Question

Part of SHM is graphed in the figure. Here, y is the displacement from mean position. The correct equation describing the SHM is:

• A. $y=4\cos \left(0.6t\right)$
• B. $y=2\sin (\frac{10}{3}t-\frac{\pi }{2})$
• C. $y=2\sin (\frac{\pi }{2}-\frac{10}{3}t)$
• D. $y=2\cos (0.6t+\frac{\pi }{2})$

Answer 1

Answer: (B)

$y=2\sin (\frac{10}{3}t-\frac{\pi }{2})$

$T=0.6\pi$
$\therefore$ $\omega =\frac{2\pi }{T}=\frac{2\pi }{0.6\pi }=\frac{10}{3}$ rad/s
Particle starts from $y=-A=-2$ cm
$\therefore$ $y=-A\cos \omega t$
$=A\sin (\omega t-\pi /2)$
$=2\sin (\frac{10}{3}t-\pi /2)$