- #1
DivGradCurl
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Hello everyone!
A question came up as I was reading Chapter 24 of the Feynman Lectures book. To more specific, it's the comments after Eq. (24.2) on the first section---called "the energy of an oscillator". I don't quite get it.
Thank you very much!
"Now let us consider the energy in a forced oscillator. The equation for the forced oscillator is
[tex]m\frac{d^2 x}{dt^2}+\gamma m \frac{dx}{dt}+m\omega _0 ^2 x = F(t). [/tex] (24.1)
In our problem, of course, [tex]F(t)[/tex] is a cosine function of [tex]t[/tex]. Now let us analyse the situation: how much work is done by the outside force [tex]F[/tex]? The work done by the force per second, i.e., the power, is the force times the velocity. [tex]\Big([/tex]We know that the differential work in a time [tex]t[/tex] is [tex]F dx[/tex], and the power is [tex]F\frac{dx}{dt}[/tex].[tex]\Big)[/tex] Thus
[tex]P=F\frac{dx}{dt}=m\left[\left(\frac{dx}{dt}\right)\left(\frac{d^2x}{dt^2}\right)+\omega _0 ^2 x\left(\frac{dx}{dt}\right) \right]+\gamma m\left(\frac{dx}{dt}\right)^2 .[/tex] (24.2)
But the first two terms on the right can also be written as
[tex]\frac{d}{dt}\left[\frac{1}{2}m\left(\frac{dx}{dt}\right)^2 +\frac{1}{2}m\omega _0 ^2 x^2 \right], [/tex]
as is immediately verifyed by differentiating."
A question came up as I was reading Chapter 24 of the Feynman Lectures book. To more specific, it's the comments after Eq. (24.2) on the first section---called "the energy of an oscillator". I don't quite get it.
Thank you very much!
"Now let us consider the energy in a forced oscillator. The equation for the forced oscillator is
[tex]m\frac{d^2 x}{dt^2}+\gamma m \frac{dx}{dt}+m\omega _0 ^2 x = F(t). [/tex] (24.1)
In our problem, of course, [tex]F(t)[/tex] is a cosine function of [tex]t[/tex]. Now let us analyse the situation: how much work is done by the outside force [tex]F[/tex]? The work done by the force per second, i.e., the power, is the force times the velocity. [tex]\Big([/tex]We know that the differential work in a time [tex]t[/tex] is [tex]F dx[/tex], and the power is [tex]F\frac{dx}{dt}[/tex].[tex]\Big)[/tex] Thus
[tex]P=F\frac{dx}{dt}=m\left[\left(\frac{dx}{dt}\right)\left(\frac{d^2x}{dt^2}\right)+\omega _0 ^2 x\left(\frac{dx}{dt}\right) \right]+\gamma m\left(\frac{dx}{dt}\right)^2 .[/tex] (24.2)
But the first two terms on the right can also be written as
[tex]\frac{d}{dt}\left[\frac{1}{2}m\left(\frac{dx}{dt}\right)^2 +\frac{1}{2}m\omega _0 ^2 x^2 \right], [/tex]
as is immediately verifyed by differentiating."