Question

A particle executing SHM has a maximum speed of 30 cm s–1 and a maximum acceleration of 60 cm s–1. The period of oscillation is

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

Answer 1

Answer: (A)

$\pi$ s

Answer 2

Maximum speed, $v_{\max}$… (i)

Maximum acceleration ${a2}_{\max}$.. (ii)

Divide (ii) by (i), we get

$\frac{a_{\max}}{v_{\max} = \frac{\omega^{2}A}{\omega A} = \omega}$

$\therefore\frac{a_{\max}}{v_{\max} = \frac{2\pi}{T}}$

$T = 2\pi\left( \frac{a_{\max}}{v_{\max}} \right)$

Here, ${v - {1 - 2}_{\max}}_{\max}$

$\therefore T = 2\pi\left( \frac{30cms^{- 1}}{60cms^{- 1}} \right) = \pi s$