Question

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

Answer 1

When the elevator is stationary the spring is stretched to support the block If the extension is x the tension is kx which should balance the weight of the block

Thus x = mg/k As the cable breaks the elevator starts falling with acceleration 'g' We shall work in the frame of reference of the elevator Then we have to use a pseudo force mg upward on the block This force will 'balance' the weight Thus the block is subjected to a net force kx by the spring when it is at a distance x from the position of unstretched spring Hence its motion in the elevator is simple harmonic with its mean position corresponding to the unstretched spring Initially the spring is stretched by = mg/k where the velocity of the block (with respect to the elevator) is zero Thus the amplitude of the resulting simple harmonic motion is mg/k