Q. A point mass $m$ is suspended at the end of a massless wire of length $L$ and cross-section area $A$. If $Y$ is the Young’s modulus for the wire, then the frequency of oscillations for the SHM along the vertical line is
A) $\frac{1}{2\pi}\sqrt{\frac{YA}{mL}}$
B) $2\pi\sqrt{\frac{mL}{YA}}$
C) $\frac{1}{\pi}\sqrt{\frac{YA}{mL}}$
D) $\pi\sqrt{\frac{mL}{YA}}$
Explanation
(a) In this case,
Stress = $\frac{mg}{A}$
Strain = $\frac{l}{L}$ (where $l$ is extension)
Now, Young’s modulus $Y$ is given by
$Y = \frac{stress}{strain} = \frac{\frac{mg}{A}}{\frac{l}{L}}$
$mg = \frac{YAl}{L}$
So, $kl = \frac{YAl}{L}$ ($\because\ \ mg = kl$)
($k$ is force constant)
Now, frequency is given by
$n = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$
$= \frac{1}{2\pi}\sqrt{\left( \frac{YA}{mL} \right)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.