Question

A ball of mass $m$ kept at the centre of a long string of length $L$ is pulled from the centre in a perpendicular direction and released. The time period of ball for small oscillations is
($T^{\prime }$ is tension in the string)

• A. $2\pi \sqrt{\frac{4T^{\prime }}{mL}}$
• B. $\frac{\pi }{4}\sqrt{mL{T}^{\prime }}$
• C. $\frac{\pi }{2}\sqrt{\frac{mL}{T^{\prime }}}$
• D. $2\pi \sqrt{\frac{mL}{4T^{\prime }}}$

Answer 1

The correct option is D $2\pi \sqrt{\frac{mL}{4T^{\prime }}}$

The forces acting on ball has been shown in FBD, after ball has been pulled by a distance $x$ in perpendicular direction and relased.

Net force towards the mean position (M.P)
$\Rightarrow F=-2T^{\prime }\sin \theta$
(Force is opposite to displacement 'x') considering $\theta$ to be small,
$T^{\prime }\to$ Tension in string
$\sin \theta \approx \theta \approx \tan \theta \approx \left(\frac{x}{L/2}\right)$
$\Rightarrow ma=-2T^{\prime }\left(\frac{x}{L/2}\right)$
$\Rightarrow m\frac{d^{2}x}{d{t}^2}=\frac{-4T^{\prime }x}{L}$
$\Rightarrow \frac{d^{2}x}{d{t}^2}=\frac{-4T^{\prime }x}{mL}(a\propto -x)$
Companing it with equation of SHM:
$a=-{\omega }^{2}x$
$\Rightarrow \omega =\sqrt{\frac{4T^{\prime }}{mL}}$
Hence, time period of oscillation will be
$T=\frac{2\pi }{\omega }$
$\therefore T=2\pi \sqrt{\frac{mL}{4T^{\prime }}}$