Question

Repeat the previous exercise if the angle between each pair of springs is ${120}^{\circ }$ initially.

Answer 1

In this, if the particle 'm' is pushed against 'C' a by distance 'x'. Total resultant force acting on man 'm' is given by

$F=kx+\frac{kx}{2}=\frac{3kx}{2}$

[Because net force by A and B]

$\sqrt{{\left(\frac{kx}{2}\right)}^2+{\left(\frac{kx}{2}\right)}^2+2\left(\frac{kx}{2}\right)\left(\frac{kx}{2}\right)}cos{120}^{\circ }$

$=\frac{kx}{2}$

$\therefore a=\frac{F}{m}=\frac{2kx}{2m}$

$\Rightarrow \frac{a}{x}=\frac{3k}{2m}=w^2$

$\Rightarrow \omega =\sqrt{\frac{3k}{2m}}$

$\therefore$ Time period = $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{2m}{3k}}$