Amplitudes of a harmonic oscillation

Discussion in 'Homework Help' started by Niles, Jan 6, 2009.

  1. Niles

    Thread Starter Active Member

    Nov 23, 2008
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    Hi all.

    I have the following differential equation, which describes the voltage across a capacitor in an RCL-circuit, which is discharging:

    <br />
V_C(t)=(A\cos(\omega t)+B\sin(\omega t))e^{-\frac{R}{2L}t}<br />

    I need to find out at what time t the amplitude of the oscillations are half of the original amplitude. This seems to be a boring and long exercise; is there no easy way of solving it?

    The original differential equation (which has the above solution) is given by:

    <br />
0=L\frac{d^2V_C(t)}{dt^2}+\frac{1}{C}V_C(t)+R\frac{dV_C(t)}{dt}<br />

    Thanks in advance.

    Best regards,
    Niles.
     
    Last edited: Jan 6, 2009
  2. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
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    It resolves to a simple natural logarithmic decay, same as an RC time constant.

    eric
     
  3. Niles

    Thread Starter Active Member

    Nov 23, 2008
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    I'm not sure I understand you correctly. First of all, I'm not even sure what the amplitude in my expression for V_C(t) is; second of all, I tried rewriting it all in terms of exponentials, but it is a mess.

    You say logarithmic decay; do you mean exponential?

    Thanks.
     
  4. steveb

    Senior Member

    Jul 3, 2008
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    The amplitude is  \sqrt{A^2+B^2} at  t=0

    Whenever you see a linear combination of sine and cosine at the same frequency like that, you can rewrite it as a cosine wave with amplitude as given above and phase shift given by  atan{{{B} \over {A}} . This is a basic trigonometric identity you can look up in a math handbook.

     A\; \cos x+B\; \sin x=\sqrt{A^2+B^}\; cos(x+atan{{{B} \over {A}}} .

    In your solution, the phase shifted sine wave decays exponentially.
     
    Last edited: Jan 6, 2009
  5. Niles

    Thread Starter Active Member

    Nov 23, 2008
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    So to sum it, what I want to do is to solve the following equation with respect to t?

    <br />
\frac{1}{2}\sqrt{A^2+B^2} = \sqrt{A^2+B^2}\,\,\,\cos(\omega t + atan\frac{B}{A})<br />

    And thanks for replying to all.
     
  6. steveb

    Senior Member

    Jul 3, 2008
    2,433
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    Nope. The amplitude versus time is as follows:

     \sqrt{A^2+B^2}\; e^{-t/\tau}

    where  \tau is the time constant.

    The implicit definition of amplitude relates to how big the sinewave is.
     
  7. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    It seems to me that this is a lot easier than has so far been indicated. The problem asks when the amplitude is 1/2 of it's original amplitude. I assume that "original amplitude" means the amplitude when t=0. So all we have to do is calculate when the multiplicative exponential factor is equal to 1/2, since when t=0 it's equal to 1.
     
  8. steveb

    Senior Member

    Jul 3, 2008
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    LoL, well yes, but we didn't want to just give the answer away. It's a homework problem.
     
  9. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Both KL7AJ and you gave him the key word, "decay", but he didn't seem to be getting it. I thought a tad bit more hint might be in order. :)

    He still has to do the computation!
     
  10. Niles

    Thread Starter Active Member

    Nov 23, 2008
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    I think the reason why I didn't get it was that I don't see why there is an exponential decay. How should I realize this by just looking at the solution for the potential?
     
  11. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Because the solution has two parts. The first part is an expression which simply generates a constant amplitude sinusoidal wave:

    <br />
V_C(t)=(A\cos(\omega t)+B\sin(\omega t))<br />

    But this is multiplied by a decaying exponential function:

    <br />
e^{-\frac{R}{2L}t}<br />

    This function starts out with a value of 1 at t=0, and decays toward a value of zero as time progresses. Since it multiplies the sinusoid, it causes the amplitude of the sinusoid to decrease with time. It provides an "envelope" within which the sinusoid decays.

    You should plot the solution function and you will see what I'm talking about. Or, have a look at this:

    http://www.motionscript.com/mastering-expressions/simulation-basics-2.html
     
  12. Niles

    Thread Starter Active Member

    Nov 23, 2008
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    Ahh, I get it now. So the potential V decays within the exponential term, so we want the time when the exponential equals 1/2.

    Very nice. Thanks to all!
     
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