Simple Harmonic Motion Problem

Question

If the equation of motion of a particle is given by s(t)=Acos(wt+d), the particle is said to undergo simple harmonic motion. Assume 0≤d<π.

(a) Find the velocity of the particle at time t.

(b) What is the smallest positive value of t for which the velocity is 0? Assume that w and d are positive.

I already found part (a)

I don't know what (b) is asking or how to find it.

Any help is much appreciated. Thanks!

Answer 1

Since velocity = ds/dt, it is found by taking the derivative of the displacement function so:

v(t) = s'(t) = -Aωsin(ωt + d). [but you already knew that]

Question b is asking for a value of t such that -Aωsin(ωt + d) = 0. We know that sin(0) = 0 and sin(π) = 0 so, since these are the two conditions that will make the velocity zero, we can start out considering when ωt + d = 0

ωt + d = 0

ωt = - d

t = -d/ω however, this is NOT a positive value (since we are told that both d and ω are positive.

So let's consider the next "zero" which is sin(π) = 0 meaning ωt + d = π

ωt + d = π

ωt = π - d

t = (π - d)/ω