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JEE Main 2007
Exam
JEE Main
Year
2007
Subject
Physics
Chapter
Oscillations

JEE Main 2007 · Physics · Oscillations

A point mass oscillates along the x -axis according to the law...

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

Options

  1. AA=x0,δ=−π/4\mathrm{A}=\mathrm{x}_0, \delta=-\pi / 4
  2. BA=x0ω2,δ=−π/4\mathrm{A}=\mathrm{x}_0 \omega^2, \delta=-\pi / 4
  3. CA=x0ω2,δ=−π/4\mathrm{A}=\mathrm{x}_0 \omega^2, \delta=-\pi / 4
  4. DA=x0ω2,δ=3π/4\mathrm{A=x_0 \omega^2, \delta=3 \pi / 4}

Correct answer

D. A=x0ω2,δ=3π/4\mathrm{A=x_0 \omega^2, \delta=3 \pi / 4}

Step-by-step solution

\mathrm{v=-x_0 \omega \sin (\omega t-\pi / 4)} \mathrm{a=-x_0 \omega^2 \cos \left(\omega t+\pi-\frac{\pi}{4}\right)} a=Acos⁡(ωt+δ)\mathrm{a=A \cos (\omega t+\delta)} A=x0ω2;δ=3π4\mathrm{A=x_0 \omega^2 ; \quad \delta=\frac{3 \pi}{4}}
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