A particle executes a linear S.H.M. In two of its positions the velocities are V₁, V₂ and accelerations are a₁ and a₂ respectively (0 < a₁ < a₂). The distance between the positions is (a) (V₁²-V₂²)/(a₁-a₂) (b) (V₂²-V₁²)/(a₁-a₂) (c) (V₁²-V₂²)/(a₁+a₂) (d) (v₂²-v₁²)/((a₁²+a₂²))

Answer: (c). V² = ω²(A² - x²) and a = ω² x (magnitudes). V₁² - V₂² = ω²(x₂² - x₁²) = ω²(x₂ - x₁)(x₂ + x₁) a₁ + a₂ = ω²(x₁ + x₂) Dividing: x₂ - x₁ = (V₁² - V₂²)/(a₁ + a₂) So the answer is (c).

A particle executes a linear S.H.M. In two of its positions the… | MHT CET 2025 (20 Apr, Shift 2)

Q19MHT CET 2025 · 20 APR · SHIFT 2Difficult

Velocity & Acceleration in SHM

A particle executes a linear S.H.M. In two of its positions the velocities are $V_1, V_2$ and accelerations are $a_1$ and $a_2$ respectively $(0 < a_1 < a_2)$ . The distance between the positions is

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Answer: (c)

$V^2 = \omega^2(A^2 - x^2)$ and $a = \omega^2 x$ (magnitudes).
$V_1^2 - V_2^2 = \omega^2(x_2^2 - x_1^2) = \omega^2(x_2 - x_1)(x_2 + x_1)$
$a_1 + a_2 = \omega^2(x_1 + x_2)$
Dividing: $x_2 - x_1 = \dfrac{V_1^2 - V_2^2}{a_1 + a_2}$
So the answer is (c).

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