2kg ball in simple harmonic motion

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  • #1
clh7871
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a 2-kg ball is supended from a spring. when disturbed in a vertical direction, the ball moves up and down in simple harmonic motion with a period of 0.25s.

a. what is the upward force exerted on the ball by the spring when the ball is at the midpoint of its up-and-down path?

b. how much did the spring stretch when the ball was first attached to its end (before the oscillatory motion that started)?
 

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  • #2
HallsofIvy
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I presume you have, or know how to find (depending upon what level this course is), that x(t)= C cos(&radic(k)t) where x(t) is the height at time t, C is the amplitude and k is the spring constant. The period of such motion is given by √(k)T= 2π which, here, is 0.25 so √(k)= 2π/0.25= 8π and k= 64π2.

a. what is the upward force exerted on the ball by the spring when the ball is at the midpoint of its up-and-down path?
That's easy- there is NO total force on the ball at the midpoint so the upward force exerted on the ball by the spring is exactly the downward force exerted by gravity on the ball- its weight.

b. how much did the spring stretch when the ball was first attached to its end (before the oscillatory motion that started)?

You know the weight of the ball and you know the spring constant, k, so use Hooke's law to determine the amount of stretch
 

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