If the equation of motion of a particle is given by s = A cos(ωt + δ),

Question

the particle is said to undergo simple harmonic motion. s'(t) = -Aω sin(ωt + δ)

When is the velocity 0? (Use n as the arbitrary integer.)

t= _____________

I thought it was 0, but it said it was wrong.

Answer 1

The solution to t in the following equation is the right answer.

s'(t) = -Aω sin(ωt + δ) = 0

Since it is an SHM (simple harmonic motion), there are many solutions.

Think of riding on a swing. When you reach the highest point, the velocity is zero. But every time you ride to the top (on each side of the swing), the velocity is zero. So this happens every 2 seconds or so. The arbitrary integer n is used to denote the sequence.

Back to the equation:

s'(t) = -Aω sin(ωt + δ) = 0

Since we know the amplitude A is not zero, and the frequency ω is not zero, so sin(ωt + δ)=0
The solution of
sin(ωt + δ)=0
is
ωt + δ)= nπ
(see link at end of post)
So solve for t to get:
t=(nπ-δ)/ω

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