Question #61576

Derive expressions for average energy of a body executing SHM.
1

Expert's answer

2016-09-01T07:49:04-0400

Answer on Question #61576-Physics-Mechanics

Derive expressions for average energy of a body executing SHM.

Solution

U=12kx2=12kA2cos2(ωt+ϕ)U = \frac{1}{2} k x^{2} = \frac{1}{2} k A^{2} \cos^{2}(\omega t + \phi)K=12mv2=12mω2A2sin2(ωt+ϕ)=12kA2sin2(ωt+ϕ)K = \frac{1}{2} m v^{2} = \frac{1}{2} m \omega^{2} A^{2} \sin^{2}(\omega t + \phi) = \frac{1}{2} k A^{2} \sin^{2}(\omega t + \phi)


Since k=mω2k = m\omega^2.


E=K+U=12kA2cos2(ωt+ϕ)+12kA2sin2(ωt+ϕ)=12kA2E = K + U = \frac{1}{2} k A^{2} \cos^{2}(\omega t + \phi) + \frac{1}{2} k A^{2} \sin^{2}(\omega t + \phi) = \frac{1}{2} k A^{2}


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Comments

Assignment Expert
25.08.16, 17:39

Dear bharat, please use panel for submitting new questions

bharat
25.08.16, 17:09

The steady-state amplitude of a weakly damped forced oscillator is given by 2 2 2 2 2 1/ 2 0 0 [( ) 4 ] ( ) w −w + w w = b f a Depict it as a function of frequency and obtain expression for resonance frequency. Also obtain expression for peak value of steady-state amplitude

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