The equation is $$ m\ddot x =-k x -\gamma x$$ Multiply by $1/m$ we get:

$$ \ddot x=-\omega_0^2x - \beta x $$

We use the ansatz $x(t)=e^{\lambda t}$

So for the $\lambda_{1,2}$ we get:

$$ \lambda_{1,2}=-\frac{\beta}{2} \pm \sqrt{\beta^2/4-\omega_0^2} $$

Under-damping means $\beta<2\omega_0$, so we have an imaginary term under the $\sqrt{.... }$

So: $$ \lambda_{1,2}= -\frac{\beta}{2} \pm \sqrt{i^2(\beta^2/4-\omega_0^2)} = -\frac{\beta}{2} \pm i \sqrt{(\beta^2/4-\omega_0^2)}= -\frac{\beta}{2} \pm i \omega$$

Where $\omega=\sqrt{(\beta^2/4-\omega_0^2)}$

The solution for $x(t)$: $$ x(t)=A_{+} e^{-\frac{\beta t}{2}} e^{i\omega t} + A_- e^{-\frac{\beta t}{2}}e^{-i\omega t} = e^{-\frac{\beta t}{2}}(A_+ e^{i\omega t} + A_-e^{-i\omega t})$$

The question is, how can I get the following for $x(t)$: $$ x(t)=e^{-\frac{\beta t}{2}}(A_1 \cos(\omega t)+A_2 \sin(\omega t)) $$

I don't see it, and it's bugging me very much. It stops me from going on with my study.


1 Answer 1


The question is easy.

The key points of your problem are that $A_+$ and $A_-$ are both complex numbers and $A_+e^{i\omega t}+A_-e^{-i\omega t}= \text{Real Number}$, because we can't have a imaginary displacement $x(t)$.

Next, we start to solve your problem from the equation $$ x(t)=e^{-\frac{\beta t}{2}}(A_+e^{i\omega t}+A_-e^{-i\omega t}) $$ by using Euler's formula $e^{i\theta}=\cos\theta+i\sin\theta$, we have $$ A_+e^{i\omega t} =A_+\cos\omega t+iA_+\sin\omega t\\ A_-e^{-i\omega t} =A_-\cos\omega t-iA_-\sin\omega t $$

Then we add above two equations and get $$ A_+e^{i\omega t}+A_-e^{-i\omega t}=(A_++A_-)\cos\omega t+i(A_+-A_-)\sin\omega t $$ Because I told you the left side of the above equation is real previously, and $\cos\omega t$, $\sin\omega t$ are both real, we can get $$ A_1 =A_++A_-=\text{Real Constant}\\ A_2 =i(A_+-A_-)=\text{Real Constant} $$

Finally, we get the result, $$ x(t)=e^{-\frac{\beta t}{2}}(A_1\cos\omega t+A_2\sin\omega t) $$


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