A string vibrates in its fundamental mode whe | vedclass

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(D) The fundamental frequency of a vibrating string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$ is the tension,

Vedclass · Accepted Answer

$\frac{9}{4}$

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(D) The fundamental frequency of a vibrating string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.Initially, $f_1 = \frac{1}{2L_1} \sqrt{\frac{T_1}{\mu}}$.Given that the length is decreased by $25 \%$, the new length $L_2 = L_1 - 0.25L_1 = 0.75L_1 = \frac{3}{4}L_1$.The new frequency $f_2$ increases by $100 \%$, so $f_2 = f_1 + 1.00f_1 = 2f_1$.Using the frequency formula for the new state: $f_2 = \frac{1}{2L_2} \sqrt{\frac{T_2}{\mu}}$.Substituting $L_2$ and $f_2$: $2f_1 = \frac{1}{2(\frac{3}{4}L_1)} \sqrt{\frac{T_2}{\mu}} = \frac{4}{6L_1} \sqrt{\frac{T_2}{\mu}} = \frac{2}{3L_1} \sqrt{\frac{T_2}{\mu}}$.Dividing the expression for $f_2$ by $f_1$: $\frac{2f_1}{f_1} = \frac{\frac{2}{3L_1} \sqrt{\frac{T_2}{\mu}}}{\frac{1}{2L_1} \sqrt{\frac{T_1}{\mu}}}$.$2 = \frac{2}{3} \cdot 2 \cdot \sqrt{\frac{T_2}{T_1}} = \frac{4}{3} \sqrt{\frac{T_2}{T_1}}$.$\sqrt{\frac{T_2}{T_1}} = 2 \cdot \frac{3}{4} = \frac{3}{2}$.Squaring both sides, $\frac{T_2}{T_1} = (\frac{3}{2})^2 = \frac{9}{4}$.

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