Question

A steel wire of length $1m$, mass $0.1kg$ and uniform cross-sectional area ${10}^{-6}{m}^2$ is rigidly fixed at both ends. The temperature of the wire is lowered by ${20}^{o}C$. If transverse waves are setup by plucking the string in the middle, calculate the frequency of the fundamental mode of vibration.
$[y=2\times {10}^{11}N/m^2,{\alpha }_{steel}=1.21\times {10}^{-5}{/}^{o}C]$

Answer 1

For the string -

$\Rightarrow l=1m$

$m=0.1kg$

Area of X-section $={10}^{-6}{m}^2$

$\Rightarrow △T=20°C$

$\Rightarrow y$( young's modules ) $=2\times {10}^{11}N/m^2$

$\Rightarrow \alpha$ steel $=1.21\times {10}^{-5}/℃$

For construction of wire due to lowering of temperature -

$\Rightarrow lf=li(1+\alpha △T)$

$|lf-li|=li\alpha △T$

$\Rightarrow △l=li\alpha △T=(1\times 1.21\times {10}^{-5}\times 20)$

Now, $y=\frac{stress}{strain}=\frac{\frac{T}{A}}{\frac{△l}{l}}$ ( T= tension produced )

$\Rightarrow 2\times {10}^{11}=\frac{T\left(1\right)}{\left({10}^{-6}\right)(1.21\times 20\times {10}^{-5})}$

$\Rightarrow T=2\times {10}^{11}\times 1.21\times 20\times {10}^{-11}$

$=48.4N$

Fundamental mode frequency $=\frac{v}{2l}=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$ [ $\mu =$ mass per unit length ]

$=\frac{1}{2\left(1\right)}\sqrt{\frac{48.4\times 10}{0.1\times 10}}=\frac{1}{2}\sqrt{484}$

$=11Hz.$

Hence, the answer is $11Hz.$