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.
• B.
• C.
• D.

Answer 1

The correct option is D

Let l be the increase in the length of the wire due to the force F = m g (see Fig. 9.31). Then
$Stress=\frac{Force}{Area}=\frac{F}{A}$
$Strain=\frac{changeinlength}{Originallength}=\frac{l}{L}$
By definition, Yoiung’s modulus is
$Y=\frac{Stress}{Strain}=\frac{mgL}{lA}$
or $F=mg=\frac{YAl}{L}$
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 in the wire acting upwards will be
=$\frac{YA}{L}(l+x)$
And downward force is F = m g. The restoring force is the net downward force. Hence,
Restoring force =$mg-\frac{YA}{L}(l+x)$
$=\frac{YAl}{L}-\frac{YA}{L}(l+x)=-\frac{YAx}{L}$
$\therefore$ Acceleration (a) of the mass = $\frac{force}{mass}$
$=\frac{YAx}{L}=-{\omega }^{2}x$
Where $\omega =\sqrt{\frac{YA}{mL}}$ is the angular frequency of the resulting motion which is simple harmonic.
Thus,
Frequency of oscillation (n) = $\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{YA}{mL}}$