Simple Harmonic Motion Displacement

[Qube]

Homework Statement

A particle is in simple harmonic motion with period T and with position as a function of time given by x(t) = A cos(wt+ø).

At time t = 0 the particle is at x = A/2 with positive velocity. The next time it is at the same position is ___?

Homework Equations

w/(2pi)=1/T

The Attempt at a Solution

So I know the cosine of the argument has to = 1/2 for the position to be half the amplitude. The first time this happens is at t = 0. This implies the phase constant ø is pi/3.

The next time the cosine argument is a positive 1/2 is when the argument equals 300 degrees. Therefore:

wt + pi/3 = 5pi/3
wt = 4pi/3
t = 4pi/(3w)

w = 2pi/T

t = 4pi/[3(2pi/T] = 4pi/[6pi/T] = 2T/3.

The answer however is t = T/3.

 

[SammyS]

Homework Statement

A particle is in simple harmonic motion with period T and with position as a function of time given by x(t) = A cos(wt+ø).

At time t = 0 the particle is at x = A/2 with positive velocity. The next time it is at the same position is ___?

Homework Equations

w/(2pi)=1/T

The Attempt at a Solution

So I know the cosine of the argument has to = 1/2 for the position to be half the amplitude. The first time this happens is at t = 0. This implies the phase constant ø is pi/3.

The next time the cosine argument is a positive 1/2 is when the argument equals 300 degrees. Therefore:

wt + pi/3 = 5pi/3
wt = 4pi/3
t = 4pi/(3w)

w = 2pi/T

t = 4pi/[3(2pi/T] = 4pi/[6pi/T] = 2T/3.

The answer however is t = T/3.

If the particle has positive velocity at t =0 zero, doesn't ωt+ø have to be -π/3 (-60°) ?