baru
  • baru
damped resonance
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
what about it
baru
  • baru
\(y'' +2py'+\omega_0^2y=cos\omega t\) show max amplitude occurs when \(\omega =\sqrt{\omega_0 -2p^2}\)
baru
  • baru
https://www.youtube.com/watch?v=Y9_zrupnz0Q skip to 41:10 the prof says, that amplitude does not build up indefinitely with time (unlike the undamped case) but looking at the exponential response formula, if \(p(\omega) =0\) then we have a solution where the amplitude has 't' in it (it grows with time)

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anonymous
  • anonymous
\(y_p = \mathfrak{R}\dfrac{e^{i\omega t}}{(i\omega)^2+2p(i\omega)+\omega_0^2}\) can you simplify ?
baru
  • baru
=\(R\frac{e^{i\omega t}( (w_0^2-\omega^2)-2p\omega i)}{(w_0^2-\omega^2)^2-4p^2\omega^2}\)
baru
  • baru
http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-ii-second-order-constant-coefficient-linear-equations/exponential-response/MIT18_03SCF11_s14_4btext.pdf
baru
  • baru
^exponential response formula @casey2 @Astrophysics @ganeshie8
baru
  • baru
ok... i think i figured out why amplitude does not grow indefinitely for a solution with 't' in the amplitude \(p(i\omega)\)=0 , not \(p(\omega)=0\) which means i\(\omega \) is a purely imaginary solution to \(D^2+2pD +\omega_0^2=0\) a purely imaginary solution occurs only when p=0 (no damping)
anonymous
  • anonymous
I think you need to find the derivative of that amplitude, set it equal to 0, and solve \(\omega \)
baru
  • baru
@zepdrix
ganeshie8
  • ganeshie8
you want to show that the max amplitude occurs when \(\omega =\sqrt{\omega_0 -2p^2}\) right ?
baru
  • baru
yes
ganeshie8
  • ganeshie8
then just do as casey2 is saying, it works nicely
ganeshie8
  • ganeshie8
\(y_p=R\frac{e^{i\omega t}( (w_0^2-\omega^2)-2p\omega i)}{(w_0^2-\omega^2)^2-4p^2\omega^2}\) write out the real part and differentiate the amplitude
ganeshie8
  • ganeshie8
do we get \[y_p=\frac{\cos(\omega t) (w_0^2-\omega^2)+2p\omega \sin(\omega t)}{(w_0^2-\omega^2)^2-4p^2\omega^2}\]
baru
  • baru
ok...
ganeshie8
  • ganeshie8
looks we have a sign mismatch, remember \((a+ib)(a-ib)=a^2+b^2\ne a^2-b^2\)
ganeshie8
  • ganeshie8
lets fix that quick
ganeshie8
  • ganeshie8
see if below looks good \[y_p=\frac{\cos(\omega t) (w_0^2-\omega^2)+2p\omega \sin(\omega t)}{(w_0^2 - \omega^2)^2\color{red}{+}4p^2\omega^2}\]
baru
  • baru
yep...
ganeshie8
  • ganeshie8
Next, recall the trig identity \(a\cos t+b\sin t = \sqrt{a^2+b^2}\sin(t-\phi)\)
ganeshie8
  • ganeshie8
you must have seen that identity before... if so, you can directly infer what the amplitdue would be by staring at the \(y_p\)
baru
  • baru
yes..i know that identiy
ganeshie8
  • ganeshie8
good, using that, lets see if we can write out just the amplitude of \(y_p\)
ganeshie8
  • ganeshie8
\[y_p=\frac{\cos(\omega t) (w_0^2-\omega^2)+2p\omega \sin(\omega t)}{(w_0^2 - \omega^2)^2\color{red}{+}4p^2\omega^2}\] \(\text{amplitude} =\sqrt{\dfrac{(\omega_0^2-\omega^2)^2 + (2p\omega)^2}{[(w_0^2 - \omega^2)^2\color{red}{+}4p^2\omega^2]^2}}\\~\\ =\dfrac{1}{\sqrt{(w_0^2 - \omega^2)^2\color{red}{+}4p^2\omega^2}} \)
ganeshie8
  • ganeshie8
differentiate that thing with respect to \(\omega \), set it equal to 0 and solve \(\omega\)
baru
  • baru
awesome!! thanks :)
ganeshie8
  • ganeshie8
if psble, may i see rest of the work :)
baru
  • baru
in progress :)
baru
  • baru
-1/2 {.....}^-3/2 \(\times ( 4\omega^3-4\omega_0^2 \omega+8p^2\omega\))=0 \(( 4\omega^3-4\omega_0^2 \omega+8p^2\omega=0\)) \(( \omega^2-\omega_0^2 +2p^2 =0\))
baru
  • baru
i hope thats understandable...didnt want to write the whole thing out xD
ganeshie8
  • ganeshie8
Ahh that was simpler than I thought.. nice

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