Question

A particle is executing SHM along a straight line. It velocities at distances $x_1$ and $x_2$ mean position are $V_1$ and $V_2$, respecilvely. Its time period is :

• A. $2 \pi \sqrt{\frac{V_1^2 + V_2^2}{x_1^2 + x_2^2}}$
• B. $2 \pi \sqrt{\frac{V_1^2 - V_2^2}{x_1^2 - x_2^2}}$
• C. $2 \pi \sqrt{\frac{x_1^2 + x_2^2}{V_1^2 + V_2^2}}$
• D. $2 \pi \sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}$

Answer 1

Answer: (D)

$2 \pi \sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}$

Answer 2

For particle undergoing SHM
${V = \omega \sqrt{A^2 - X^2}}$
so {V_1 = \omega \sqrt{A^2 - x^2_}} & V_2 = $\omega \sqrt{ A ^2 - x^2_2}$
solving these two equation we get
$\omega = \sqrt{\frac{V^2_1 -V^2_2}{x^2_2 - x^2_1}}= \frac{2\pi}{T}$
$\Rightarrow T = 2\pi \sqrt{\frac{x^2_2 -x^2_1}{V^2_1 -V^2_2}}$.