Simple harmonic Motion problem ??for 40 pts?

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Simple harmonic Motion problem ??for 40 pts?

An object is undergoing simple harmonic motion with period 1.24 s and amplitude 0.610 m. At t=0 the object is at x=0. How far is the object from the equilibrium position at time t = 0.485 s?

2006-06-28 21:27:33 · 3 answers · asked by Sagely 4 in Science & Mathematics ➔ Physics

well done, its not so hard really uncle sam......

2006-06-28 21:42:09 · update #1

3 answers

y = A sin (2pi * t/T)
y= 0.610* sin(2 pi * 0.485/1.24)
y= 0.610 * 0.63
y= 0.3843 metres = 3.843 cm

2006-06-28 21:36:29 · answer #1 · answered by Anonymous · 0⤊ 0⤋

t=.485

period = 1.24

.485 = ?/.610

? = .794

.794x.610 m = .485

answer = .485 m.

2006-06-29 04:41:10 · answer #2 · answered by Me in Canada eh 5 · 0⤊ 0⤋

you kiddin?

2006-06-29 04:29:37 · answer #3 · answered by Lard Cherrybakins 4 · 0⤊ 0⤋