Question

A particle is making simple harmonic motion along the X-axis. If at a distances \(x_1\) and \(x_2\) from the mean position the velocities of the particle are \(v_1\) and \(v_2\) respectively. The time period of its oscillation is given as :

• A. \(T = 2\pi\sqrt{\dfrac{x_2^2 + x_1^2}{v_1^2 - v_2^2}}\)
• B. \(T = 2\pi\sqrt{\dfrac{x_2^2 + x_1^2}{v_1^2 + v_2^2}}\)
• C. \(T = 2\pi\sqrt{\dfrac{x_2^2 - x_1^2}{v_1^2 + v_2^2}}\)
• D. \(T = 2\pi\sqrt{\dfrac{x_2^2 - x_1^2}{v_1^2 - v_2^2}}\)

Hint:

Use \(v^2 = \omega^2(A^2 - x^2)\) at the two positions and subtract to eliminate the amplitude.

Answer 1

The Correct Option is D

For SHM, \(v^2 = \omega^2(A^2 - x^2)\).
At \(x_1\): \(v_1^2 = \omega^2(A^2 - x_1^2)\). At \(x_2\): \(v_2^2 = \omega^2(A^2 - x_2^2)\).
Subtract: \(v_1^2 - v_2^2 = \omega^2(x_2^2 - x_1^2)\), so \(\omega^2 = \dfrac{v_1^2 - v_2^2}{x_2^2 - x_1^2}\).
\(T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{x_2^2 - x_1^2}{v_1^2 - v_2^2}}\).
Answer: (4) \(T = 2\pi\sqrt{\dfrac{x_2^2 - x_1^2}{v_1^2 - v_2^2}}\)