Wave Equation - Example 2
Wave Equation - Example 3
Wave Equation - Example 4
University of North Carolina at Chapel Hill
Wave Equation - Example 1


Comments

Comments are currently disabled.

Video Transcript

welcome to our first example video. Considering a wave on a string in this video, we're going to ask ourselves if we have a string that is 1.5 m long and it has a mass of 10 g 10. What we're going to do then, is we're going to pluck this string, and we're going to send a wave through it and we want to find is the wave speed. Now, um, looking at this, we see we have 1.5 m 10 g so we could find mu. That's 10 times 10 to the negative 3 kg divided by 1.5 meters. But wave speed is equal to the square root of tension divided by muse. So we have mu but no tension, and we want to find wave speed. So this equation is not going to vice. So what we can do instead is we can pluck this thing and say, Count the number of times it goes down and back, Say so, say it goes down and back 10 times in eight seconds. We'll notice that V is also equal to distance divided by time, and this is traveling at a constant speed even though it's not a constant velocity because it's changing direction. But if we ask then for what is the wave speed that is the distance notice, not the displacement, but the distance divided by time. So distance then will be 10 times down and back is 3 m because down once is 1.5 do that in eight seconds, so that's gonna end up as 30 divided by 8 m per second or if you prefer 15 4th meters per second. So this is how fast the wave is traveling back and forth on this strength. We could then go through and solve for F t finding that f T equals V squared times mu and be able to use this in order to solve for the tension in the string.