A block of mass m is suspended from the ceiling of a stationary standig elevator through a spring of spring constant k. Suddenly, the cable breaks and the elevator starts falling freely. Show that the bklock now executes a simple harmonic motion of amplitude `mg/k` in the elevator

Explanation

When the elevator is stationary , the spring is streched to support the block . If the exterension is x , the tension is kx which should balance the weight of the block.
Thus, `x = mh//k`. As the cable breakes , the elevoctor starts falling with acceleration 'g' . We shall work in the fram of reference of the elevoctor . Then we have to use a psuedo force mg upwadr on the vlock . This force will 'balance' the weight . Thus , the block is subjected to a net force ky by the spring when it is distance x from the position of unstreched spring . Hence its motion in the elevotor is simple harmonic with its mean position corresponding to the unstretched spring . Initially , the spring is streched by `x = mg//k`, where the velocity of the block (with respect to the elevator) is zero. Thus, the amplitude of the resulting simple harmoni motion is `mg//k`.