Question

A block of mass $m$ is released from rest on an inclined plane of angle $\theta$ (as shown). Initially spring is at its natural length. Find the time period and the amplitude of oscillation.
(The inclined plane is smooth)

• A. $2\pi \sqrt{\frac{m}{k}},\frac{mg\sin \theta }{k}$
• B. $2\pi \sqrt{\frac{m\sin \theta }{k}},\frac{2mg\sin \theta }{k}$
• C. $2\pi \sqrt{\frac{m}{k}},\frac{mg\cos \theta }{k}$
• D. None of these

Answer 1

The correct option is A $2\pi \sqrt{\frac{m}{k}},\frac{mg\sin \theta }{k}$

As time period for the spring block system is independent of inclination,
$T=2\pi \sqrt{\frac{m}{k}}$
When '$m$' is released from rest the force $mg\sin \theta$ acting on the block (along inclined plane) will lead to extension in spring.


Applying equilibrium condition,
$kx=mg\sin \theta$
$\Rightarrow x=\frac{mg\sin \theta }{k}$
Initially spring is at its natural length , $x_i=0$
As amplitude is maximum displacement of particle from equilibrium position, we get $A=x-x_i=\frac{mg\sin \theta }{k}$