Question

question 13
problem reference 10.2
a simple harmonic oscillator vibrates back and forth and its displacement as a function of time is given by
$x(t) = 0.500\\ m\\ \cos\left(\frac{\pi}{3} t\
ight)$
what is the period of the oscillator?
\\(\bigcirc\\) 1.50 s
\\(\bigcirc\\) 6.00 s
\\(\bigcirc\\) 3.00 s
\\(\bigcirc\\) 0.667 s

Explanation:

Step1: Identify angular frequency

Standard SHM form: $x(t)=A\cos(\omega t)$. Here $\omega=\frac{\pi}{3}$ rad/s.

Step2: Calculate period

Period formula: $T=\frac{2\pi}{\omega}$. Substitute $\omega$: $T=\frac{2\pi}{\pi/3}=6$ s.

Answer:

B. 6.00 s