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Simple Harmonic Motion question

2007 · Shift 0 · Q87
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Simple Harmonic Motion question

2007 · Shift 0 · Q87

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
A point mass oscillates along the xxx-axis according to the law x=x0 cos⁡(ωt−π/4).x = {x_0}\,\cos \left( {\omega t - \pi /4} \right).x=x0​cos(ωt−π/4). If the acceleration of the particle is written as a=A cos⁡(ωt+δ),a = A\,\cos \left( {\omega t + \delta } \right),a=Acos(ωt+δ), then
  1. A
    A=x0ω2,  δ=3π/4A = {x_0}{\omega ^2},\,\,\delta = 3\pi /4A=x0​ω2,δ=3π/4
  2. B
    A=x0,  δ=−π/4A = {x_0},\,\,\delta = - \pi /4A=x0​,δ=−π/4
  3. C
    A=x0ω2,  δ=π/4A = {x_0}{\omega ^2},\,\,\delta = \pi /4A=x0​ω2,δ=π/4
  4. D
    A=x0ω2,  δ=−π/4A = {x_0}{\omega ^2},\,\,\delta = - \pi /4A=x0​ω2,δ=−π/4
View written solutionFree

Correct answer: A

  1. The displacement is given by x=x0cos⁡(ωt−π4).x = x_0\cos\left(\omega t - \frac{\pi}{4}\right).x=x0​cos(ωt−4π​).

  2. In SHM, acceleration is the second derivative of displacement: a=d2xdt2=−ω2x.a = \frac{d^2x}{dt^2} = -\omega^2 x.a=dt2d2x​=−ω2x. So, a=−ω2x0cos⁡(ωt−π4).a = -\omega^2 x_0\cos\left(\omega t - \frac{\pi}{4}\right).a=−ω2x0​cos(ωt−4π​).

  3. Now write the minus sign as a phase shift of π\piπ: −cos⁡θ=cos⁡(θ+π).-\cos\theta = \cos(\theta + \pi).−cosθ=cos(θ+π). Hence, a=x0ω2cos⁡(ωt−π4+π).a = x_0\omega^2 \cos\left(\omega t - \frac{\pi}{4} + \pi\right).a=x0​ω2cos(ωt−4π​+π).

  4. Simplify the phase: −π4+π=3π4.-\frac{\pi}{4} + \pi = \frac{3\pi}{4}.−4π​+π=43π​. Therefore, a=x0ω2cos⁡(ωt+3π4).a = x_0\omega^2 \cos\left(\omega t + \frac{3\pi}{4}\right).a=x0​ω2cos(ωt+43π​).

  5. Comparing with a=Acos⁡(ωt+δ),a = A\cos(\omega t + \delta),a=Acos(ωt+δ), we get A=x0ω2,δ=3π4.A = x_0\omega^2, \qquad \delta = \frac{3\pi}{4}.A=x0​ω2,δ=43π​.

  6. Check options:

  • A: A=x0ω2, δ=3π/4A = x_0\omega^2,\ \delta = 3\pi/4A=x0​ω2, δ=3π/4 ✅
  • B: wrong amplitude
  • C: wrong phase
  • D: wrong phase

Therefore, the correct option is A.

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