Simple harmonic motion - damping introduced

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  • #2
Charles Link
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The damping coefficient is ## b ## in your formula. Do you know how to solve ## e^{-\frac{b}{2m}t}=\frac{1}{2} ## for ## t ## ? . ## \\ ## Edit: A google of the topic calls ## \gamma=\frac{b}{2m} ## the damping coefficient, but it is really a choice of terminology. By the units they give you, they are giving you ## b ##.
 
  • #3
The damping coefficient is ## b ## in your formula. Do you know how to solve ## e^{-\frac{b}{2m}t}=\frac{1}{2} ## for ## t ## ? .
I don't know how much is x(t)
 
  • #4
Charles Link
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I don't know how much is x(t)
The sinusoidal oscillation is assumed to happen at a much higher frequency with small damping, so that the period of the oscillation ## T ## is quite short, and you don't need to consider the term ## \cos(\omega t ) ##. The amplitude is ## A e^{- \frac{b}{2m} t} ##.
 
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  • #5
The sinusoidal oscillation is assumed to happen at a much higher frequency with small damping, so that the period of the oscillation ## T ## is quite short, and you don't need to consider the term ## \cos(\omega t ) ##. The amplitude is ## A e^{- \frac{b}{2m} t} ##.
Ok, thank you and what about x(t)
 
  • #6
Charles Link
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Your (first equation) that you posted in part 2 of the OP for ## x(t) ## is correct. (You didn't list an equation for ## \omega' ## though). And your second equation, which I think is a damping ratio number is really not of prime interest here. It would help if you would state what the parameter is. I had to go googling for the second equation you posted, to see if it was correct.## \\ ## When they say, how long does it take for the amplitude to "halve" that means ## A e^{-\frac{b}{2m} t}=\frac{1}{2} Ae^{-\frac{b}{2m} 0} ##, so that ## e^{-\frac{b}{2m} t}=\frac{1}{2} ##.
 
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  • #7
Oh, I am sorry I didn't think an equation for w' would be helpful here. Thank you very much for your help!
 
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  • #8
Charles Link
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Now, the next step is take the natural log of both sides of this last equation in order to solve for ## t ##. (It may be worthwhile for me to mention that, because I'm not sure how advanced you may be).
 
  • #9
Now, the next step is take the natural log of both sides of this last equation in order to solve for ## t ##. (It may be worthwhile for me to mention that, because I'm not sure how advanced you may be).
Yeah, I know how to proceed from here ## A e^{-\frac{b}{2m} t}=\frac{1}{2} Ae^{-\frac{b}{2m} 0} ##,. I got t=3.036s. I hope that this is correct
 
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  • #10
Charles Link
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Suggest you use ## \ln{2}=.693 ## and you get ## t=3.05 ## seconds. (I see you must have used ## \ln{2}=.690 ##).
 
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  • #11
That's true. I did plug 0.69. Thank you!
 
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