Question

The velocities of a particle in SHM at positions $x_1$ and $x_2$ are $v_1$ and $v_2$ respectively, its time period will be -

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

Answer 1

The correct option is C $2\pi \sqrt{(x_1^2-x_2^2)/(v_2^2-v_1^2)}$

Velocity of particle in SHM
$v=\omega \sqrt{A^2-X^2}$
Given that, the velocities of a particle in SHM at positions $x_1$ and $x_2$ are $v_1$ and $v_2.$
Hence
$v_1^2={\omega }^2(A^2-x_1^2)......\left(i\right)$
$v_2^2={\omega }^2(A^2-x_2^2).......\left(ii\right)$
Equation (ii) - (i)
$v_2^2-v_1^2={\omega }^2(x_1^2-x_2^2)$
${\omega }^2=\frac{v_2^2-v_1^2}{x_1^2-x_2^2};{\left(\frac{2\pi }{T}\right)}^2=\frac{v_2^2-v_1^2}{x_1^2-x_2^2}[\omega =\frac{2\pi }{T}]$
$\frac{T^2}{4{\pi }^2}=\frac{x_1^2-x_2^2}{v_2^2-v_1^2};T=2\pi \sqrt{\frac{x_1^2-x_2^2}{v_2^2-v_1^2}}$.