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 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 elevatore starts falling with accelertion `'g'`. We shall work in the frame of reference of the elevator. Then we have to use pseudo force `mg` upward on the blod 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 inthe elevatore simple harmonic with its mean position corresponding to the unstretched spring. initially, the spring is stretched by `x = 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`.