Question

A particle executing SHM has a maximum speed of $30cm{s}^{-1}$ and a maximum acceleration of $60cm{s}^{-1}$. What is the period of oscillation in seconds is ?

• A. $\mathrm{\pi }$
• B. $\frac{\mathrm{\pi }}{2}$
• C. $2\mathrm{\pi }$
• D. $\frac{2\mathrm{\pi }}{15}$

Answer 1

The correct option is A

$\mathrm{\pi }$

Step 1: Given data

The given particle in SHM has,

Maximum speed, $v_{max}=30cm{s}^{-1}$

Maximum acceleration, $a_{max}=60cm{s}^{-2}$

The period of oscillation, $T=?$

Step 2: Formula used

Displacement, $x=A\sin \omega t$

Velocity, $v=\frac{dx}{dt}=A\omega \cos \omega t$

Maximum velocity, $v_{max}=A\omega$

Acceleration, $a=\frac{dv}{dt}=-A{\omega }^2\sin \omega t$

Maximum acceleration, $a_{max}=A{\omega }^2$

Step 3: Calculating time

$$\begin{gathered} v_{max}=A\omega =30cm{s}^{-1} \\ \Rightarrow A\omega =30----------\left(1\right) \end{gathered}$$

$$\begin{gathered} a_{max}=A{\omega }^2=60cm{s}^{-1} \\ \Rightarrow A{\omega }^2=60----------\left(2\right) \end{gathered}$$

$$\begin{gathered} \frac{\left(2\right)}{\left(1\right)}\Rightarrow \frac{A{\omega }^2}{A\omega }=\frac{60}{30} \\ \Rightarrow \omega =2rad/s \end{gathered}$$

We know,

$$\begin{gathered} \omega =\frac{2\mathrm{\pi }}{\mathrm{T}} \\ 2=\frac{2\mathrm{\pi }}{\mathrm{T}}\left(\because \omega =2rad/s\right) \\ T=\mathrm{\pi }\sec \end{gathered}$$

The period of oscillation in seconds is $\mathrm{\pi }$.

Hence, option A is correct.