes 	ext{density} = \frac{\pi d^2}{4} ho$,we have $f = \frac{1}{2l} \sqrt{\frac{T}{\pi d^2 ho / 4}} = \frac{1}{ld} \sqrt{\frac{T}{\pi ho}}$.Given that $T$,$l$,and $ho$ are constant,we have $f \propto \frac{1}{d}$.Therefore,$\frac{f_2}{f_1} = \frac{d_1}{d_2} = \frac{0.90}{0.93} \approx 0.9677$.The percentage change in frequency is given by $\frac{f_2 - f_1}{f_1} 	imes 100 = \left( \frac{f_2}{f_1} - 1 ight) 	imes 100$.Substituting the values: $\left( \frac{0.90}{0.93} - 1 ight) 	imes 100 = \left( \frac{0.90 - 0.93}{0.93} ight) 	imes 100 = \left( \frac{-0.03}{0.93} ight) 	imes 100 \approx -3.22 \%$.Rounding to the nearest option,the percentage change is $-3.2 \%$.

$A$ piano wire with a diameter of $0.90 \ mm$ is replaced by another wire of diameter $0.93 \ mm$ of the same material. If the tension of the wire is kept the same,then the percentage change in the frequency of the fundamental tone is

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