The displacement of a particle undergoing SHM | vedclass

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Question

(B) Given the displacement equation $x(t) = x_m \cos(\omega t + \phi)$.At $t = 0$,$x = -x_m$,so $-x_m = x_m \cos(\phi)$,which implies $\cos(\phi) = -1

Vedclass · Accepted Answer

$t = 0.50\, T$

Vedclass · Answer

(B) Given the displacement equation $x(t) = x_m \cos(\omega t + \phi)$.At $t = 0$,$x = -x_m$,so $-x_m = x_m \cos(\phi)$,which implies $\cos(\phi) = -1$,so $\phi = \pi$.The equation becomes $x(t) = x_m \cos(\omega t + \pi) = -x_m \cos(\omega t)$.We want to find $t$ when $x = +x_m$.So,$x_m = -x_m \cos(\omega t)$,which means $\cos(\omega t) = -1$.This occurs when $\omega t = \pi$ (for the first time).Since $\omega = \frac{2\pi}{T}$,we have $\frac{2\pi}{T} \cdot t = \pi$.Solving for $t$,we get $t = \frac{T}{2} = 0.50\, T$.

The displacement of a particle undergoing $SHM$ with time period $T$ is given by $x(t) = x_m \cos(\omega t + \phi)$. The particle is at $x = -x_m$ at time $t = 0$. The particle is at $x = +x_m$ when:

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