Question

A particle is executing SHM along a straight line. Its velocities at distances $2\text{cm}$ and $4\text{cm}$ from the mean position are $3\text{cm/s}$ and $2\text{cm/s}$ respectively. Find the time period of SHM.

• A. $2\pi \sqrt{\frac{12}{5}}\text{s}$
• B. $\pi \sqrt{\frac{12}{5}}\text{s}$
• C. $2\pi \sqrt{\frac{6}{5}}\text{s}$
• D. $\pi \sqrt{\frac{8}{5}}\text{s}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{12}{5}}\text{s}$

Given,
$x_1=2\text{cm};x_2=4\text{cm}$
$v_1=2\text{cm/s};v_2=3\text{cm/s}$
Since, the particle is executing SHM. we use
$v^2={\omega }^2(A^2-x^2)$
From which,
$v_1^2={\omega }^2(A^2-x_1^2)$ ... (1)
$v_2^2={\omega }^2(A^2-x_2^2)$ ... (2)
On subtracting (1) and (2)
$v_1^2-v_2^2={\omega }^2(x_2^2-x_1^2)$
$\Rightarrow \omega =\sqrt{\frac{v_1^2-v_2^2}{x_2^2-x_1^2}}$
We know that,
Time period $\left(T\right)=\frac{2\pi }{\omega }$
$=2\pi \sqrt{\frac{x_2^2-x_1^2}{v_1^2-v_2^2}}$
$=2\times \pi \times \sqrt{\frac{(4{)}^2-(2{)}^2}{(3{)}^2-(2{)}^2}}$
$=2\times \pi \times \sqrt{\frac{12}{5}}$