Question

On a horizontal table lies two identical mass m fastened by a spring of rigidity K. To the right mass we connect another identical mass by a thread thrown through the pulley, Friction is not present anywhere.

Answer 1

Answer:

if the spring is compressed mg/2k and the system released from rest, the hanging mass will initially fall freely.

Answer 2

The complete question is:

On a horizontal table lies two identical mass m fastened by a spring of rigidity K. To the right mass we connect another identical mass by a thread thrown through the pulley, Friction is not present anywhere. Choose the correct option:

1. If the spring is extended by mg/3K and the system is released from rest, all three will start moving with the same magnitude of acceleration.
2. If the spring is extended by mg/K and the system is released from rest. The hanging will start accelerating upwards initially.
3. If the spring is compressed by mg/2K and the system released from the rest, the hanging mass will initially fall freely.
4. If the spring is extended  by 2mg/K and the system released from rest, all three masses will initially accelerate in different directions.

Given:

Rigidity of spring = K

Mass = m

To find:

The correct option.

Solution:

When the spring is compressed, the equation will be:

 x = mg/4K

so when x will be greater than mg/4K then the hanging mass will initially fall freely.

The correct option is 3.