Question

A 30.0-cm-long wire having a mass of 10.0 g is fixed at the two ends and is vibrated in its fundamental mode. A 50.0-cm-long closed organ pipe, placed with its open end near the wire, is set up into resonance in its fundamental mode by the vibrating wire. Find the tension in the wire. Speed of sound in air = $340m{s}^{-1}$.

Answer 1

Given, m=10 gm

$=10\times {10}^{-3}kg,$

I=30 cm =0.3 m

Let the tension in the string will be =T.

$m=\frac{\text{Mass}}{\text{Unit length}}$

$=33\times {10}^{-3}kg/m$

The fundamental frequency,

$\Rightarrow n_0=\frac{1}{2L}\sqrt{\frac{T}{m}}........\left(1\right)$

The fundamental frequency of closed pipe.

$\Rightarrow n_0=\left(\frac{v}{4I}\right)$

$=\frac{340}{2\times 30\times {10}^{-2}}............\left(2\right)$

According to equations (1)and (2)we get

$170=\frac{1}{2\times 30\times {10}^{-2}}\times \frac{T}{33\times {10}^{-3}}$

$\Rightarrow T=347$ Newtons