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

Options

A. the block executes simple harmonic motion with time period `2pisqrt((m)/(k))`
B. the block excutes simple harmonic motion with amplitude `(mg)/(k)`
C. the block executes simple harmonic motion about its mean position and the mean position is the position When the spring acquires its natural length.
D. all of the above.

Answer

D

Explanation

When the elevator is at rest, the elongation in spring is given by `ky_0=mg` or `y_0=(mg)/(k)`
As the instant the elevator starts falling down with acceleration g, the block is at rest w.r.t elevator and the net force acting on it is `ky_0` in the upward direction w.r.t. lift frame of reference. Due to this force, the block moves up and as a result elongation in the spring decreases, and the force experienced by the block becomes zero when spring and stops momentarily till compression in the spring becomes `(mg)/(k)`.
Hence the block will always have net force towards relaxed position of spring and the block will perform simple harmonic notion with time period `T=2pisqrt((m)/(k))` and with amplitude `(mg)/(k)`