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)
$V_{1}^{2}-V_{2}^{2}=\left(\omega \right)^{2}\left(x_{2}^{2} - x_{1}^{2}\right)$
$\omega =\sqrt{\frac{V_{1}^{2} - V_{2}^{2}}{x_{2}^{2} - x_{1}^{2}}}$
$T=2\pi \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{V_{1}^{2} - V_{2}^{2}}}$