Proof of oscillation about the equilibrium

  • #1
Bonnie
18
1

Homework Statement



The problem is question 2(a) in the attached pdf. I seem to find myself at a dead end and am not sure where to go from here - I will attach my working in a separate file, but basically I need to show that the oscillator passes/crosses over the x = 0 boundary at a positive time, but I don't seem to be able to do that (or have enough information to) without ending up with nothing, or an impossible equation, like an exponential term = 0.

Homework Equations




The Attempt at a Solution

 

Attachments

  • PHYS205 Assignment 8.pdf
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  • 20180514_101513.jpg
    20180514_101513.jpg
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Answers and Replies

  • #2
BvU
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Your picture is half illegible. Seems you do find an ##\omega## though.
However: your try with an intial displacement of 0 of course doesn't let the thing oscillate !

free tip: work with symbols, not with numbers. Only at the last moment, if/when a value is needed, subtitute values.
 
  • #3
Bonnie
18
1
Your picture is half illegible. Seems you do find an ##\omega## though.
However: your try with an intial displacement of 0 of course doesn't let the thing oscillate !

free tip: work with symbols, not with numbers. Only at the last moment, if/when a value is needed, subtitute values.

ω is 20 rads-1, from the equation (I have shown it only with values substituted):
p = -γ/2 +/- √[ω2 - (γ/2)2]
where ω2 = 400 and γ = 5\

And I set x(0) = D, not 0, as x is a function of t, is that incorrect?
Thanks
 
  • #4
BvU
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And I set x(0) = D, not 0, as x is a function of t, is that incorrect?
No, that's correct. But it looks a lot like a zero on my screen.

Your Solution
upload_2018-5-14_0-44-14.png

has ##x=Ce^{-{5\over 2}t} e^{\rm illegible}## when originally it was
upload_2018-5-14_0-45-25.png
( so probably ##x=Ce^{pt} ## ?) and you found two ##p##. What happened to the second ?
 

Attachments

  • upload_2018-5-14_0-44-14.png
    upload_2018-5-14_0-44-14.png
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  • upload_2018-5-14_0-45-25.png
    upload_2018-5-14_0-45-25.png
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  • #5
Bonnie
18
1
No, that's correct. But it looks a lot like a zero on my screen.

Your Solution View attachment 225681
has ##x=Ce^{-{5\over 2}t} e^{\rm illegible}## when originally it was View attachment 225682 ( so probably ##x=Ce^{pt} ## ?) and you found two ##p##. What happened to the second ?
I'll try to attach a better photo, but the second p is included in the e+/- 393.75j
 

Attachments

  • 20180514_101513.jpg
    20180514_101513.jpg
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  • #6
Bonnie
18
1
I'll try to attach a better photo, but the second p is included in the e+/- 393.75j
Ah, I've just realized that the photo quality is significantly decreased by uploading it here. Apologies for that
 
  • #7
BvU
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You need to rethink your solution. A second order differential equation needs two integration constants (one for ##x## and one for ##\dot x## if you want to put it that way).

If there are two ##p## to solve the characteristic equation, then ##C_1 \,e^{p_1 t}## is a solution and so is ##C_2 \,e^{p_2 t}##.
 

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