Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Energy of a damped oscillator

  1. Jan 28, 2008 #1
    [SOLVED] Energy of a damped oscillator

    1. The problem statement, all variables and given/known data
    I simply need to show that the rate of energy for a damped oscillator is given by:

    dE / dT = -bv^2, where b is the dampening coefficient

    2. Relevant equations

    I am instructed to differentiate the formula:

    E = 1/2 mv[tex]^{2}[/tex] + 1/2 kx[tex]^{2}[/tex] (1)

    and use the formula: -kx - b dx/dt = m d[tex]^{2}[/tex]x/dt[tex]^{2}[/tex] (2)


    3. The attempt at a solution

    I differentitiate

    E = 1/2 mv[tex]^{2}[/tex] + 1/2 kx[tex]^{2}[/tex]

    to get dE/dT = m d[tex]^{2}[/tex]x/dt[tex]^{2}[/tex] + k dx/dt

    the only thing I can see to do here is sub in the above formula (2), to get

    dE / dt = -kx - b dx/dt + k dx/dt

    or

    dT / dt = -kx - bv + kv


    I must be missing something here, or maybe I made a mistake somewhere, but this question has been bugging me since yesterday. If anyone could steer me in the right direction I would definitely appreciate it.

    Thanks alot
     
    Last edited: Jan 28, 2008
  2. jcsd
  3. Jan 28, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You've got dE/dt completely wrong. The derivative of (1/2)mv^2 is mv*dv/dt. Do you see why? What's the derivative of (1/2)kx^2? Also what you are trying to prove should be dE/dt=-bv^2.
     
  4. Jan 28, 2008 #3
    Thanks alot for the quick reply Dick. You're right, I corrected the typo above.

    If I use mv*dv/dt as the derivative of (1/2)mv^2, and kx*dx/dt as the deriviative of (1/2)kx^2, the solution is very straight forward. I suppose I need to go back in my textbook and see how you derived that.

    Thanks alot for your help, I can already tell that this forum is going to be a huge resource for me for the next few years.

    Cheers!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook