Amplitude of damped mass-spring system

[lampCable]

Homework Statement

Given the IVP
[tex]
\ddot{r}+\dot{r}+20r=0\\
r(0) = 0.8\\
\dot{r}(0) = 0
[/tex]
for the length of an oscillating spring (damped), we find that the general solution is
[tex]
r=e^{-0.5t}[0.8\cos(\sqrt{19.75}t)+\frac{0.4}{\sqrt{19.75}}\sin(\sqrt{19.75}t)]
[/tex]
and I wish to find the curve bounding the upper part of the solution, i.e. the amplitude of the oscillations.

Homework Equations

The Attempt at a Solution

I understand that if
[tex]
r=e^{-0.5t}0.8\cos(\sqrt{19.75}t)
[/tex]
for example, then the amplitude would be obtained by removing the cosine factor, since [itex]e^{-0.5t}0.8[/itex] correspond to the amplitude and [itex]\cos(\sqrt{19.75}t)[/itex] corresponds to how fast the spring is oscillating. But how does one do when there are two oscillating terms as above?

 

[LCKurtz]

Homework Statement

Given the IVP
[tex]
\ddot{r}+\dot{r}+20r=0\\
r(0) = 0.8\\
\dot{r}(0) = 0
[/tex]
for the length of an oscillating spring (damped), we find that the general solution is
[tex]
r=e^{-0.5t}[0.8\cos(\sqrt{19.75}t)+\frac{0.4}{\sqrt{19.75}}\sin(\sqrt{19.75}t)]
[/tex]
and I wish to find the curve bounding the upper part of the solution, i.e. the amplitude of the oscillations.

Homework Equations

The Attempt at a Solution

I understand that if
[tex]
r=e^{-0.5t}0.8\cos(\sqrt{19.75}t)
[/tex]
for example, then the amplitude would be obtained by removing the cosine factor, since [itex]e^{-0.5t}0.8[/itex] correspond to the amplitude and [itex]\cos(\sqrt{19.75}t)[/itex] corresponds to how fast the spring is oscillating. But how does one do when there are two oscillating terms as above?

Use$$A\cos\theta + B\sin\theta = \sqrt{A^2+B^2}\left(\frac A {\sqrt{A^2+B^2}}\cos\theta + \frac B {\sqrt{A^2+B^2}}\sin\theta \right)$$ $$=\sqrt{A^2+B^2}\cos(\theta-\alpha)$$