Q 11-13-125JEE MainJEE Main 2020 (6 Sep, Shift 1)Easy
An object of mass $m$ is suspended at the end of a massless wire of length $L$ and area of cross-section $A$. The Young's modulus of the material of the wire is $Y$. If the mass is pulled down slightly, its frequency of oscillation along the vertical direction is:
Answer: (B) $f = \dfrac{1}{2\pi}\sqrt{\dfrac{YA}{mL}}$
The wire behaves like a spring. From $Y = \dfrac{F/A}{\Delta L/L}$, the restoring force for extension $x$ is $F = \dfrac{YA}{L}x$, so $k = \dfrac{YA}{L}$.
$$f = \frac{1}{2\pi}\sqrt{\frac km} = \frac{1}{2\pi}\sqrt{\frac{YA}{mL}}$$
Solution by Sreeraj P, M.Sc Physics