\(f=\frac{v}{2 \ell}=\frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}=\frac{1}{2 \ell} \sqrt{\frac{T}{A \rho}}\left[\because v=\sqrt{\frac{T}{\mu}} \text { and } \mu=\frac{m}{\ell}\right]\)
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, \(f \approx 178.2 \mathrm{Hz}\)
(હવામાં ધ્વનિનો વેગ $330\;m/sec$ છે)