Question

The acceleration of a particle performing S.H.M. is 12 cm/sec² at a distance of 3 cm from the mean position. Its time period is

• A. 0.5 sec
• B. 1.0 sec
• C. 2.0 sec
• D. 3.14 sec

Answer 1

Answer: (D)

3.14 sec

Answer 2

$A = \omega^{2}y$ ⇒ $\omega = \sqrt{\frac{A}{y}} = \sqrt{\frac{12}{3}} = 2$;

but $T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi = 3.14$