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jlew
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[SOLVED] Energy of a damped oscillator
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
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)
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 + kvI 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
Homework Statement
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
Homework 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)
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 + kvI 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
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