Question

A block of mass 1 kg hangs without vibrating at the end of a spring with a force constant 1N/m attached to the ceiling of an elevator. The elevator is rising with an upward acceleration of g/4.The acceleration of elevator suddenly ceases. What is the amplitude of the resulting oscillations?

Answer 1

When the lift suddenly stops accelerating, then the inertia of the mass of the block comes into play. The block tries to remain in the initial state of accelerated motion and hence, that becomes the driving force for the oscillations. Now,

F_driving  = - F_restoring

A spring has a restoring force of -k*x, where k = spring constant and x = length by which it is stretched or compressed. Also, x will be the amplitude of resultant oscillations.

So, F_restoring = -k*x

Again F_driving= M*(g+a) , where a = acceleration of the elevator.

Finally, after substituting the values of F_driving and F_restoring, we get :

=> M*(g+a) = - (-k*x)

=> M*(g+a) = k*x

=> x = M*(g+a) / k

In our case, M = 1kg , g = 9.8 m/s² , a = g/4 = 2.45 m/s² and k = 1 N/m .Substituting all these values, we get:

=> x = 12.25 metres

Hence, the resultant oscillations will have an amplitude of 12.25 metres.