A spring of spring constant K is fixed to the ceiling of a lift. The other end of the spring is attached to a block of mass m. The mass is in equilibrium. Now the lift accelerates downwards with an acceleration 2g.

Engineering

Physics

Basics of Simple Harmonic Motion

Newtons First Law of Motion

SHM of Spring Block System

Question

A spring of spring constant K is fixed to the ceiling of a lift. The other end of the spring is attached to a block of mass m. The mass is in equilibrium. Now the lift accelerates downwards with an acceleration 2g.

The block will not perform SHM and it will stick to the ceiling.

The block will perform SHM with time period $2 π \sqrt{m / K}$ .

The amplitude of the block will be 2mg/K if it perform SHM.

The minimum potential energy of the spring during the motion of the block will be 0.

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Now as lift starts descending by acceleration ‘g’ of downward, in the frame of lift

Net force = – kx

$\frac{F}{m} = \frac{- k}{m} \cdot k$

$t = 2 π \sqrt{\frac{m}{k}}$ ;

minimum potential energy is at the mean position = 0 when x = 0