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Question

The Bay of Fundy, Nova Scotia, has the highest tides in the world. Assume in midocean and at the mouth of the bay the Moon's gravity gradient and the Earth's rotation make the water surface oscillate with an amplitude of a few centimeters and a period of $12 \mathrm{h} 24 \mathrm{min}$. At the head of the bay, the amplitude is several meters. Assume the bay has a length of $210 \mathrm{km}$ and a uniform depth of $36.1 \mathrm{m}$. The speed of long-wavelength water waves is given by $v=\sqrt{g d}$, where $d$ is the water's depth. Argue for or against the proposition that the tide is magnified by standing-wave resonance.

Sarah Mccrumb · Accepted Answer

Hi, everybody. So, for this one, um, we need to argue are for or against the proposition that the tide is nine fired by the standing wave. Um, resonance. So we need to find the speed of water as we have speed of water as g divide about times D. And that&#x27;s the grab These in 9.81 meters per second squared and the old things were routed times on the distance, which was 36.1 years, which is what was given to us for the Nova Scotia uniformed death so that he&#x27;s been depth and we get 18 a point eight meters per second. OK, so now we need a wavelength of the water wave, and that is gonna be Landau over four. Okay. And again, that&#x27;s just has to dio where, um, sine waves. And that&#x27;s a terrible sign has to do with sine waves. And, um, this one is just the fourth of it. Okay. And, um um, you guessed race this, We don&#x27;t need it. You do, Erica. And because of that, um, to me, land a before and then it. So you&#x27;re lamp lambda is ableto four of the O, which is gonna be four times two tied at times. Time to three nurturers. Okay, Which equals 8 40 times 10 to 3. Nurse. Okay. And so we have our t equals one over. Frequency and frequency is been equal to be over Lambda. So that means you&#x27;re t is gonna equal Teoh, uh, lambda over velocity and we get eight 40 times 10 to 3 meters, divided by the velocity that we found above 18 point year for eight meters per second. And then we get 4.47 times, 10 to the fourth seconds. And if you want to convert this two hours, we have one hour over three, 66 seconds and we get 12 points for one hours. Okay? And so now what we can dio is it&#x27;s giving us, um for this. It says that centimeter period is 12 hours with 24 minutes. So what we have here is 12 hours, people. Full hours. Plus, we separate the 0.41 that we have here hours and we&#x27;re income for that little bit. Two minutes, minutes, hours. Okay. And with this wind up getting in at 12 hours and 24 minutes and This is just using everything they gave us in the problem. And so this does agree that with the period of the wiener expectations of initial high tide is amplitude of the residence. Okay, guys, and let me just make that 60 just a little bit better for everybody. Okay? Thank you, guys.

Mark Mathison · Answer

All right, So in this question, we are deciding whether or not, um, we should support the argument that Ah, the tides in the Bay of Fundy are so amplified because due to standing wave resonance or not, um and so there is, you know, quite a few ways that we can do this, but let&#x27;s start with what we&#x27;re given in the problem. Um, so the period Ah, that&#x27;s observed for the tides in the bear fund, a Bay of Fundy, uh, is 12 hours and 24 minutes. Um, we&#x27;re told that the length of the bay is 210 kilometers. Say that l is equal to 210 kilometers, and we are supposed to assume that it has a uniform depth of 3 36.1 meters. So we can say D is equal to 36 0.1 meters. Um then the last thing that we&#x27;re given is we should assume that, um, waves in the ocean. Ah, the speed of these waves can be given by the formula V. Is he holding the square root of g de where D is the depth of the water at that point? So the same day that we just used to define the depths of the baby. Um, so then, ah, this is all we&#x27;re given. So the first step is to kind of compare the Bay of Fundy to some sort of physical symptoms system that we have experience with, um, and one that we can compare it to is a pipe organ that is open at one end. Um, so for this one for this kind of system, um, yep, that would mean that wavelength, uh, the wavelength of waves in a pipe organ or that&#x27;s open up one end is equal to four l divided by end. So it&#x27;s a little different from what we&#x27;ve been doing with, like, waves on a string or something, where they&#x27;re bound at both ends. If it&#x27;s only bound at one end. Um, then the wavelength is four time for l divided by N instead to elder. Bye bye. So that&#x27;s the important distinction to make here. And that&#x27;s kind of where the first logical step is that you need to do for this problem. Um, so next we have the period. So let&#x27;s relate that to the frequency. So the period of the fundamental, um will be won over the frequency, the fundamental frequency. So this argument that the tides in the Bay of Fundy are so large due to standing wave resonance the way that we can prove or disprove that or support it or cast doubt on it would be to see if the periods that would be created, theoretically by standing wave rather excuse me, standing wave resonance. It is similar to are nearly equal to, um, the actual period that is experienced by the system. Because if those periods are the same than, um, modeling it as a standing wave system does show that this period would create resonance. So we want to see if the theoretical period for a standing wave in the Bay of Fundy would be the same as the actual period. So what we&#x27;re doing is we&#x27;re calculate. We&#x27;re trying to calculate the theoretical period if the Bay of Fundy you were experiencing standing way resonance. So to do this, we need to look at the physical characteristics years. So we have that the theoretical period would be won over the fundamental frequency. Um, now F one, when it&#x27;s related to wave, speed and wavelength will be V over Lambda. And as we&#x27;ve said, V is equal to the square of G D. And Lambda is equal to four l over end. Now, in this case, we&#x27;re dealing with the fundamental so we can just say that the wavelength is then for help. So if we rewrite our equation for the period, we will get that substituting for ah, substituting V and Linda in for F one. We&#x27;ll get that t one is equal to Lambda over v and we have expressions for Linda and be so we can say that Linda is then equal to for L and V is equal to the square root of G d. And we have values for all these. So we can go ahead and put these in so we&#x27;ll get that four times. Hell is four times 210 kilometers, but our depth is in meters, so we&#x27;ll go ahead and convert that to 210,000 meters. Divide that by the square root of G d. Who that was sloppy. Here we go 9.80 times 36.1. So if you plug this into your calculator, you will get that the period T one is equal to something along the lines of ah, four point 47 times 10 to the fourth seconds. But our enters an hour and minute or the the given result for the period is in hours and minutes. So we need to convert that in two hours. Ah, and we do this by multiplying it by one over 3600 cause there&#x27;s 60 minutes in an hour. 60 seconds in a minute, 60 times sixties, 3600. If you plug that in your calculator, you will get that, uh, T one is equal to 12 point for hours, and if you double check, 0.4 hours is equal 24 minutes. So the theoretical ah period is 12 hours, well, approximately equal to 12 hours and 24 minutes. There&#x27;s some decimals after our t one, but it is approximately 12 hours 24 minutes. And since the um, theoretical period is very close to the actual period, we can say that, um, the argument that the Bay Fund these tides are so large due to standing with business has some physical significance to it, and so we can support that argument. And that&#x27;s how you do this problem

Jonathan Ibarra · Answer

were given that the waves at Earth surface water has a time period of 12 hours and 24 minutes. Only teeter argue for or against this proposition. 14. So to solve this problem, we&#x27;re gonna use the fact that a time period t is equal 21 over f, which is also equal do the wavelength divided by its velocity. So what we&#x27;re gonna have to do to solve for this wavelength and also solved for this velocity sent? These are the two unknowns that we need to find. That&#x27;s all for this time period. So now we&#x27;re gonna start by gang this velocity and by the problem, this velocity is going to be equal to the square root of G time, Steve. So plug in 9.8 Fergie and then plug in an 36.1 40. And so this will give us a velocity of 18.8 meters per second. Next we can say that length, though, is equal to a wavelength divided by four. And so we plug in junior 10 times 10 to the three meters for ill and said it equal to Lambda, divided by four. And now take this solve it for Lambda and we&#x27;ll get away fling. That&#x27;s equal 2 840 kilometers. So now we can say that a time period t is equal to Lambda divided by. So we just plug in Lambda and then we plug in B and this will be you cool to 4.47 times 10 to the four seconds which we can also say is equal to 12 hours and 24 minutes. So we can say that we agree with the time period that is given by the problem.

Khoobchandra Agrawal · Answer

I agree. One. Here it is given the way off Funding Novaes. Scotty A has the highest types in the world at the history. In the photograph assume that in the mid ocean on at the mouth off the the the multi gripped ingredient on earth rotation Make the water service oscillate with an amplitude off few centimeters on period off. 12 hours, 24 minutes at the head. Off the way. The ambulatory, several meters. You have tow argo in favor or against off. That site is amplified by standing big graduates as soon the way has a lend. 200 kilometers uniformed death off 36 point perimeter. The speed off long wavelength water wave is given by route Off de into the Let us see him The wave experience 18.8 m per second. You may calculate by this No, the way has one and open and one on closed. It&#x27;s the simplest regiments. Is this unknown? Off original velocity bit are not off. Our gentle Valerie, which is also an is also and okay. And the note off Vertical displacement as the head off way and and anti note. Okay. Off. Felicity written no off displacement at the mouth. The vibration off water in the way is one day, 10 kilometers. It&#x27;s called Lambda by foot. So we have a lint off. Vibration off. Water is 8. 40. Yeah, and tend to the power 3 m. Therefore, a period off. Our solution, a solution or vibration is the ratio call of your country. That is, they placed upon its speed. The violence is 8. 40 in tow. 10 to the power three speed. We have measured 18.8. So it will be 4.47 Yeah, that is 12 hours, 24 minutes. This agrees bit period off. This agrees with. Yeah, Beat it up. No, not excitation. So we identify that extra high, tight as amplified by region. It&#x27;s so we identified the extra height stripes as amplified by regiments. That&#x27;s all for it. Thanks for watching it.

Heather Zimmers · Answer

this problem is about high tides, and we have this equation D of tea, which represents the depth of the water and tea representative, represents the number of hours since midnight, and we&#x27;re asked for how fast the tide is rising or falling at different times. And when you see the phrase how fast you want to think about rate of change and when you think about rate of change, that means derivative in calculus. And so we want to start by finding the derivative of the function so deep prime of tea, the derivative of 70 And then we move on to the other term and we can leave the five and take the derivative of co sign, which is negative sign. So we have negative sign of the entire inside. And then, according to the chain rule, we&#x27;re going to multiply by the derivative of the inside. And so we would be multiplying by 0.503 Okay, so now that we have that derivative, we could simplify it if we wanted to. But chances are we&#x27;re just going to be using a calculator from this point. So the calculators not going to mind if you don&#x27;t simplify it for part A. We want to find the speed of the tide changing how fast the tide is rising or falling at 3 a.m. And 3 a.m. is going to be three hours since midnight, so we&#x27;re going to use T equals three. We&#x27;re going to substitute that into the derivative, use a calculator and you get approximately 2.39 meters per hour for part B. Three more hours have gone by, and now it&#x27;s 6 a.m. So T equals six. We&#x27;re going to substitute six into the derivative, use a calculator. Make sure your calculators and radiance, by the way, and we get 60.9 to 6 meters per hour. And for part C, it&#x27;s now 9 a.m. So we&#x27;re going to use T equals nine. Substitute that into the derivative. Put it in your calculator on you get approximately negative 2.28 Now, why is it negative? If the derivative is negative, that means you&#x27;re quantity is decreasing. So that&#x27;s telling us that the tide is going down and then finally, for party, it&#x27;s noon, so T is 12 12 hours since midnight, and the rate of change AT T equals 12 is approximately negative. 1.21 Again, the tide must still be going down at this point in time.

Adnan Gill · Answer

so looking at the data given in the question, we can draw this craft so this would be a sign craft starting A T is equal to zero. The repeated off 12 and the maximum height reaches 21 lost his negative 21. So this would be the graph that we can draw for this question, given the information. So the aptitude is 21. The period it is 12 hours, and that means we can count the B, which is two pi divided. Find a period. So that&#x27;s a two pi divided by 12 which is equal to pi or six. So the equation finding that beacon right would be equal Why is equal to 21 sign high over six times T?

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On a marimba (Fig. P14.53), the wooden bar that s…

Problem 52 Hard Difficulty

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$12.4 \mathrm{h} \approx 12 \mathrm{h} 24 \mathrm{min}$

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$12.4 \mathrm{h} \approx 12 \mathrm{h} 24 \mathrm{min}$

Principles of Physics

Farnaz M.

Zachary M.

Aspen F.

Meghan M.

Electric Charge

The Electric Force

Raymond A. Serway, John W. Jewett, Jr.Principles of Physics

Farnaz M.

Zachary M.

Aspen F.

Meghan M.

Raymond A. Serway, John W. Jewett, Jr.

Principles of Physics