Question

A particle of mass $m$ is attached with four identical springs, each of length $l.$ Initially, each spring has tension $F_0$. Neglecting gravity, calculate the time period for small oscillations of the particle along a line perpendicular to the plane of the figure.

• A. $T=2\pi \sqrt{\frac{ml}{F_0}}$
• B. $T=2\pi \sqrt{\frac{ml}{2F_0}}$
• C. $T=2\pi \sqrt{\frac{ml}{4F_0}}$
• D. $T=2\pi \sqrt{\frac{4ml}{F_0}}$

Answer 1

The correct option is C $T=2\pi \sqrt{\frac{ml}{4F_0}}$

Let us displace the particle slightly by distance $x$ along a line perpendicular to the plane of figure. Since the springs are identical, increase in tension in each spring is the same. Let the increase in tension be $d{F}_0$. The figure shows the forces exerted by spring $AP$ and $CP$:


Restoring force produced by these two springs is given by
$2(F_0+d{F}_0)\sin \theta$
in the direction opposite to displacement $x$.
For small oscillations, we can neglect $d{F}_0$. Also, we know $\theta$ is small, so $\sin \theta \approx \tan \theta \approx \frac{x}{l}$
$\therefore F_{rest}^{\prime }=2F_0\frac{x}{l}$
Similarly, Restoring force produced by remaining two springs $BP$ and $DP$ is given by $F_{rest}^{\prime \prime }=2F_0\frac{x}{l}$
Net Restoring Force $F=F_{rest}^{\prime }+F_{rest}^{\prime \prime }=\frac{4F_{0}x}{l}$
Comparing the above equation with $F=-K_{eff}x$, we get $K_{eff}=\frac{4F_0}{l}$
Time period of oscillation $T=2\pi \sqrt{\frac{m}{K_{eff}}}=2\pi \sqrt{\frac{ml}{4F_0}}$
Thus, option (c) is the correct answer.