# Feynman Lectures, Chapter 24: Transients

1. Sep 3, 2004

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."

2. Sep 3, 2004

### pervect

Staff Emeritus
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.

3. Sep 3, 2004

I see... Thanks!

Great... I now understand it.

Thank you so much!

4. Sep 3, 2004

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.

5. Sep 4, 2004

### HallsofIvy

Staff Emeritus
$$\left(\frac{dx}{dt}\right)\left(\frac{d^2x}{dt^2}\right)$$ 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, $$\left(\frac{dx}{dt}\right)\left(\frac{d^2x}{dt^2}\right)= \frac{1}{2}\frac{d\left(\frac{dx}{dt}\right)^2}{dt}$$.

Last edited: Sep 4, 2004
6. Sep 4, 2004

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 $$\frac{dx}{dt}\frac{d\left(\frac{dx}{dt}\right)}{dt}$$?

7. Sep 4, 2004

### speeding electron

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

Last edited: Sep 4, 2004
8. Sep 5, 2004

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!!!!