A point mass oscillates along the x -axis according to the law x = x₀ cos (ω t - π/4). If the acceleration of the particle is written as a = Acos (ω t + δ), then (a) A = x₀ω², δ = 3π/4 (b) A = x₀, δ = -π/4 (c) A = x₀ω², δ = π/4 (d) A = x₀ω², δ = -π/4

Answer: (a). a = (d²x)/(dt²) = -ω² x₀cos (ω t - π/4). Use -cos θ = cos (θ + π): a = x₀ω²cos (ω t - π/4 + π) = x₀ω²cos (ω t + 3π/4). So A = x₀ω², δ = 3π/4. So the answer is (a).

A point mass oscillates along the x -axis according to the law x = x₀… | JEE Main 2007 (AIEEE)

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Periodic Motion & SHM Basics

A point mass oscillates along the $x$ -axis according to the law $x = x_0 \cos(\omega t - \pi/4).$ If the acceleration of the particle is written as $a = A\cos(\omega t + \delta),$ then

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Answer: (a)

$a = \frac{d^2x}{dt^2} = -\omega^2 x_0\cos(\omega t - \pi/4)$ .
Use $-\cos\theta = \cos(\theta + \pi)$ : $a = x_0\omega^2\cos(\omega t - \pi/4 + \pi) = x_0\omega^2\cos(\omega t + 3\pi/4)$ .
So $A = x_0\omega^2$ , $\delta = 3\pi/4$ .
So the answer is (a).

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