Question 1
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.
- T/8 (11/32) (34%)
- More than T/8 (15/32) (47%)
- Less than T/8 (4/32) (12%)
- 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
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T
|
|
If x = A/2 we have:
A
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2
|
|
= A cos(ωt) |
1
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2
|
|
= cos |
( |
2πt
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T
|
|
) |
What angle has a cosine of 1/2?
Taking the inverse cosine of both sides gives:
π
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3
|
|
= |
2πt
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T
|
|
Solving for time gives: t = |
T
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6
|
|