A particle performs linear S.H.M. At a particular instant, velocity of the particle is ' u ' and acceleration is ' a₁ ' while at another instant velocity is ' v ' and acceleration is ' a₂ ' (0 < a₁ < a₂). The distance between the two positions is (a) (v²-u²)/(a₁-a₂) (b) (v²+u²)/(a₁+a₂) (c) (u²+v²)/(a₁-a₂) (d) (u²-v²)/(a₁+a₂)

Answer: (d). u² = ω²(A² - x₁²), v² = ω²(A² - x₂²), a₁ = ω² x₁, a₂ = ω² x₂. u² - v² = ω²(x₂ - x₁)(x₂ + x₁) and a₁ + a₂ = ω²(x₁ + x₂) x₂ - x₁ = (u² - v²)/(a₁ + a₂) So the answer is (d).

A particle performs linear S.H.M. At a particular instant, velocity… | MHT CET 2024 (11 May, Shift 1)

Q38MHT CET 2024 · 11 MAY · SHIFT 1Difficult

Velocity & Acceleration in SHM

A particle performs linear S.H.M. At a particular instant, velocity of the particle is ' $u$ ' and acceleration is ' $a_1$ ' while at another instant velocity is ' $v$ ' and acceleration is ' $a_2$ ' $(0 < a_1 < a_2)$ . The distance between the two positions is

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Answer: (d)

$u^2 = \omega^2(A^2 - x_1^2)$ , $v^2 = \omega^2(A^2 - x_2^2)$ , $a_1 = \omega^2 x_1$ , $a_2 = \omega^2 x_2$ .
$u^2 - v^2 = \omega^2(x_2 - x_1)(x_2 + x_1)$ and $a_1 + a_2 = \omega^2(x_1 + x_2)$
$x_2 - x_1 = \dfrac{u^2 - v^2}{a_1 + a_2}$
So the answer is (d).

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Velocity & Acceleration in SHM question from MHT CET 2024 (11 May, Shift 1)

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