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

Vedclass · Accepted Answer

$4: 9$

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(D) The fundamental frequency $f$ of a stretched string is given by the formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.Given $f_1 = 300 \ Hz$, $L_1 = L$, and $T_1 = T_1$. Thus, $300 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}}$.Given $f_2 = 100 \ Hz$, $L_2 = 2L$, and $T_2 = T_2$. Thus, $100 = \frac{1}{2(2L)} \sqrt{\frac{T_2}{\mu}} = \frac{1}{4L} \sqrt{\frac{T_2}{\mu}}$.Dividing the two equations: $\frac{300}{100} = \frac{\frac{1}{2L} \sqrt{\frac{T_1}{\mu}}}{\frac{1}{4L} \sqrt{\frac{T_2}{\mu}}}$.$3 = \frac{4L}{2L} \sqrt{\frac{T_1}{T_2}} = 2 \sqrt{\frac{T_1}{T_2}}$.$1.5 = \sqrt{\frac{T_1}{T_2}}$.Squaring both sides: $2.25 = \frac{T_1}{T_2}$, which is $\frac{9}{4} = \frac{T_1}{T_2}$.Therefore, $\frac{T_2}{T_1} = \frac{4}{9}$.

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