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

The displacement of the particle is given by $x = 2.0 \cos(50 \pi t + \tan^{-1} 0.75)$.

The acceleration of a particle in simple harmonic motion is $a = -\omega^2 x$. The magnitude of acceleration is maximum when the displacement is at its extreme positions, i.e., $x = \pm 2.0$.

This condition is satisfied when $\cos(50 \pi t + \tan^{-1} 0.75) = \pm 1$.

Let $\phi = \tan^{-1}(0.75) = \tan^{-1} \left(\dfrac{3}{4}\right). The angle$\phi$is approximately$37^{\circ} or \dfrac{37\pi}{180} radians.

At $t = 0$, the phase is $\phi \approx 37^{\circ}$. The first time the magnitude of acceleration reaches its maximum is when the phase equals $\pi$ (or $180^{\circ}$).

Equating the phase to $\pi$:

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

$50 \pi t = \pi - \phi$

Converting to degrees for simpler calculation:

$50 \times 180^{\circ} \times t = 180^{\circ} - 37^{\circ}$

$9000 t = 143$

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

Rounding to appropriate significant figures, we get $t \approx 1.6 \times 10^{-2} \text{ s}$.