A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean position are V1 and V2, respectively. Its time period is

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) \( T = 2\pi \sqrt{\frac{V_{1}^2 + V_{2}^2}{x_{1}^2 + x_{2}^2}} \)

(b) \( T = 2\pi \sqrt{\frac{V_{1}^2 - V_{2}^2}{x_{1}^2 - x_{2}^2}} \)

(c) \( T = 2\pi \sqrt{\frac{x_{1}^2 + x_{2}^2}{V_{1}^2 + V_{2}^2}} \)

(d) \( T = 2\pi \sqrt{\frac{x_{2}^2 - x_{1}^2}{V_{1}^2 - V_{2}^2}} \)

Detailed Answer:

The particle is undergoing Simple Harmonic Motion (SHM). The relationship between velocity \(v\), angular frequency \(\omega\), amplitude \(A\), and displacement \(x\) in SHM is given by: \[ v = \sqrt{\omega^2 (A^2 - x^2)}. \] For the two given velocities \(V_1\) and \(V_2\) at distances \(x_1\) and \(x_2\) from the mean position, we can write: \[ V_1 = \sqrt{\omega^2 (A^2 - x_1^2)} \quad \text{and} \quad V_2 = \sqrt{\omega^2 (A^2 - x_2^2)}. \] Squaring these equations: \[ V_1^2 = \omega^2 (A^2 - x_1^2) \quad \text{and} \quad V_2^2 = \omega^2 (A^2 - x_2^2). \] Subtract the second equation from the first: \[ V_1^2 - V_2^2 = \omega^2 (x_2^2 - x_1^2). \] Rearrange to solve for \(\omega^2\): \[ \omega^2 = \frac{V_1^2 - V_2^2}{x_2^2 - x_1^2}. \] The angular frequency \(\omega\) is related to the time period \(T\) by: \[ \omega = \frac{2\pi}{T} \quad \Rightarrow \quad T = \frac{2\pi}{\omega}. \] Substituting \(\omega^2\): \[ T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}. \] Therefore, the correct option is: \[ \boxed{T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}.} \] This formula provides the time period of the particle based on the given velocities and distances from the mean position.

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