An object attached to a spring is pulled a distance A from the equilibrium position and released from rest. It then experiences simple harmonic motion with a period T. The time taken to travel between the equilibrium position and a point A from equilibrium is T/4. How much time is taken to travel between points A/2 from equilibrium and A from equilibrium? Assume the points are on the same side of the equilibrium position, and that mechanical energy is conserved.

1. T/8 (11/32) (34%)
2. More than T/8 (15/32) (47%)
3. Less than T/8 (4/32) (12%)
4. It depends whether the object is moving toward or away from the equilibrium position (2/32) (6%)

The further the object gets from equilibrium the slower it gets, so the average speed moving 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

=

2π T

If x = A/2 we have:

A 2	= A cos(ωt)

1 2

= cos

(

2πt T

)

What angle has a cosine of 1/2?

Taking the inverse cosine of both sides gives:

π 3

=

2πt T

Solving for time gives:     t =

T 6