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What happens to the c2sin(t) part of the worked solution?
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[QUOTE="StillAnotherDave, post: 6350530, member: 669331"] [B]Homework Statement:[/B] Why does the final solution of the equation of motion for damped oscillation not have a sine function? [B]Relevant Equations:[/B] $$x(t)=Ae^{-bt/2m} cos(ωt+φ)$$ Hello folks, So the solution of the equation of motion for damped oscillation is as stated above. If we were to take an specific example such as: $$\frac{d^2x}{dt^2}+4\frac{dx}{dt}+5x=0$$ then the worked solution to the second order homogeneous is: $$x\left(t\right)=e^{-2t}\left(c_1cos(t)+c_2sin(t)\right)$$ What happens to the $$c_2sin(t)$$ part of the worked soution? Why is it not part of the actual solution of the equation of motion? Or does the sine function give the phase constant phi? [/QUOTE]
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Introductory Physics Homework Help
What happens to the c2sin(t) part of the worked solution?
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