Question

A wire of density $9\times 10kg{m}^{-3}$ stretched between two clamps $1m$ apart is subjected to an extension of $4.9\times {10}^{-4}m$. If Young's modulus of the wire is $9\times {10}^{-4}m$. The lowest frequency of transverse vibrations in the wire is

• A. $35Hz$
• B. $70Hz$
• C. $105Hz$
• D. $140Hz$

Answer 1

The correct option is A $35Hz$

Fundamental frequency $\nu =\frac{1}{2l}\sqrt{\frac{T}{m}}$.
Now, Young's modulus $Y=\frac{T/A}{\Delta l/l}=\frac{Tl}{A\Delta l}\Rightarrow T=\frac{YA\Delta l}{l}$.
So, $\nu =\frac{1}{2l}\sqrt{\frac{\frac{YA\Delta l}{l}}{A\rho }}=\frac{1}{2l}\sqrt{\frac{YA\Delta l}{lA\rho }}=\frac{1}{2}\sqrt{\frac{9\times {10}^{10}\times 4.9\times {10}^{-4}}{9\times {10}^3}}=35$ Hz.
Ans: A