A question on spring extension

When body is lowered to mean position, its extension is x.

According to Hooke's law:

 

F = kx = mg

Thus, we have

 

x = mg/k                        ...... (1)

 

Now, when the spring is allowed to fall under gravity, it oscillates about its mean position.

 

Let y be the amplitude of vibration of spring.

 

Then, by conservation of energy at the lowest point

 

½ky² = mgy

 

Thus, we have

 

y = 2mg/k                        ...... (2)

 

Hence, from (1) and (2), we can say that

 

y = 2x

 

Hence, correct option is b.