Question

What is the value of the phase angle ϕ if the initial velocity is positive and the initial displacement is negative?

Answer 1

Answer:

-2π ≤ ∅ ≤ 3π/2

0 ≤ ∅ ≤ π/2

Explanation:

In simple harmonic motion, the displacement and velocity vectors are as follows:

$x(t) = A\cos(\omega t + \phi)\\v(t) = -\omega A\sin(\omega t + \phi)$

where ∅ is the phase angle. At t = 0, the initial displacement and velocity are

$x(0) = A\cos(\phi)\\v(0) = -\omega A\sin(\phi)$

According to two equations, we need to find the ∅ values which makes both cosine and sine terms positive.

So, ∅ is in the range: [-2π, -3π/2] and [0,π/2]