Simple harmonic motion question

Question

I was working through my textbook and came across these 2 questions:

An object is undergoing SHM with period 0.900 s and amplitude 0.320 m. At t = 0 the object is at x = 0.320 m and is instantaneously at rest. Calculate the time it takes the object to go

(a) from x = 0.320 m to x = 0.160 m and

(b) from x = 0.160 m to x = 0

The answers are 0.150 seconds and 0.0750 seconds respectively. I am not able to figure out part b, although I got part a. Can someone show me how to solve this? In particular I cannot figure out how to calculate the phase angle.

Answer 1

Hi Ajit S.,

Since you found the first part of the problem, I assume you got the equation x = .32*cos(20πt/9). At t = 0, x = .32

When x = .16:

.16 = .32cos(20πt/9)

1/2 = cos(20πt/9)

cos^-1(1/2) = 20πt/9

t₁ = 9cos^-1(1/2)/(20π) = 9(π/3)/(20π) = 3/20 = .15

When x = 0, 0 = cos(π/2), so this is where:

20πt/9 = π/2

t₂ = 9/40 = .225

From x = .16 to x = 0: t₂ - t₁ = .225 - .15 = .075

I hope this helps, Joe.