Question

A mass m is suspended at the end of a massless wire of length L and cross-sectional area A. If Y is the Young's modulus of the material of the wire, the frequency of oscillations along the vertical line is given by

• A. $v=\frac{1}{2\pi }\sqrt{\frac{mL}{YA}}$
• B. $v=\frac{1}{2\pi }\sqrt{\frac{YL}{mA}}$
• C. $v=\frac{1}{2\pi }\sqrt{\frac{AL}{Ym}}$
• D. $v=\frac{1}{2\pi }\sqrt{\frac{YA}{mL}}$

Answer 1

The correct option is D

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

Let l be the increase in the length of the wire due to the force F = mg.
So, stress $=\frac{Force}{Area}=\frac{F}{A}=\frac{mg}{A}$
and strain $=\frac{Changeinlength}{Originallength}=\frac{l}{L}$
By definition, Young's modulus, $Y=\frac{Stress}{Strain}=\frac{mgL}{lA}$
$\Rightarrow F=mg=\frac{YAl}{L}$

This is the force acting upwards in the equilibrium state. If the mass is pulled down a little through a distance $x$, so that the total extension in the string is $(l+x)$, then the force acting upwards will be $=\frac{YA}{L}(l+x)$ and that acting downwards will be $mg$.
So, net force in downward direction $=mg-\frac{YA}{L}(l+x)$
$=\frac{YAl}{L}-\frac{YA}{L}(l+x)$
$=\frac{-YAx}{L}$
$\therefore$ Acceleration, $a=\frac{Force}{Mass}$
$=-\frac{YAx}{mL}=-{\omega }^{2}x$
where $\omega =\sqrt{\frac{YA}{mL}}$ is the angular frequency of the resulting motion which is simple harmonic.
Thus, frequency of oscilltion, $v=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{YA}{mL}}$ and the correct choice is (d).