Feynman Lectures, Chapter 24: Transients

[DivGradCurl]

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
m\frac{d^2 x}{dt^2}+\gamma m \frac{dx}{dt}+m\omega _0 ^2 x = F(t). (24.1)

In our problem, of course, F(t) is a cosine function of t. Now let us analyse the situation: how much work is done by the outside force F? The work done by the force per second, i.e., the power, is the force times the velocity. \Big(We know that the differential work in a time t is F dx, and the power is F\frac{dx}{dt}.\Big) Thus

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 . (24.2)

But the first two terms on the right can also be written as
\frac{d}{dt}\left[\frac{1}{2}m\left(\frac{dx}{dt}\right)^2 +\frac{1}{2}m\omega _0 ^2 x^2 \right],
as is immediately verifyed by differentiating."

 

[pervect]

OK, let u = dx/dt. Considering that 1/2 and m are constants, applying the chain rule we can write.

\frac{d}{dt}\left[\frac{1}{2}m\left(\frac{dx}{dt}\right)^2 \right] = \frac{1}{2}m\frac{d}{dt}u^2 = \frac{1}{2}m 2 u\frac{du}{dt}

but \frac{du}{dt} = \frac{d^2x}{dt^2}

so we wind up with

m \frac{dx}{dt} \frac{d^2x}{dt^2}

The other half of the problem is similar, substitute as needed for clarity and then apply the chain rule.

 

[DivGradCurl]

I see... Thanks!

Great... I now understand it.

Thank you so much!

 

[DivGradCurl]

Oh... one more thing

Now I see how the last expression can be expanded. But, How did he get it out of Eq. (24.2)?

Thank you.

 

[HallsofIvy]

\left(\frac{dx}{dt}\right)\left(\frac{d^2x}{dt^2}\right) is fairly standard.

If we write u= \frac{dx}{dt}, this is just uu'.. but that's just \frac{1}{2}u^2'.

In other words, \left(\frac{dx}{dt}\right)\left(\frac{d^2x}{dt^2}\right)= \frac{1}{2}\frac{d\left(\frac{dx}{dt}\right)^2}{dt}.

 

[DivGradCurl]

Thanks for your tip, HallsofIvy...

I can understand that it comes down to writing u\cdot u' . The only consequence I can draw from it is that \frac{dx}{dt}\frac{d\left(\frac{dx}{dt}\right)}{dt}. My point is... How does the \frac{1}{2} factor and \left(\frac{dx}{dt}\right)^2 appear in \frac{1}{2}\frac{d\left(\frac{dx}{dt}\right)^2}{dt} if we only have \frac{dx}{dt} and \frac{d^2x}{dt^2} to begin with? How do you find it if you initially have \frac{dx}{dt}\frac{d\left(\frac{dx}{dt}\right)}{dt}?

 

[speeding electron]

Let u = \frac{dx}{dt}. Now
\frac{d(\frac{1}{2} u^2)}{du} = u
\frac{1}{2} \frac{d({\frac{dx}{dt})^2}}{dt} = \frac{du}{dt} \frac{d({\frac{1}{2} u^2)}{du} = \frac{du}{dt} u = \frac{\frac{dx}{dt}}{dt} \frac{dx}{dt} = \frac{d^2{x}}{{dt}^2} \frac{dx}{dt}

 

[DivGradCurl]

I finally get it!

The pattern g \frac {dg}{dt} = \frac {1}{2} \frac {d g^2}{dt} fits for both terms. I finally worked it out. Thanks to HallsofIvy, speeding electron, and Tide for all the help!