Question

A point mass $m$ is suspended at the end of a massless wire of length $L$ and cross-section area $A$. If $Y$ is 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. $\pi \sqrt{\frac{mL}{YA}}$

Answer 1

Answer: (A)

$\frac{1}{2\pi }\sqrt{\frac{YA}{mL}}$

Frequency depends on spring factor and inertia factor.
In this case, stress $=\frac{mg}{A}$
Strain $=\frac{l}{L}$ (where $l$ is extension)
Now, Young's modulus $Y$ is given by
$Y=\frac{stress}{strain}=\frac{mgA}{lL}$
$mg=\frac{YAl}{L}$
So, $kl=\frac{YAl}{L}(\because mg=kl)$ ($k$ is force constant)
$\therefore k=\frac{YA}{L}$
Now, frequency is given by
$n=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$
$=\frac{1}{2\pi }\sqrt{\left(\frac{YA}{mL}\right)}$
Note: In linear SHM the spring factor stands for force per unit displacement and inertia factor for mass of the body executing SHM.