Calculate Particle's Speed in SHM with Period of 2.4 s and Amplitude of .1 m

[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(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

 

[Ambitwistor]

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)

 

[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?

 

[Ambitwistor]

Yes.