The general formla is $mx''+cx'+kx=0$. Solving these mass spring systems is difficult for me when it comes to some of the position functions and period, frequency, and amplitude. This first problem does not require me to find the amplitude,period…etc just the damped and undamped position function (when $c = 0$). On the first problem the damped postion function was easy to find but the undamped was not. I really need a thorough explanation behind the ideas here and some of the algebra.
For example I am given the problem: $x{''} + 6x'+8x=0$ with initial values of position and velocity being, $x_0=2, v_0=0$. The general solution comes to $y(x)=4e^{-2x}-2e^{-4x}$ with $2,4$ being the constants that came about by plugging in the initial conditions.
Clearly the un-damped position function is $x{''}+8x=0$. But we are given the following formulas. The undamped position function denoted $u(t)$ is $u(t)=R\cos(\mathbb{w}_ot-\delta)$ where $R=\sqrt{(c_1)^2+(c_2)^2}$ and $w_0 = \sqrt{\frac{k}{m}}$ and $\delta = \arctan(\frac{c_2}{c_1})$
This means I would gent $R=\sqrt{4^2+2^2}=\sqrt{20}=2\sqrt{5}$ and $w_0=\sqrt{8}=2\sqrt{2}$ Also $\delta=\arctan(\frac{-2}{4})=\arctan(-\frac{1}{2})=-.46036$ Combining all of these properties gives me:
$2\sqrt{5}\cos(2\sqrt{2}t+.46036)$ Some how the answer is just $2\cos(2\sqrt{2}t)$ I don't know why I am off. If someone could show me the steps, reasoning, and algebra that would be great.
The second problem is $x{''}+6x'+25=0$. I need to find the damped position function, the undamped position function, period, phase angle, amplitude. The roots come out to be complex roots: $-3 \pm 4i$ So the damped position function is $e^{-3x}[c_1\cos(4x)+c_2\sin(4x)]$ After plugging in the initial conditions $x_0=0$ and $v_0=-8$ I get $c_1=8$ and $c_2=-8$. However the back of the book gives a different answer..supposedly the damped position function is $2e^{-3t}\cos(4t-\frac{3\pi}{2})$ I can see where the exponential came from but to completely throw out the sin function makes no sense to me at all. Using the formulas for $R, \delta, w_0$ to get:
$R= \sqrt{8^2+(-8)^2} = \sqrt{128}= 8\sqrt{2}$ and $w_0=\sqrt{25}=5$ and $\delta=\arctan(\frac{c_1}{c_2})=\arctan(\frac{-8}{8})=\arctan(-1)=\frac{-\pi}{4}$ Putting this all together would give an undamped position function of $u(t)=8\sqrt{2}\cos(5t+\frac{\pi}{4})$ but the answer is $\frac{8}{5}\cos(5t-\frac{3\pi}{2})$ How do I find the period amplitude and phase angle? I see that some of these things like the amplitude is in the final solution for the damped position function.
Clearly I am missing some key points and I need them broken down and shown to me. I need to see where the algebra came from, how the operations work, when to apply them. Why is the damped position function as a solution, $e^{ax}[c_1\cos(bx)+c_2\sin(bx)]$ reduced to just a $\cos$ function with a shift?
Best Answer
For some reason, you seem to believe that the constants that you've calculated for the damped motion are the same for the undamped motion. It's not true. Just ignore what you have done before. Plug in the solution for undamped motion into the initial conditions, and you will get the result. So, in the first problem $$x(t)=R\cos(\omega t-\delta)$$ Calculating $v$, you get $$v(t)=-R\omega\sin(\omega t-\delta)$$ If you use $v(0)=0$, you get $\delta=0$. Then $x(0)=2=R$ from the first equation. So $$x(t)=2\cos(\sqrt 8 t)$$ Similarly, in the second problem, if you plug in $x(t=0)=0$, you should get $c_1=0$.