Diff EQ. Suppose that a mass and spring system with a mass m = 1, spring constant k = 26, damping coefficient c unknown had the vibration following the trajectory x(t) = e^(−t)cos(5t)+2e^(-t)sin(5t) What is the damping constant?
What's your progress? what equations are you using? what are you struggling with?
I don't understand my professor explanation for the problems part A. Especially why he's not using c^2-4km when solving for part A.
A short run through of what he did for part A in simplest terms would be great.
To be honest, I don't really follow your professor's work either, so I'll just do it my own way and see what i wind up with.
So we know that the general formulation for a damped mass spring system is going to yield a differential equation of the form: \(mu''(t)+\gamma u'(t)+ku(t)=0\) given that we have a sinusoidal solution with exponential decay, we know that \(\gamma\) must be non zero, and that our discriminant \(\gamma^2-4mk\) must be less than zero
When that is the case, we get a solution of \(y(t)=e^{-\gamma/(2m)t}(c_1 \cos(\frac{\sqrt{\gamma^2-4mk}}{2m}t)+c_2\sin(\frac{\sqrt{\gamma^2-4mk}}{2m}t))\) We're looking for the damping constant gamma here.
Well, we can just take the term \(-\gamma/(2m)=-1\) as I'm simply matching the coefficients from your problem \(e^{-t}\) vs \(e^{-\gamma/(2m)t}\)
One moment, I'm trying to match up what's going on here lol
Oh boy, I'm getting damping coefficient of 2 lol
Also my correction to my general solution: should actually be \(y(t)=e^{-\gamma/(2m)t}(c_1 \cos(\frac{\sqrt{4mk-\gamma^2}}{2m}t)+c_2\sin(\frac{\sqrt{4mk-\gamma^2}}{2m}t))\)
Wait, where did you get that value 2m from?
Sorry, I skipped some steps I probalby should've explained. Because our equation is linear homogenous with constant coefficients, we can write a characteristic equation: \(mr^2+\gamma r+k=0\) The solution to this is \(\Huge \frac{-\gamma\pm\sqrt{\gamma^2-4mk}}{2m}\)
OHHH I see what the problem is
Okay, so I think I finally get what's going on lol. your professor wants to express the solution as a single cosine wave, and then obtain the coefficient in front of that cosine wave.
so we know that \(y=e^{−t}\cos(5t)+2e^{-t}\sin(5t)\) we want to express our answer in the form of this \(y=e^{−t}\cos(\omega_0t-\delta)\) To do so we'll need to use some trigonometric identities.
*\(y=Re^{-t}\cos(\omega_0t-\delta)\) sorry
\(y=Re^{−t}\cos(\omega_0t-\delta)=Re^{-t}(\cos(\omega_0t)\cos(\delta)+\sin(\omega_0t)\sin(\delta))\) Call \(R\cos(\delta)=A\) and call \(R\sin(\delta)=B\) then, \((R\sin(\delta))^2+(R\cos(\delta))^2=A^2+B^2\) but can be simplified to \(R=\sqrt{A^2+B^2}\)
So I would find the values of A and B like I would for the average non-homogeneous equation?
Well, they're already presented in the solution.
A=1 and B=2 so \(\sqrt{A^2+B^2}=\sqrt{1+4}=\sqrt{5}\)
Notice that B/A is also equal to \(\frac{R\sin(\delta)}{R\cos(\delta)}=R\tan(\delta)\) which means \(B/A=\tan(\delta)\) Since B and A are both positive (in this example ONLY), we're safe taking the inverse tangent of both sides and thus getting \(\delta=\tan^{-1}(B/A)\) which is around 1.107
Here's a verification of the result.
Wow, the process was easy, I just couldn't understand his question or explanation. I really appreciate your help!
They do indeed seem to be the same function. http://prntscr.com/ahwkwm
No problem... I took this from my own class's notes; here is a pdf if you want more references. Do note that in the section where they solve for R, they are doign it for a UNDAMPED problem so the e^(-x) term is not there, but similar reasoning applies.
Thanks, your a real Gem!! Does your professor have a website with notes similar to this?
no, but if you want I can upload all the notes... I'll send you a pm.
Yes! That would be much appreciated.
I showed the solution to my professor and he said I would have to prove my solution. So for c=sqrt(5) dictates that t''+sqrt(5)t'+26t = e^(-t)cos(5t)+2Be^(-t)sin(5t) Problem is I've tried a multiple times and don't get that the two sides are equal.
No, that c value I got 2...
Solving for \(-\gamma/(2m)=-1\) with m=1 you obtain \(\gamma=2\)
so your original differential equation is \((1)u''(t)+2u'(t)+26t=0\)
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