A particle executing linear S.H.M. has velocities V₁ and V₂ at distance x₁ and x₂ respectively, from the mean position, its angular velocity is (a) √((V₂² - V₁²)/(x₁² - x₂²)) (b) √((V₂² - V₁²)/(x₂² - x₁²)) (c) √((V₂² - V₁²)/(x₁ x₂)) (d) √((V₁ V₂)/(x₁ + x₂))

Answer: (a). V₁² = ω²(A² - x₁²) and V₂² = ω²(A² - x₂²). Subtract: V₂² - V₁² = ω²(x₁² - x₂²). ω = √((V₂² - V₁²)/(x₁² - x₂²)). So the answer is (a).

A particle executing linear S.H.M. has velocities V₁ and V₂ at… | MHT CET 2026 (19 Apr, Shift 2)

Q12MHT CET 2026 · 19 APR · SHIFT 2Medium

Velocity & Acceleration in SHM

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

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Answer: (a)

$V_1^2 = \omega^2(A^2 - x_1^2)$ and $V_2^2 = \omega^2(A^2 - x_2^2)$ .
Subtract: $V_2^2 - V_1^2 = \omega^2(x_1^2 - x_2^2)$ .
$\omega = \sqrt{\frac{V_2^2 - V_1^2}{x_1^2 - x_2^2}}$ .
So the answer is (a).

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Velocity & Acceleration in SHM question from MHT CET 2026 (19 Apr, Shift 2)

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