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 is

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

Answer 1

Answer: (A)

$2\pi \sqrt{\frac{x_2^2-x_1^2}{v_1^2-v_2^2}}$

Let $A$ be the amplitude of oscillation, then,

$v_1^2={\omega }^2(A^2-x_1^2)$ .....(1)

$v_2^2={\omega }^2(A^2-x_2^2)$ ....{2}

Subtracting eqn (1) from eqn (2), we get

$v_1^2-v_2^2={\omega }^2(x_2^2-x_1^2)$

$⟹\omega =\sqrt{\frac{v_1^2-v_2^2}{x_2^2-x_1^2}}$

$⟹\frac{2\pi }{T}=\sqrt{\frac{v_1^2-v_2^2}{x_2^2-x_1^2}}$

Time Period, $T=2\pi \sqrt{\frac{x_2^2-x_1^2}{v_1^2-v_2^2}}$