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Periodic Motion

  • Thread starter prophet05
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Homework Statement


A 500g block is attached to a spring on a frictionless horizontal surface. The block is pulled to stretch the spring by 10cm, then gently released. A short time later, as the block passes through the equilibrium position, its velocity is 1m/s.
>A)What is the block's period of oscillation?
>B)What is the block's speed at the point where the spring is compressed by 5cm?

Homework Equations


T = 2(pi)sqrt(I/mgd)

The Attempt at a Solution


I'm having troubles starting this problem. I'm thinking since it's 1m/s at it's equilibrium position (5cm) then it take 20seconds for half a cycle? That sounds completely wrong. I need some help.
 
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Answers and Replies

  • #2
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Homework Statement


A 500g block is attached to a spring on a frictionless horizontal surface. The block is pulled to stretch the spring by 10cm, then gently released. A short time later, as the block passes through the equilibrium position, its velocity is 1m/s.
>A)What is the block's period of oscillation?
>B)What is the block's speed at the point where the spring is compressed by 5cm?

Homework Equations


T = 2(pi)sqrt(I/mgd)

The Attempt at a Solution


I'm having troubles starting this problem. I'm thinking since it's 1m/s at it's equilibrium position (5cm) then it take 20seconds for half a cycle? That sounds completely wrong. I need some help.
Have you encountered an equation that looks like x=Asin(w*t) ??

This is a general eqn for describing simple harmonic motion, which is what this problem is about. It also describes the motion of a pendulum.

w*t (angular velocity * time) in the sine expression above, is the key to answering this problem as it also determines the period thru the relation,

T=2*pi/w.

So how to determine w?

Well if we were given a value of time and position we could do so as we are given A, the amplitude, as equal to 10cm.

But all we are told is that a "short time later", its velocity is 1m/s as it passes thru the equalibrium point. That is when x=0.

We can differentiate the above expression with respect to time to get,

dx/dt=v=w*A*cos(wt). Since we know that x=0, it follows sin(wt)=0 at that time, and most importantly for the purposes of this problem cos(wt)=1 at the same time, so we can substitute 1 for cos(wt).

Can you finish from here?
 

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