This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Lecture 4 Notes: 06 / 30 Energy carried by a wave We want to find the total energy (kinetic and potential) in a sine wave on a string. A small segment of a string at a fixed point x behaves as a harmonic oscillator with amplitude A and angular frequency : The mass of this segment is equal to m = dx , where is the mass per unit length of the string, and dx is the (small) length of the segment. What is the energy of this oscillator? Recall that the total energy of a harmonic oscillator is We are using capital K for the spring constant to avoid confusion with the wave number k . The spring constant is related to the frequency as follows: Thus, the energy of the segment of string is equal to The length of the segment is dx , so the energy per unit length (energy density) is At what rate is the energy carried by the traveling wave? Consider a traveling wave of finite length L . At time t = 0 , this wave begins to cross from region 1 into region 2. After a time t has elapsed, a length of ct will have moved into region 2: So, the energy in Region 2 is equal to E 2 = E ct and the energy remaining in Region 1 is equal to E 1 = E (L  ct) . Region 2 therefore gains energy at the rate I = E c joules per second, while Region 1 loses energy at the same rate. This rate of energy transport is known as the energy flux of the wave. Plugging in our result for the energy density E , the energy flux is We can also express this quantity in terms of the tension in the string, rather than the mass per unit length. Recall that Plugging this into our equation for the energy flux, we obtain Example: How much energy is transmitted per second by a wave given by the following function, assuming that the tension in the string is F T = 100N ? We can read off the amplitude and the angular frequency from the function, and are given the tension. We need to calculate the wave speed. It is c = / k = 30 m/s. Given all this, we can calculate the energy flux: The wave carries 17.5 watts of power. Reflection and transmission from an interface Now suppose that we join together two strings, each with a different mass per unit length. Tension is applied to the combined string, so that the tension force is the same everywhere, but because the mass per unit length is different, the wave speed will be different between the two segments. Let the interface be located at x = 0 : Let y 1 (x,t) be the displacement of the string to the left of the interface, and y 2 (x,t) be the displacement of the string to the right of the interface. displacement of the string to the right of the interface....
View
Full
Document
This note was uploaded on 09/19/2011 for the course PHYS 1C taught by Professor Smith during the Spring '07 term at UCSD.
 Spring '07
 Smith
 Energy

Click to edit the document details