Question

A particle executing SHM has a maximum speed of $30\, cm\, s^{-1}$ and a maximum acceleration of $60 \,cm \,s^{-2}$. 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, $v_{\text{max}} = \omega A\quad ...(i)$ Maximum accelaration, $a_{\text{max}} = \omega^2 A \quad ...(ii)$ Divide $(ii)$ by $(i)$, we get $\frac{a_{max}}{v_{max}} = \frac{\omega^{2}A}{\omega A} = \omega$ $\therefore \frac{a_{max}}{v_{max }} = \frac{2\pi}{T} ,$ $T = 2\pi \left(\frac{v_{max}}{a_{max}}\right)$ Here, $v_{max} = 30 \, cm \,s^{-1},$ $a_{max} = 60 \, cm \,s^{-2}$ $\therefore T =2\pi\left(\frac{30 \,cm \, s^{-1}}{60 \, cm \,s^{-2}}\right)$ $= \pi s$