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

The Correct Option is D

 To determine the time period of a particle executing Simple Harmonic Motion (SHM) given its velocities at two distances from the mean position, we can use the basic properties of SHM.

In SHM, the velocity \( V \) of the particle at a displacement \( x \) from the mean position is given by: \(V = \pm \omega \sqrt{A^2 - x^2}\), where \( \omega \) is the angular frequency and \( A \) is the amplitude of motion.

For the given distances \( x_1 \) and \( x_2 \) with corresponding velocities \( V_1 \) and \( V_2 \), we can write:

1. From displacement \( x_1 \), \(V_1 = \omega \sqrt{A^2 - x_1^2}\)
2. From displacement \( x_2 \), \(V_2 = \omega \sqrt{A^2 - x_2^2}\)

 

Squaring both equations, we have:

1. \( \omega^2 (A^2 - x_1^2) = V_1^2 \)
2. \( \omega^2 (A^2 - x_2^2) = V_2^2 \)

 

Subtract the second equation from the first to eliminate \( \omega^2 A^2 \): \(\omega^2 (x_2^2 - x_1^2) = V_1^2 - V_2^2\)

Solve for \( \omega^2 \): \(\omega^2 = \frac{V_1^2 - V_2^2}{x_2^2 - x_1^2}\)

Since the time period \( T \) is related to angular frequency by \( T = \frac{2\pi}{\omega} \), substituting for \( \omega \), we get: \(T = 2 \pi \sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}\)

This matches the correct answer. The given options include the correct expression for the time period of the SHM based on the provided conditions.