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

The given equation of simple harmonic motion is $x = 2.0 \cos(50 \pi t + \phi)$, where $\phi = \tan^{-1}(0.75)$.

The particle comes to rest when its velocity is zero:

$v = \dfrac{dx}{dt} = -100 \pi \sin(50 \pi t + \phi) = 0$

This implies $\sin(50 \pi t + \phi) = 0$, which gives $50 \pi t + \phi = n\pi$ for $n = 1, 2, 3, \dots$

For the particle to come to rest for the second time, we set $n = 2$:

$50 \pi t + \phi = 2\pi$

$t = \dfrac{2\pi - \phi}{50 \pi} = \dfrac{2}{50} - \dfrac{\phi}{50 \pi}$

Since $\tan \phi = 0.75 = \dfrac{3}{4}$, the angle \phi \approx 37^{}^{{}^{\circ}}. Converting to radians, $\phi \approx \dfrac{37\pi}{180} \approx 0.205\pi$.

Substituting this value:

$t \approx \dfrac{2\pi - 0.205\pi}{50 \pi} = \dfrac{1.795}{50} = 0.0359 \text{ s}$

Rounding to two significant figures, we get $t \approx 3.6 \times 10^{-2} \text{ s}$.