SHM Question

timtng

A particle vibrates in SHM with the period of 2.4 s and the amplitude of .1 m.

a) How long is required for the particle to move from one end of its path to the nearest point .05 m from the equilibrium position?
b) What is the particle's speed at this position?

for a) this is what I did
x=Acos(&omega;t), 5=10cos(&omega;t), &omega;=2pi/T=2pi/2.4=2.62
t=arccos(.5)/2.62= 0.4 s

I'm a little confused on part b.

Please see if I did a correctly, and show me how to answer b.

Thanks

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Ambitwistor

Part (a) looks fine. For part (b), you have to take the time derivative of position to get the velocity,

v = dx/dt = -A&omega; sin(&omega;t)

timtng

so v=-(.1)(2.62)(sin(2.62*.4))= .00479 m/s

Is that correct?

Ambitwistor

Originally posted by timtng
so v=-(.1)(2.62)(sin(2.62*.4))= .00479 m/s

Is that correct?
Almost; you have to switch your calculator to radians to take the sine.

timtng

Okay, so it should be .227 m/s?

Yes.

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