Question

Acceleration-displacement graph of a particle executing SHM is as shown in the figure. The time period of oscillation is (in sec)

• A. $\frac{\pi }{2}$
• B. $2\pi$
• C. $\pi$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (B)

$2\pi$

In SHM
$a=-{\omega }^{2}x$
or
$a=\frac{-K}{m}x$
so, from graph
$-\frac{K}{m}=-1$ $(\because slopeis-1)$
$\frac{K}{m}=1$
Time period $=2\pi \sqrt{\frac{m}{K}}$
$=2\pi \sqrt{\frac{1}{1}}$
$=2\pi$