Question

A disc of mass $3m$ and a disc of mass $m$ are connected by a massless spring of stiffness $k$. The heavier disc is placed on the ground with the spring vertical and lighter disc on top. Both discs are horizontal. From its equilibrium position the upper disc is pushed down by a distance $\delta$ and released. Then,

• A. If $\delta >\frac{3mg}{k}$, the lower disc will bounce up
• B. If $\delta =\frac{2mg}{k}$, maximum normal reaction from ground on lower disc $=6mg$
• C. If $\delta =\frac{2mg}{k}$, maximum normal reaction from ground on lower disc $=4mg$
• D. If $\delta >\frac{4mg}{k}$, the lower disc will bounce up

Answer 1

Answer: (B), (D)

If $\delta >\frac{4mg}{k}$, the lower disc will bounce up

$F$ is the force by hand on upper disc

$=3mg+kl+2mg$
$=3mg+mg+2mg$
$=6mg(\text{for}\delta =\frac{2mg}{k})$
$\therefore$ Correct option is (b)

When $F$ is removed, the upper disc accelerates upwards and when it attains the position as in figure 2, its acceleration reduces to zero and the velocity gained by it takes it further upwards. Restoring force on the upper plate now acts downwards and that on the lower plate acts in the upward direction and would lift it (lower plate) if
$k(\delta -l)>3mg$
i.e., $k\delta >3mg+kl$
or $k\delta >3mg+mg$
or $\delta >\frac{4mg}{k}$
$\therefore$ Correct option is (d).