Question

A particle is describing simple harmonic motion. If its velocities are $v_1$ and $v_2$ when the displacements from the mean position are $y_1$ and $y_2$ respectively, then its time period is

• A. $2\pi \sqrt{\frac{y_1^2+y_2^2}{v_1^2+v_2^2}}$
• B. $2\pi \sqrt{\frac{v_2^2-v_1^2}{y_1^2-y_2^2}}$
• C. $2\pi \sqrt{\frac{v_2^2+v_1^2}{y_1^2+y_2^2}}$
• D. $2\pi \sqrt{\frac{y_1^2-y_2^2}{v_1^2-v_2^2}}$

Answer 1

Answer: (D)

$2\pi \sqrt{\frac{y_1^2-y_2^2}{v_1^2-v_2^2}}$

In simple harmonic motion,
velocity $v=\omega \sqrt{A^2-y^2}$
$\therefore v_1=\omega \sqrt{A^2-y_1^2}\Rightarrow v_1^2={\omega }^2{A}^2-{\omega }^2{y}_1^2....\left(i\right)$
and $v_2=\omega \sqrt{A^2-y_2^2}\Rightarrow v_2^2={\omega }^2{A}^2-{\omega }^2{y}_2^2.....\left(ii\right)$
Solving equations (i) and (ii), we get
$v_2^2-v_1^2={\omega }^2(y_1^2-y_2^2)$
$\omega =\sqrt{\frac{v_2^2-v_1^2}{y_1^2-y_2^2}}$
$\Rightarrow T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{y_1^2-y_2^2}{v_1^2-v_2^2}}$.