The maximum velocity and maximum acceleration of a particle performing a linear S.H.M. is ' α ' and ' β ' respectively. Then the path length of the particle is (a) (α²)/(β) (b) (βα²)/(2α²) (c) (2α²)/(β) (d) (2β)/(α²)

Answer: (c). vmax = Aω = α, amax = Aω² = β (α²)/(β) = (A²ω²)/(Aω²) = A Path length = 2A = (2α²)/(β) So the answer is (c) (2α²)/(β).

The maximum velocity and maximum acceleration of a particle… | MHT CET 2024 (4 May, Shift 2)

Q25MHT CET 2024 · 4 MAY · SHIFT 2Easy

Velocity & Acceleration in SHM

The maximum velocity and maximum acceleration of a particle performing a linear S.H.M. is ' $\alpha$ ' and ' $\beta$ ' respectively. Then the path length of the particle is

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Answer: (c)

$v_{max} = A\omega = \alpha$ , $a_{max} = A\omega^2 = \beta$
$\frac{\alpha^2}{\beta} = \frac{A^2\omega^2}{A\omega^2} = A$
Path length $= 2A = \frac{2\alpha^2}{\beta}$
So the answer is (c) $\frac{2\alpha^2}{\beta}$ .

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