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Wbco Trulcf #zte tavels on lakcThe water panticlcs more Up and dowtiThe Wave docs HO( carry any cncrgyTbc frcquercy aluayt cqual (0 thc WarelererThie WAtet patticle...

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Wbco Trulcf #zte tavels on lakcThe water panticlcs more Up and dowtiThe Wave docs HO( carry any cncrgyTbc frcquercy aluayt cqual (0 thc WarelererThie WAtet patticlee nove right along

Wbco Trulcf #zte tavels on lakc The water panticlcs more Up and dowti The Wave docs HO( carry any cncrgy Tbc frcquercy aluayt cqual (0 thc Warelerer Thie WAtet patticlee nove right along



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A water wave has frequency $0.40 \mathrm{~Hz}$ and wavelength $2.0 \mathrm{~m} .$ Its speed is (a) $5.0 \mathrm{~m} / \mathrm{s}$ (b) $1.3 \mathrm{~m} / \mathrm{s}$ (c) $1.0 \mathrm{~m} / \mathrm{s}$ (d) $0.80 \mathrm{~m} / \mathrm{s}$.

In this problem. We're going to kill. Create the referent off. Somewhere in what? But this reverend is our Linda. In order to calculate this, Reverend, let's write the formula for the speed of sound as a week or two. Linda. So what is holding the squaring off? Orlando, you can write Lambda equals two really wanted by if their school and equating about one here. This with the speed of sound in water. No way. In setting, weathered into the square we can write Lambda is equals to We have the very off for this we as 1000 493 meter Boskin, Mr Wanted by the frequencies. 261 point six hours. So this would be would be writing for this land, Eiselin vehicles to 5.7 meters. So this is very quiet. Answer. Thank you

In number eight were given the velocity of waves and Jell O water, The wavelength of the waves and the depth of the water when we're supposed to find speed on the period of the waves with the speed of the waves were going to use the equation Lee gave us G is 9.8 acceleration due to gravity and times the depth, it's 0.75 centimeters. I'm gonna put that in meters. I get point to seven meters per second. Then they asked me the period. Well, the period is the reciprocal of the frequency. And I know the frequencies in the wave equation so I can just plug in My velocity had just found but wavelength I was given I'm gonna put that in meters, so point 0 to 6. And I want to find period, not the frequency. And the period will be just the reciprocal. The reciprocal period is the frequency. So now I just, um, divide and then I'll take the reciprocal both sides to get tea by itself and they get a period of 0.96 seconds

In this problem, we have given the frequency of the Web. Ad let it be f. is equal to 0.50. I. Just Also we have given the amplitude value, let it be a 0.35 meet. And where plant is given edge linda is equal to 2.6 m. As we all know that the formula for equivalent blamed I is equal to we buy F as it is independent of amplitude, of Web, as it is independent of amplitude of the given Web. So when amplitude double the Dublin become unchanged, the we planned to become unchained, the web land become untamed when the amplitude doubles.

All right. I wrote down the pertinent information. We're given an equation for V of deep water waves. We're told that it applies when the death is greater than 1/4 of a wavelength. We're given the depth and width of a pull. And then I drew a diagram of what this would look like. Um, looks like I actually chose the 1234 the fifth mode to draw. Um, nevertheless, part a asks us for the 1st 3 standing wave modes. Well, first mode is V over to L where l is the width of the pool. Although I don't want to use that yet. The 1st 3 standing waves are going to look like this. Okay, that would be the 1st 1 And so the wavelength would be the entire length of the pool, times two. So that would be 20 meters. Um, the 2nd 1 is gonna look like this. All right. Um, the 2nd 1 the wavelength would be the full length of the pole. 10 meters, and the 3rd 1 is gonna look like, um, do this again. Oops. Wavelength is going to be here. Um, 2/3 of the length of the pole, which is going to be 6.67 meters. All right, moving on to be Wait a minute. I can't move on to be yet. They didn't answer the whole question. So, um, are these deep water waves? Um, and I have to draw a graph. Reach Well, I did draw a graph. Uh, right here. Whoops. Right there. I do know my goodness. There. I drew the graph. Um, are these deep water waves? Well, the depth, which is five, must be greater than 1/4 of the wavelength, So the wavelength must be less than 20 meters. Um, so, yes, except for the 1st 1 Because the 1st 1 is exactly, Ah, 20 meters. So, yes, except for the first Ah, now one to be just have to calculate the wave's speed. Now, I think this is really going to apply to the 1st 1 even because it's just on the edge of applying. So wave speed is the square root of G wavelength over two pi g over two pi and the wavelength for number one is 10. So let me put these into a calculator. Square root of 9.81 is G wavelength one was 20 over to FYI, that gives me 5.59 meters per second v two square root of G wavelength to over two pi wavelength to with 10. That gives me 3.95 meters per second and the third wave with 6.67 So I just put in a 6.67 there, 3.22 or 3.23 meters per second. All right, see frequencies of standing waves. So frequencies of standing waves would be some number m times v over to l, which would be some number m times V which well, em over to l Times V, which is the square root of G lambda, uh, over two pi efs of em. And I have G i m and l just thinking about why Lambda is in there. But ah, I think that's fine. All right. De the periods of the first three standing wave modes. So frequency first is one over to El Square root of G lambda. Um, Lambda for the 1st 1 was 20. I'm gonna write lambda one there over two pi. All right, let me put this in the calculator 1/2 times. 10 because l with 10 square root of 9.81 times Lambda one, which is 20 over to high. That gives me a frequency, uh, 0.28 But we're not actually asked for frequency were asked for period. So one over the frequency is the period, which is three point five. And I'm looking for how many significant digits? Perhaps three significant digits. 3.5 eight seconds. Okay, so let's just go for a period to immediately. All I have to do is invert this. This will be to L. A times the square root of two pi over g limbed to So to all. The only thing that's changing is Lambda, which is now 10. Actually, I think I'd like to proceed in the way that I had done for the 1st 1 My goodness, what is going on here? F two is one over to L. The square root of G lambda to over two pi. Um, so Lambda Two was 10 one over to L Square root of 9 21 times 10/2 pi. That gives me 0.1. Actually, 0.20 hurts hurts. This is giving me a period of five 0.6 But that's not the answer that's in the back of the book. Unless I'm mistaken, I'm gonna go check says two point 53 And the last one says 2.7 The 2.53 would be hair of that frequency that I just calculated. So where did I make a mistake? Here? I did one over to L Square root of G lambda over two pi. All right, well, let's do the 3rd 1 Oh, I forgot about em. Ahem. Is to now right to over to l times that. And then when I divide by two Okay, that's going to give me the 2.53 seconds. God 3/3 l square. Root of G limbed. Ah, 3/2. Pi is okay. Three, not three l in the denominator. That should be a to their That doesn't change over to l. Ah. This time Lambda is 6.67 All right? That gives me an answer for T three one over F two of 2.7 seconds, which is correct. That's what's written in the back. Oh, my goodness. It was written in the back of the book. Um, let me figure out what that frequency Waas 48 hurts. All right, So I made a little mistake Where I forgot to put em in em is one to three. And once I did that gave me the correct answers.


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