Question

(a) If the coordinate of a particle varies as $\mathit{x}\mathbf{=}\mathbf{-}\mathit{A}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\omega }\mathit{t}$, what is the phase constant in Equation 15.6? (b) At what position is the particle at t=0?

Short answer

(a) $\varphi =\pi$, or $\varphi =-\pi$

(b) At t = 0, the particle is at its turning point on the negative side of equilibrium, at x = -A.

Step-by-step solution

Step 1: The general equation of position

The general equation of position is described by

$x\left(t\right)=A\cos \left(\omega t+\varphi \right)$

A is the amplitude of the wave

$\omega$is the angular frequency

t is the time

$\varphi$is the phase angle

Step 2(a): Find the phase constant

The general equation of position is $x\left(t\right)=A\cos \left(\omega t+\varphi \right)$ If $x=-A\cos \omega t$, then $\varphi =\pi$, or equally well, $\varphi =-\pi$

Step 3(b): Find the position at t = 0

At t = 0, the particle is at its turning point on the negative side of equilibrium, at x = -A.