Some help with oscillations and damping

  • #1
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Consider damped harmonic oscillations. Let the coeffient of friction gamma be half the value of the one that just gies critical damping.

How many times is the period T larger than it would be for gamma = 0??

WHen gamma is zero -

[tex] T = \frac{2 \pi}{\omega} [/tex]

When gamma is half of the value for critical damping

now for critical damping

[tex] \frac{\gamma}{2} = \omega_0 [/tex]

So then for the question, (half of the value for gamma) then

that yields omega / 2

and then gives a period [tex] T = \frac{2 \pi}{\frac{\omega}{2}} [/tex]

and that gives [tex] T = \frac{4 \pi}{\omega} [/tex]

which is half the period for the case when gamma is zero

My text book says the answer - the ratio between T(damped) and T(undamped) = 2 / root(3)

Determine the ratio between two successive swings on teh same side.

It doesn't quite give the case for WHICH case it wants us to consider but it is definitely related to the previous question

I am completely baffled as to how to go about this

do i plug this into the equation for the damping that is

[tex] = x = x_0 e^{\frac{\gamma t}{2}} Cos(\omega_d t + \theta) [/tex]
Answer of the text is X2/X1 = exp (-2pi / 3) i dont know how
i am not sure

your help is greatly appreciated!!
 
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Answers and Replies

  • #2
dextercioby
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I think i can help u with the first part
[tex] \omega_{damping}=\sqrt{\omega_{0}^{2}-\frac{\gamma^{2}}{4}} [/tex] (1)

,okay??

Now find the value for gamma for which [itex] \omega_{damping} [/itex] is zero...That means "critical value"...

Take half of that value and plug it in (1).Then compare the angular freqeancies omega zero and omega damped...U'll find the answer.

Daniel.
 
  • #3
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thats helpful

however for the second part is it somehow related to the first part in a sense that for one the time is t=0 and the other is time t= 2 / root 3?

and then find the ratio betweeen the two x values??
 
  • #4
dextercioby
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Take in the equation of movement (x=x(t)) t=0 and T=T_{damped} and take the ratio of the 2 values...You might take the initial phase arbitrary,since anyway it won't matter...

Daniel.
 
  • #5
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dextercioby said:
Take in the equation of movement (x=x(t)) t=0 and T=T_{damped} and take the ratio of the 2 values...You might take the initial phase arbitrary,since anyway it won't matter...

Daniel.
and the equation of movement is
[tex] = x = x_0 e^{\frac{-\gamma t}{2}} Cos(\omega_d t + \theta) [/tex]

so when i substitute the value of gamma = omega (which i got from the part 1)

then for the equation for damped turns to [tex] = x = x_0 e^{\frac{-\omega_{0} t}{2}} Cos(\frac{\sqrt{3}}{2}\omega_d t + \theta) [/tex]

lets seay we start looking at swings starting from t = t and then the next swing o nteh same side is t = (root3 / 2) +t??

and then sub into the x0 quation and find the ratio between the two>??
 
  • #6
dextercioby
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No,that number involving the sqrt is to be put in the exponential olny.When taking the ratio of the 2 "x"-s,the cosine part vanishes (is simplified through),so it doesn't matter whether u put or not in the cosine...

Daniel.
 

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