Question

The system shown lies on a horizontal smooth surface. The mass '$m$' is given a small displacement along the x - axis. The time period will be:

• A. $2\pi \sqrt{\frac{2m}{5k}}$
• B. $2\pi \sqrt{\frac{5m}{2k}}$
• C. $2\pi \sqrt{\frac{3m}{2k}}$
• D. None

Answer 1

The correct option is A $2\pi \sqrt{\frac{2m}{5k}}$


$x^{\prime }=xcos{60}^{\circ }=\frac{x}{2}$

restoring force $\left(F_R\right):$
$F_R=-[\frac{kx}{2}\text{cos}{60}^{\circ }+\frac{kx}{2}\text{cos}{60}^{\circ }+2kx]$
$F_R=-\frac{5kx}{2}$
Now, time - period
$T=2\pi \sqrt{\frac{m}{k_{eq}}}$
where $k_{eq}$ is the equivalent spring constant
$T=2\pi \sqrt{\frac{2m}{5k}}$
Hence, option (A) is correct.