A particle performs linear S.H.M. At a particular instant, velocity of the particle is ' u ' and acceleration is ' α ' while at another instant, velocity is ' v ' and acceleration is ' β ' (0 < α < β). The distance between the two positions is (a) (u²-v²)/(α+β) (b) (u²+v²)/(α+β) (c) (u²-v²)/(α-β) (d) (u²+v²)/(α-β)

Answer: (a). u² = ω²(A² - x₁²), v² = ω²(A² - x₂²), α = ω² x₁, β = ω² x₂. u² - v² = ω²(x₂ - x₁)(x₂ + x₁) and α + β = ω²(x₁ + x₂) x₂ - x₁ = (u² - v²)/(α + β) So the answer is (a).

A particle performs linear S.H.M. At a particular instant, velocity… | MHT CET 2024 (16 May, Shift 2)

Q37MHT CET 2024 · 16 MAY · SHIFT 2Difficult

Velocity & Acceleration in SHM

A particle performs linear S.H.M. At a particular instant, velocity of the particle is ' $u$ ' and acceleration is ' $\alpha$ ' while at another instant, velocity is ' $v$ ' and acceleration is ' $\beta$ ' $(0 < \alpha < \beta)$ . The distance between the two positions is

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Answer: (a)

$u^2 = \omega^2(A^2 - x_1^2)$ , $v^2 = \omega^2(A^2 - x_2^2)$ , $\alpha = \omega^2 x_1$ , $\beta = \omega^2 x_2$ .
$u^2 - v^2 = \omega^2(x_2 - x_1)(x_2 + x_1)$ and $\alpha + \beta = \omega^2(x_1 + x_2)$
$x_2 - x_1 = \dfrac{u^2 - v^2}{\alpha + \beta}$
So the answer is (a).

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Velocity & Acceleration in SHM question from MHT CET 2024 (16 May, Shift 2)

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