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

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The discussion focuses on calculating a particle's speed in simple harmonic motion (SHM) with a period of 2.4 seconds and an amplitude of 0.1 meters. For part (a), the user correctly determined the time required for the particle to move to a position 0.05 meters from the equilibrium, calculating it to be approximately 0.4 seconds. In part (b), the user initially calculated the speed using the formula v = -Aωsin(ωt) but expressed confusion over the final result. The correct speed at the specified position should be approximately 0.227 m/s, indicating a need for clarification on the calculations. Overall, the thread highlights the process of applying SHM formulas to find time and speed.
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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
 
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so v=-(.1)(2.62)(sin(2.62*.4))= .00479 m/s

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