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A point mass oscillates along the x-axis according to the law x=x_0 cos (ω t-π / 4). If the acceleration of the particle is written as a=A cos (ω t+δ)

JEE Main 2007 — Physics Waves & Oscillations

2007mcqmedium

A point mass oscillates along the xxx-axis according to the law x=x0cos⁡(ωt−π/4)\mathrm{x=x_0 \cos (\omega t-\pi / 4)}x=x0​cos(ωt−π/4). If the acceleration of the particle is written as a=Acos⁡(ωt+δ)\mathrm{a=A \cos (\omega t+\delta)}a=Acos(ωt+δ)

Official previous-year question

Held on 30 Apr 2007 · Verified 6 Jul 2026.

Options

  1. A

    A=x0,δ=−π/4\mathrm{A}=\mathrm{x}_0, \delta=-\pi / 4A=x0​,δ=−π/4

  2. B

    A=x0ω2,δ=−π/4\mathrm{A}=\mathrm{x}_0 \omega^2, \delta=-\pi / 4A=x0​ω2,δ=−π/4

  3. C

    A=x0ω2,δ=−π/4\mathrm{A}=\mathrm{x}_0 \omega^2, \delta=-\pi / 4A=x0​ω2,δ=−π/4

  4. D

    A=x0ω2,δ=3π/4\mathrm{A=x_0 \omega^2, \delta=3 \pi / 4}A=x0​ω2,δ=3π/4

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