Question

damped resonance

Answer 1

what about it

Answer 2

\(y'' +2py'+\omega_0^2y=cos\omega t\)

show max amplitude occurs when \(\omega =\sqrt{\omega_0 -2p^2}\)

Answer 3

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)

Answer 4

\(y_p = \mathfrak{R}\dfrac{e^{i\omega t}}{(i\omega)^2+2p(i\omega)+\omega_0^2}\)

can you simplify ?

Answer 5

=\(R\frac{e^{i\omega t}( (w_0^2-\omega^2)-2p\omega i)}{(w_0^2-\omega^2)^2-4p^2\omega^2}\)

Answer 6

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

Answer 7

^exponential response formula

@casey2 @Astrophysics @ganeshie8

Answer 8

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)

Answer 9

I think you need to find the derivative of that amplitude, set it equal to 0, and solve \(\omega \)

Answer 10

@zepdrix

Answer 11

you want to show that the max amplitude occurs when \(\omega =\sqrt{\omega_0 -2p^2}\) right ?

Answer 12

yes

Answer 13

then just do as casey2 is saying, it works nicely

Answer 14

\(y_p=R\frac{e^{i\omega t}( (w_0^2-\omega^2)-2p\omega i)}{(w_0^2-\omega^2)^2-4p^2\omega^2}\)

write out the real part and differentiate the amplitude

Answer 15

do we get
\[y_p=\frac{\cos(\omega t) (w_0^2-\omega^2)+2p\omega \sin(\omega t)}{(w_0^2-\omega^2)^2-4p^2\omega^2}\]

Answer 16

ok...

Answer 17

looks we have a sign mismatch,
remember \((a+ib)(a-ib)=a^2+b^2\ne a^2-b^2\)

Answer 18

lets fix that quick

Answer 19

see if below looks good

\[y_p=\frac{\cos(\omega t) (w_0^2-\omega^2)+2p\omega \sin(\omega t)}{(w_0^2 - \omega^2)^2\color{red}{+}4p^2\omega^2}\]

Answer 20

yep...

Answer 21

Next, recall the trig identity \(a\cos t+b\sin t = \sqrt{a^2+b^2}\sin(t-\phi)\)

Answer 22

you must have seen that identity before... if so, you can directly infer what the amplitdue would be by staring at the \(y_p\)

Answer 23

yes..i know that identiy

Answer 24

good, using that, lets see if we can write out just the amplitude of \(y_p\)

Answer 25

\[y_p=\frac{\cos(\omega t) (w_0^2-\omega^2)+2p\omega \sin(\omega t)}{(w_0^2 - \omega^2)^2\color{red}{+}4p^2\omega^2}\]


\(\text{amplitude} =\sqrt{\dfrac{(\omega_0^2-\omega^2)^2 + (2p\omega)^2}{[(w_0^2 - \omega^2)^2\color{red}{+}4p^2\omega^2]^2}}\\~\\
=\dfrac{1}{\sqrt{(w_0^2 - \omega^2)^2\color{red}{+}4p^2\omega^2}}
\)

Answer 26

differentiate that thing with respect to \(\omega \), set it equal to 0 and solve \(\omega\)

Answer 27

awesome!! thanks :)

Answer 28

if psble, may i see rest of the work :)

Answer 29

in progress :)

Answer 30

-1/2 {.....}^-3/2 \(\times ( 4\omega^3-4\omega_0^2 \omega+8p^2\omega\))=0

\(( 4\omega^3-4\omega_0^2 \omega+8p^2\omega=0\))

\(( \omega^2-\omega_0^2 +2p^2 =0\))

Answer 31

i hope thats understandable...didnt want to write the whole thing out xD

Answer 32

Ahh that was simpler than I thought.. nice