A particle is executing SHM along a straight line. Its velocities at distances $x_1$ and $x_2$ from the mean p

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:

Accepted Answer

1.  **Velocity-Displacement Formula:** The velocity $v$ of a particle performing Simple Harmonic Motion at a displacement $x$ from the mean position is given by:
    $$ v = \omega \sqrt{A^2 - x^2} $$
    where $\omega$ is the angular frequency and $A$ is the amplitude.
2.  **Formulate Equations:** Squaring the relation gives $v^2 = \omega^2 (A^2 - x^2)$. Applying this to the two given cases:
    $$ v_1^2 = \omega^2 (A^2 - x_1^2) \quad 	ext{--- (i)} $$
    $$ v_2^2 = \omega^2 (A^2 - x_2^2) \quad 	ext{--- (ii)} $$
3.  **Eliminate Amplitude ($A$):** Subtract equation (ii) from equation (i) to eliminate $A$:
    $$ v_1^2 - v_2^2 = \omega^2 (A^2 - x_1^2) - \omega^2 (A^2 - x_2^2) $$
    $$ v_1^2 - v_2^2 = \omega^2 (x_2^2 - x_1^2) $$
4.  **Solve for $\omega$:**
    $$ \omega^2 = \frac{v_1^2 - v_2^2}{x_2^2 - x_1^2} \implies \omega = \sqrt{\frac{v_1^2 - v_2^2}{x_2^2 - x_1^2}} $$
5.  **Calculate Time Period ($T$):** The time period is $T = \frac{2\pi}{\omega}$. Substituting the expression for $\omega$:
    $$ T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}} $$

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