Question

A particle is executing SHM along a straight line. Its velocities at distances $x_1$ and $x_2$ from the mean position are $v_1$ and $v_2$ respectively. Its time period of oscillation is

• A. $2\pi \sqrt{\frac{x_2^2-x_1^2}{v_1^2-v_2^2}}$
• B. $2\pi \sqrt{\frac{v_1^2+v_2^2}{x_1^2+x_2^2}}$
• C. $2\pi \sqrt{\frac{v_1^2-v_2^2}{x_1^2-x_2^2}}$
• D. $2\pi \sqrt{\frac{x_1^2+x_2^2}{v_1^2+v_2^2}}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{x_2^2-x_1^2}{v_1^2-v_2^2}}$

For SHM, the velocity of particle at a distance $x$ from mean position is given by,
$v=\omega \sqrt{A^2-x^2}$
$\Rightarrow v_1=\omega \sqrt{A^2-x_1^2}$
$\Rightarrow v_1^2={\omega }^2(A^2-x_1^2)......\left(i\right)$
Similarly,
$v_2^2={\omega }^2(A^2-x_2^2).....\left(ii\right)$
Subtracting $\left(ii\right)$ from equation $\left(i\right)$;
$v_1^2-v_2^2={\omega }^2(x_2^2-x_1^2)$
$\Rightarrow \omega =\sqrt{\frac{v_1^2-v_2^2}{x_2^2-x_1^2}}$
So,
$T=\frac{2\pi }{\omega }$
$\Rightarrow T=2\pi \sqrt{\frac{x_2^2-x_1^2}{v_1^2-v_2^2}}$