A particle travels in simple harmonic motion described by the...

The position of the particle is given by:

$x = 2.0 \cos(50 \pi t + \tan^{-1} 0.75)$

The velocity of the particle is the rate of change of position:

$v = \dfrac{dx}{dt} = -100 \pi \sin(50 \pi t + \tan^{-1} 0.75)$

The particle comes to rest when its velocity is zero, i.e., $v = 0$:

$\sin(50 \pi t + \tan^{-1} 0.75) = 0$

For the particle to come to rest for the first time after $t = 0$, the phase must be equal to $\pi$ (since the initial phase $\tan^{-1} 0.75$ is between $0$ and $\pi/2$):

$50 \pi t + \tan^{-1} 0.75 = \pi$

We know that $\tan^{-1} 0.75 = \tan^{-1}\left(\dfrac{3}{4}\right) \approx 37^{{}^{{}^{\circ}}}$.

Converting $37^{{}^{{}^{\circ}}}$ to radians:

$\tan^{-1} 0.75 \approx \dfrac{37 \pi}{180}$

Substituting this into the phase equation:

$50 \pi t + \dfrac{37 \pi}{180} \approx \pi$

Dividing the entire equation by $\pi$:

$50 t + \dfrac{37}{180} = 1$

$50 t = 1 - \dfrac{37}{180} = \dfrac{143}{180}$

$t = \dfrac{143}{180 \times 50} = \dfrac{143}{9000} \approx 0.01588 \text{ s}$

Rounding to two significant figures, we get:

$t \approx 1.6 \times 10^{-2} \text{ s}$