# Total energy of oscillation

1. Nov 19, 2006

### lizzyb

Q: A sinusoidal wave of the form: $$y = A \sin{kx - \omega t}$$ is traveling along a string in the x direction, where A = 0.88 mm, k = 2 m^-1, omega = 25 rad/s, with x in meters and t in seconds. For this string, the mass per unit length is given by mu = 0.01 kg/m.

For a length segment delta x = 1 cm along the string, what is the total energy of oscillation? Answer in units of J.

First I tried $$E_\lambda = U_\lambda + K_\lambda = \frac{1}{2} \mu \omega^2 A^2 \lambda$$ but that was wrong so then I changed the lamda to delta x, and that was wrong as well. How do we compute the total energy of a small line segment? thanks.

2. Nov 19, 2006

### Kurdt

Staff Emeritus
The kinetic energy density is: $$\frac{dK}{dx}=\frac{1}{2}\mu(\frac{dy}{dt})^2$$and the potential energy density is $$\frac{dU}{dx}=\frac{1}{2}T(\frac{dy}{dt})^2$$

where T is the tension in the string, $$T=\mu \omega^2$$

The energy density is the energy per unit length on the string. To find the energy for a particular interval you have to integrate the equations using the interval bounds as limits.

3. Nov 19, 2006

### lizzyb

so I have to take the derivitive of the wave function with respect to t, put it into the equation, then integrate over the range of x?

$$\frac{dy}{dt} = -A \omega \cos(\omega t - k x)$$

So $$\frac{dK}{dx}=\frac{1}{2}\mu(\frac{dy}{dt})^2 = \frac{1}{2}\mu(-A \omega \cos(\omega t - k x))^2$$

and then I integrate it from 0 to .01 meters?

$$K = \int_0^{.01} \frac{1}{2}\mu(-A \omega \cos(\omega t - k x))^2 dx$$

t is unaccounted for? Thank you for your help.

4. Nov 19, 2006

### OlderDan

This is more of an intuitive observation that needs to be verified. Time will take care of itself if you look at both the kinetic and potential energy of the length interval. The equation you tried in your first post I am assuming is the energy of one wavelength of the string. If that is the case, and you can calculate the wavelength, you should be able to find the energy of any other length with a simple ratio of length to wavelength. The rationale behind that is that each bit of mass dm that is vibrating is a harmonic oscillator with amplitude A and the same k, so each little dm must have the same total energy as any other dm.

5. Nov 19, 2006

### lizzyb

ok, so $$E_\lambda = U_\lambda + K_\lambda - \frac{1}{2}\mu\omega^2A^2\lamda$$
From the problem we have: $$A = .00088 "m", k = \frac{"2"}{"m"}, \omega = 25 \frac{"rad"}{"s"}, \mu = .01 \frac{"kg"}{"m"} "and" \lambda = \frac{2\pi}{k} = \pi "m"$$
Thus $$E_\lambda = \frac{1}{2} \cdot .01 \frac{"kg"}{"m"} \cdot (25 \frac{"rad"}{"s"})^2 \cdot (.00088 "m")^2 (\pi "m") = 7.6027 \times 10^{-6}$$
but that's for the entire wavelength, hence for the length segment $$\Delta x = 1 "cm"$$ we have $$E_{.01} = \frac{.01}{\lambda} E_\lambda = \frac{.01}{\pi}E_\lambda = .003183 \cdot 7.6027 \times 10^{-6} = 2.42 \times 10^{-8}$$

does that look ok to you?

Last edited: Nov 19, 2006
6. Nov 19, 2006

### OlderDan

Looks OK to me. You have a typo problem with the equation in this post, but if you look at the equation for the energy for one full wavelength, then you can see that everyting multiplying the λ is the energy per unit length. An equivalent calculation would be to just replace the λ in that equation by the 1cm length of interest.

7. Nov 19, 2006

### lizzyb

$$E_\lambda = U_\lambda + K_\lambda = \frac{1}{2}\mu\omega^2A^2 \lambda$$
So since I'm only interested in .01 meters, I can do this:
$$E_{.01} = \frac{1}{2} \cdot \mu \cdot \omega^2 \cdot A^2 \cdot .01 = 2.42 \times 10^{-8}$$
which is the same answer; that answer was accepted by the homework server! thanx! i thought i tried that early on. Thank you.