A body oscillates with simple harmonic motion along the x axis. Its displacement varies with time according to the equation x = 5.0 cos (πt). The magnitude of the acceleration (in m/s²) of the body at t = 1.0 s is approximately A) 3.5 B) 49 C) 14 D) 43 E) 4.3

Question

Quizplus · Accepted Answer

The acceleration in simple harmonic motion (SHM) is given by $a = -\omega^2 x$, where $\omega$ is the angular frequency and $x$ is the displacement. From the given equation $x = 5.0 \cos(\pi t)$, we can identify $\omega = \pi$. At $t = 1.0$ s, $x = 5.0 \cos(\pi \times 1.0) = 5.0 \cos(\pi) = -5.0$ m (since $\cos(\pi) = -1$). Thus, $a = -(\pi)^2 \times (-5.0) = \pi^2 \times 5.0$. Given $\pi \approx 3.14$, $a \approx 3.14^2 \times 5.0 \approx 9.86 \times 5.0 \approx 49$ m/s$^2$.

A Body Oscillates with Simple Harmonic Motion Along the X