Q121 / 150 · MHT CET 2026 (19 April Shift 2) · Physics
A particle executing linear S.H.M. has velocities $V_1$ and $V_2$ at distance $x_1$ and $x_2$ respectively, from the mean position, its angular velocity is
- Correct answerA$\sqrt{\dfrac{V_2^2 - V_1^2}{x_1^2 - x_2^2}}$
- B$\sqrt{\dfrac{V_2^2 - V_1^2}{x_2^2 - x_1^2}}$
- C$\sqrt{\dfrac{V_2^2 - V_1^2}{x_1 x_2}}$
- D$\sqrt{\dfrac{V_1 V_2}{x_1 + x_2}}$
Solution
The velocity of a particle executing linear S.H.M. at a distance $x$ from the mean position is given by $V = \omega \sqrt{A^2 - x^2}$, where $A$ is the amplitude and $\omega$ is the angular velocity.
Squaring both sides, we get $V^2 = \omega^2 (A^2 - x^2)$.
For the two given positions, we can write:
$V_1^2 = \omega^2 (A^2 - x_1^2)$
$V_2^2 = \omega^2 (A^2 - x_2^2)$
Subtracting the first equation from the second, we get:
$V_2^2 - V_1^2 = \omega^2 (A^2 - x_2^2) - \omega^2 (A^2 - x_1^2)$
$V_2^2 - V_1^2 = \omega^2 (x_1^2 - x_2^2)$
$\omega^2 = \dfrac{V_2^2 - V_1^2}{x_1^2 - x_2^2}$
$\omega = \sqrt{\dfrac{V_2^2 - V_1^2}{x_1^2 - x_2^2}}$
Answer: $\sqrt{\dfrac{V_2^2 - V_1^2}{x_1^2 - x_2^2}}$
Squaring both sides, we get $V^2 = \omega^2 (A^2 - x^2)$.
For the two given positions, we can write:
$V_1^2 = \omega^2 (A^2 - x_1^2)$
$V_2^2 = \omega^2 (A^2 - x_2^2)$
Subtracting the first equation from the second, we get:
$V_2^2 - V_1^2 = \omega^2 (A^2 - x_2^2) - \omega^2 (A^2 - x_1^2)$
$V_2^2 - V_1^2 = \omega^2 (x_1^2 - x_2^2)$
$\omega^2 = \dfrac{V_2^2 - V_1^2}{x_1^2 - x_2^2}$
$\omega = \sqrt{\dfrac{V_2^2 - V_1^2}{x_1^2 - x_2^2}}$
Answer: $\sqrt{\dfrac{V_2^2 - V_1^2}{x_1^2 - x_2^2}}$
Hence, the correct answer is option: 'A'.