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  • A wire of length L and mass per unit length kg/m is put under tension of 440 N

A wire of length L and mass per unit length4\times 10 ^{-3} kg/m is put under tension of 440 N . Two consecutive frequencies that it resonates at are 320 Hz and 380Hz. Then L in metres.

Option: 1

2.75 m


Option: 2

1.75m


Option: 3

3.15m


Option: 4

3.75m


Answers (1)

best_answer

\begin{aligned} & f_n=\frac{n v}{2 l} \\ & f_n=\frac{n}{2 l} \sqrt{\frac{T}{\mu }} \end{aligned}

\begin{aligned} & f_n \propto n \\ & \frac{f_n}{f_{n+1}}=\frac{n}{n+1} \\ & \frac{320}{380}=\frac{n}{n+1} \end{aligned}

\begin{aligned} & \frac{16}{19}=\frac{n}{n+1} \\ & 16 n+16=19 n \\ & 3 n=16 \\ & n=\frac{16}{3} \end{aligned}

\begin{aligned} & 320=\frac{16}{3 \times 2 l} \sqrt{\frac{440}{4 \times 10^{-3}}} \\ & 320=\frac{8}{3l} \sqrt{110 \times 10^3} \\ & 160 =\frac{4}{3l} \sqrt{11 \times 10^4} \end{aligned}

\begin{aligned} & 40=\frac{1}{3l} \times 10^2 \sqrt{11} \\ & 40 \times 3 l=10^2 \times \sqrt{11} \\ & 120 l=10^2 \times \sqrt{11} \end{aligned}

\begin{aligned} & l=\frac{\sqrt{11} \times 10^2}{120} \\ & l=\frac{3.31 \times 10}{12} \\ & l=\frac{3.31 \times 5}{6}= \end{aligned}

\frac{331 \times 5}{600}=\frac{331}{120} = 2.75 m

Posted by

Ajit Kumar Dubey

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