SHM Question

  • Thread starter timtng
  • Start date
  • #1
25
0
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(ωt), 5=10cos(ωt), ω=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
 

Answers and Replies

  • #2
841
1
Part (a) looks fine. For part (b), you have to take the time derivative of position to get the velocity,

v = dx/dt = -Aω sin(ωt)
 
  • #3
25
0
so v=-(.1)(2.62)(sin(2.62*.4))= .00479 m/s

Is that correct?
 
  • #4
841
1
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.
 
  • #5
25
0
Okay, so it should be .227 m/s?
 
  • #6
841
1
Yes.
 

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