Question

Write mathematical expression which describes the relation between displacement and velocity.​

Answer 1

Answer:

We will first derive the expression for the velocity of the particle which executes simple harmonic motion by differentiating its displacement from the equilibrium position with respect to time. We know that the displacement from equilibrium position is given by:

x=Asin(ωt+ϕ)x=Asin⁡(ωt+ϕ)

v=dxdtv=dxdt

⇒v=ddt(Asin(ωt+ϕ))=ωAcos(ωt+ϕ)⇒v=ddt(Asin⁡(ωt+ϕ))=ωAcos⁡(ωt+ϕ)

Now, at the equilibrium, initial phase angle is zero which means ϕ=0ϕ=0

⇒v=ωAcosωt=ωA1−sin2ωt−−−−−−−−√⇒v=ωAcos⁡ωt=ωA1−sin2ωt

Also at equilibrium, 

x=Asinωt⇒sin2ωt=x2A2x=Asin⁡ωt⇒sin2ωt=x2A2

Putting this value in the equation of velocity, we get

v=Aω1−x2A2

Explanation:

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Answer 2

Answer:

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