Q 11-14-060JEE MainJEE Main 2023 (11 Apr, Shift 2)Medium
A wire of density $8\times10^3\ \text{kg m}^{-3}$ is stretched between two clamps $0.5\ \text{m}$ apart. The extension developed in the wire is $3.2\times10^{-4}\ \text{m}$. If $Y=8\times10^{10}\ \text{N m}^{-2}$, the fundamental frequency of vibration in the wire will be ______ Hz.
Numerical value type. Enter your answer.
Answer: 80
Stress $=Y\dfrac{\Delta L}{L}$, and $v=\sqrt{\dfrac{\text{stress}}{\rho}}=\sqrt{\dfrac{8\times10^{10}\times3.2\times10^{-4}}{0.5\times8\times10^3}}=\sqrt{6400}=80\ \text{m s}^{-1}$.
$f=\dfrac{v}{2L}=\dfrac{80}{1}=80\ \text{Hz}$.
Solution by Sreeraj P, M.Sc Physics