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A sonometer wire oflength $1.5\ m$ is made of steel. The tension in it produces an elastic strain of $1 \%$. What is the fundamental frequency of steel if density and elasticity of steel are $7.7 \times 10^3 $ $kg/m^3$ and $2.2 \times 10^{11}$ $N/m^2$ respectively?
$770$
$188.5$
$178.2$
$200.5$
Solution
Fundamental frequency,
$f=\frac{v}{2 \ell}=\frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}=\frac{1}{2 \ell} \sqrt{\frac{T}{A \rho}}\left[\because v=\sqrt{\frac{T}{\mu}} \text { and } \mu=\frac{m}{\ell}\right]$
Also, $Y=\frac{T \ell}{A \Delta \ell} \Rightarrow \frac{T}{A}=\frac{Y \Delta \ell}{\ell} \Rightarrow f=\frac{1}{2 \ell} \sqrt{\frac{\gamma \Delta \ell}{\ell \rho}}$ $…(i)$
Putting the value of $\ell, \frac{\Delta \ell}{\ell}, \rho$ and $\gamma$ in eq $^{n} .$ $(i)$ we get,
$f=\sqrt{\frac{2}{7}} \times \frac{10^{3}}{3}$ or, $f \approx 178.2 \mathrm{Hz}$
