Question

The time period of vertical simple harmonic motion of the block shown in the diagram is $T=\frac{x \pi}{y} s$. Find $(x +y)$. Given that the spring is massless and ideal with $k=20\, N / m$, the pulleys are massless and smooth and the block's mass is $m=5 \,kg$, the inclined plane is at $\theta=30^{\circ}$ to the horizontal and the thread is inextensible and massless.

Answer 1

At equilibrium position, if the extension of the spring is $x_{0}$
$ \Rightarrow 2 k x_{0}=m g$, on downward (or upward) displacement $s$ of the block, the net extension of the spring becomes $x=x_{0}+2 s$ (or $x=x_{0}-2 s$ ),
hence, from the dynamical equation
$2 k x-m g=m a $
$\Rightarrow 2 k\left(x_{0}+2 s\right)-m g=m a$
$\Rightarrow 4 k s=-m \frac{d^{2} s}{d t^{2}}$
$\Rightarrow T=2 \pi \sqrt{\frac{m}{4 k}}$