Vibrating String kinetic energy

[Math Jeans]

Homework Statement

The kinetic energy of a segment of length \Delta x and mass \Delta m of a vibrating string is given by \Delta K = \frac{1}{2} \Delta m (\frac{\partial y}{\partial t})^2 = \frac{1}{2} \mu (\frac{\partial y}{\partial t})^2 \Delta x, where \mu = \frac{\Delta m}{\Delta x}.

a. find the total kinetic energy of the nth mode of vibration of a string of length L fixed at both ends.

b. Give the maximum kinetic energy of the string.

c. What is the wave function when the kinetic energy has its maximum value?

d. Show that the maximum kinetic energy in the nth mode is proportional to n^2 A_n^2.

Homework Equations

Pretty much given.

The Attempt at a Solution

I have tried numerous times to get a start on this problem but I can't seem to figure it out. Neither can my older brother or other peers .

 

[qspeechc]

Ok, I can't answer them all, but here's what I can answer:

We might as well work with differentials
\frac{dK}{dx}=\frac{1}{2} \mu (\frac{\partial y}{\partial t})^2

Ok, the next bit is hand-wavy. For a vibrating string

y(x,t) = Asin(\omega_{0} (t - x/\nu ))

where omega is the angular frequency- you may have seen it in some slightly different but equivalent form.

\frac{\partial y}{\partial t} = A\omega cos(\omega_{0} (t - x/\nu ))

At t=0:
\frac{\partial y}{\partial t} = A\omega cos(\omega_{0} x/\nu )

Then:
\frac{dK}{dx}=\frac{1}{2} \mu (\frac{\partial y}{\partial t})^2 = \frac{1}{2} \mu (A \omega_{0} )^2 cos^2(\omega_{0} x/\nu )

So, in one wavelength:

\int_{0}^{\lambda } \frac{1}{2} (A\omega_{0} )^2 cos^2(\omega_{0} x/\nu )dx

You will see that the integral is equal to:

\frac{1}{4} \lamda \mu (A\omega_{0} )^2

\omega_{0} is the angular frequency of the first normal mode, and the frequency of the nth normal mode is n\omega_{0}. Some of the results follow directly from that.

As a matter of interest- what book are you using?

 

[Math Jeans]

As a matter of interest- what book are you using?

Physics for Scientists and Engineers vol.5 by Paul A. Tipler

 

[Math Jeans]

I think I understand now. Thanks for the help.

 

[mln326]

? What is the smallest possible 'unit' that can be in all of this.