Question 2
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?
 T/8
 More than T/8
 Less than T/8
 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.