A mass attached to a spring is pulled a distance A from equilibrium and released from rest. It then undergoes SHM with period T. The time to travel between the equilibrium position and A is T/4. How much time is taken to travel between A/2 and A?

1. T/8
2. More than T/8
3. Less than T/8
4. It depends whether the object is moving toward or away from the equilibrium position

The further the mass moves away from equilibrium the slower it goes, so the average speed between x = A and x = A/2 is less than the average speed between x = A/2 and x = 0. Thus it takes longer than T/8 to move between x = A and x = A/2.

Let's figure out exactly how long it takes.

The equation of motion is x = A cos ωt, where ω = 2 π/T.

At t = 0 the mass is at x = A. What is the time when x = A/2?

A 2	= A cos(ωt)

1 2

= cos

(

2πt T

)

Taking the inverse cosine of both sides gives:

π 3

=

2πt T

Solving for time gives:   t = T/6.