Oscillations

JEE Main level, single correct, medium

50 of 194

Question 50 of 194

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10s−1.10{s}^{-1}. At, t=0t=0 the displacement is 5m.5m. What is the maximum acceleration? The initial phase is π4\dfrac{\pi }{4} .

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Answer: (D) 5002ms−2500\sqrt{2}m{s}^{-2}

In a simple harmonic motion,
Maximum acceleration, fmax=ω2A{f}_{max}={\omega }^{2}A

Maximum velocity, vmax⁡=Aω{v}_{\max }=A\omega.

fmax⁡vmax⁡=ω2AAω=ω\dfrac{{f}_{\max }}{{v}_{\max }}=\dfrac{{\omega }^{2}A}{A\omega }=\omega

ω=10\omega =10

Displacement in SHM is x=Asin⁡⁡(ωt+π4)x=A\sin ⁡\left(\omega t+\dfrac{\pi }{4}\right)

At t=0t=0

5=Asin⁡⁡(π4)5=A\sin ⁡\left(\dfrac{\pi }{4}\right)

A=52A=5\sqrt{2}.

Max acceleration = ω2A{\omega }^{2}A

=100×52=100\times 5\sqrt{2}

=5002ms−2=500\sqrt{2}m{s}^{-2}

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