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

Answer 1

Answer: (B)

For particle undergoing SHM,
$V=\omega \sqrt{A^2-x^2}\Rightarrow V^2={\left(\omega \right)}^2\left(A^2-x^2\right)$
$V_1=\omega \sqrt{A^2-x_1^2}\Rightarrow V_1^2={\left(\omega \right)}^2\left(A^2-x_1^2\right)$ ...(i)
$V_2=\omega \sqrt{A^2-x_2^2}\Rightarrow V_2^2={\left(\omega \right)}^2\left(A^2-x_2^2\right)$ ...(ii)
$V_1^2-V_2^2={\left(\omega \right)}^2\left(x_2^2-x_1^2\right)$
$\omega =\sqrt{\frac{V_1^2-V_2^2}{x_2^2-x_1^2}}$
$T=2\pi \sqrt{\frac{x_2^2-x_1^2}{V_1^2-V_2^2}}$