https://youtu.be/Zn_aEoaVqI4
PHYS 1101: Lecture Twenty-Five, Part Five
Okay. Let’s do an example. Our problem is we have a person on an air mattress in the ocean, rises and falls through one complete cycle every five seconds. The crest of the wave causing the motion, the crests, are 20 meters apart. Determine, A, the frequency and, B, the speed of the wave.
Let me do a quick sketch here so we can label these features, numbers that we’re given. Here’s my object. This is a particle. I can think of a particle, if you will, that’s attached to this wave. This person is going up and down and completes one cycle every five seconds. That’s the period. If that’s the period of this spot in the wave going up and down, that’s also the same time it takes a crest to move over one wave length.
We know that the distance between the crest is 20 meters. Therefore that’s the wave length. For part A, we’re after the frequency. That’s the variable f. Frequency, as it was in the previous lecture, is related to the period. It’s just being inversely related or calculated. The frequency is one over the period. One over five seconds equals 0.2 hertz. Okay, that’s the answer to part A.
Part B asks for the speed of the wave. That’s vwave. I like remembering always the relationship that one Lambda equals vwave times the period. If I then just rearrange that for the variable I need to solve for, vwave is equal to Lambda over the period. That’s 20 meters divided by five seconds, and that’s four meters per second.
Okay. Let’s do a last example here for the lecture, something a little more complicated. The previous example was a wave, a water wave, and just took advantage of, or relied on, general knowledge of waves and what the meaning of wave length is and period and velocity. Now we’re going to do a problem specific to a wave on a string. We have a five-kilogram block. It’s hung from the ceiling on a two-meter-long metal wire with a mass of four grams. The wire is plucked at the very bottom where it connects to the block. How long does it take the pulse to reach the ceiling?
Definitely want to start with a good sketch of this to help my mind settle on the details. From the ceiling is hung a wire, and I have a block hanging from the bottom of the wire. It’s five kilograms. I know that the mass of the wire of this entire length of the wire is four grams. That’s .004 kilograms. I also know the length of the wire is two meters.
And I think I’ve added now to my drawing all the information, sketched it, and now let me try to sketch the essence of this motion that I’m following. The wire is plucked at the bottom, and then I’m interested in the time it takes for this pulse to reach the ceiling. So if I pluck it here at the bottom, I’m going to indicate that I will have added a disturbance, or caused a disturbance in the wire. This then is going to propagate and travel up the wire, and, of course, then it’ll hit the ceiling, and that’s what I want to know the time it takes to do that. This is a disturbance that’s traveling as a wave along the wire, so all of what I have learned in this lecture I can use to help me solve this.
Let me summarize here at the top with the variable that we want so we stay focused on where we need to go. How long does it take? Let’s call that the variable t, the time, and let me just think through how I would get that. The pulse is traveling at some characteristic rate. I know I used the variable vwave to indicate that. Well, the velocity of this pulse relates to the time it takes to go that distance with really the definition of velocity connection, which is the velocity of something is equal to distance. It travels over time.
So here I have my variable T I want to solve for. Distance, let me call that the variable l so it matches what I have in my drawing. I see I have l. I see the variable that I need. Let’s think about the left side of this equation. Can I independently figure out a number for vwave, in which case I’d be able to plug it in? I would have this number, and therefore I would have the ability to solve for l over T.
Well, we did learn in this lecture that the velocity of a wave is purely characteristic of the properties of the medium, meaning the characteristics of this wire that this pulse is traveling along, that the wave is traveling in, for a string the velocity depends on the properties of the string, which are the tension. I’m going to use as a variable plain T divided by this linear density, the mass divided by one. To calculate this denominator, I can take any length of the string, plug it in here, plug in the total mass of that length, and then this is the proper ratio that I would put in the denominator to calculate this.
This I have the information for. I know the wire mass, then I know the length. How about the tension? Can I get that? What’s the tension in the wire? Well, tension we know is a force. It’s a force that this wire is exerting on the ceiling and the block. Thinking of the block experiencing this force brings to mind that I probably can do a force analysis of this block to work out how big that tension must be. This block is an object that has zero acceleration, so all the forces on it, one of which is the one I need to figure out the tension, has to balance all others. The only other thing I have on the block is the force of gravity.
I’m going to go over here to my block and do a quick little force sketch to help me be sure I think through that analysis properly. This force T for the tension that I drew has to be balancing the weight, the only other force that’s on this object, mg, that’s down. Its acceleration is 0, so the tension equals mg. The forces add to zero. This mathematically rearranges to what we just said in words, the tension equals mg, so I’m set now. When I put in a number for that, I’ve got five kilograms times my 9.8 meters per second2. That tension then is equal to 49 Newtons.
I’m set now to go back into my equation and plug in all my numbers. Let’s first do the step of getting a value for vwave. Vwave will be the square root of 49 Newtons divided by the small mass of the wire and that divided by the length of the wire, two meters. This mass divided by a length is this linear density. It’s this property of the string that reflects the inertia of that material. You multiply those numbers out, and it’s a pretty high velocity. In a string, a wire, tense wire like that, I end of up with a velocity of 156.5 meters per second. That’s almost 300 miles an hour.
I can now go back to my original equation that I set up that captured the essence, the connection between the things I could figure out and what I needed to solve for, and I can carry out that calculation. Now I have a number for vwave. That equation was vwave equals l over t. If I rearrange that then, the variable I’m trying to solve for, the time, is equal to l divided by vwave. That’s equal to then two meters divided by 156.5 meters per second. I end up with the time. That’s .013 seconds, very short.