Question

A particle executing SHM has maximum acceleration of $2\pi {\text{m/s}}^2$ and maximum velocity of $6\text{m/s}$. Find its time period

• A. $\frac{{\pi }^2}{2}\text{s}$
• B. $\frac{2{\pi }^2}{3}\text{s}$
• C. $6\text{s}$
• D. $\frac{\pi }{3}\text{s}$

Answer 1

The correct option is C $6\text{s}$

Given,
Maximum acceleration $\left(a_{\text{max}}\right)=2\pi {\text{m/s}}^2$
Maximum velocity $\left(v_{\text{max}}\right)=6\text{m/s}$
Since, the particle executes SHM, we can use
$a_{\text{max}}={\omega }^{2}A$ ... (1)
$v_{\text{max}}=\omega A$ ... (2)
From (1) and (2)
$\omega =\frac{a_{\text{max}}}{v_{\text{max}}}$
$\Rightarrow T=\frac{2\pi v_{\text{max}}}{a_{\text{max}}}=\frac{2\pi \times 6}{2\pi }=6\text{s}$