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In my school textbook, I have the following explanation to Damped SHM:

[tex]\hline [/tex]

Soving the differential equation:

[tex] m \frac{d^2x}{dt^2}+\gamma m \frac{dx}{dt}+m\omega _0 ^2x = 0[/tex] (Eq. 1)

gives

[tex] x(t)=x_{m}e^{-\frac{\gamma}{2}t}\cos \left( \omega _{\gamma}t + \phi \right), [/tex] (Eq. 2)

where:

[tex] \gamma = \frac{b}{m} [/tex]

and

[tex] \omega _{\gamma}=\sqrt{\omega _0 ^2 - \frac{b^2}{4m^2}} [/tex]

and

[tex] \omega _0 = \sqrt{\frac{k}{m}}. [/tex]

[tex] \hline [/tex]

It goes as simply as that. However, I've read on the web that this is an under damped SHM, and that the answer really goes like this:

[tex]\hline [/tex]

[tex] x(t)=Ae^{-\frac{\gamma}{2}t}\cos \left( \omega _{\gamma}t + \phi _1 \right)+Be^{-\frac{\gamma}{2}t}\cos \left(- \omega _{\gamma}t + \phi _2 \right) [/tex] (Eq. 3)

[tex]\hline [/tex]

My question is: How do I go about finding (Eq. 2) out of (Eq. 3). Is there any special trick?

Thank you very much.

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# Damped SHM

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