A series of parallel linear water wave fronts are traveling directly toward the shore at 15.0 cm/s on an otherwise placid lake. A long concrete barrier that runs parallel to the shore at a distance of 3.20 m away has a hole in it. You count the wave crests and observe that 75.0 of them pass by each minute, and you also observe that no waves reach the shore at +-61.3 cm from the point directly opposite the hole, but waves do reach the shore everywhere within this distance. (a) How wide is the hole in the barrier?

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Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem in a water park. Surface waves are generated in a large swimming pool and propagate at a speed of 0.2 m per second. A 5.0 m wall stands in the pool perpendicular to the direction of the propagation of the waves and has an opening in its center using a wave profile device. You measure that every 30 seconds, 45 wave crests strike the wall. You also notice that no waves reach the opposite side of the swimming pool 4.0 m away from the wall at a position of 0.40 m from the point directly opposite from the opening. However, waves do not reach the opposite side everywhere. Within this distance, determine the width of the wall opening. So our end goal is to determine the width of the wall opening. So we're given some multiple choice answers. They're all in the same units of meters. Let's read them off to see what our final answer might be A is 0.133 B is 0.399 C is 0.447 and D is 1.34. OK. So first off, let's let us recall the equation for destructive interference. And let's call it equation one. So destructive interference states that sine theta where theta represents some angle is equal to M which represents a minimum value which is some integer value multiplied by the wavelength represented by lambda divided by a which represents the slit width. So the width of the slit. OK. So to begin, let us determine the wavelength of the water waves. So let's call that equation two. So the equation define the wavelength is wavelength is equal to the speed of the propagation of the water waves divided by the frequency of the water waves. So now we need to solve for the frequency which it's called that equation three. So to find frequency, we need to take this key piece of information in the prom which we need to use the wave profile device answer that we found in the prom. So every 30 seconds, 45 wave crests strike the wall. So this is how we find our frequency. So we take the amount of the wave crests. So there's 45 crests divided. So every 45 crests occur at 31st. So every 30 seconds, 45 crests occur. So that equals 1.5 Hertz. These are Hertz value, the units is like cycle per second. You can look up that in a, in a video to get further information on how Hertz is found. But that's how we determine our frequency. So now we can solve for lambda to the wavelength. So using equation two and plugging in all of our known variables, wavelength is equal to the speed of the propagation of waves which is 0.2 m per second divided by the frequency of the waterways which is 1.5 Hertz. OK. So now we need to consider the following diagram to help us determine the first angle at which destructive interference occurs. OK. Which is gonna be theta one in the distance Y subscript one. So theta subscript one and Y subscript one is shown here with this red arrow. OK. So we need to use the following equation, let's call it equation four. So tangent theta subscript one, theta one is equal to Y subscript one, Y one divided by X where in this case X is 4 m. OK. And why one is given to us in the problem as 0.40 m? OK. So in order to solve for theta one, we need to take the inverse of tangent multiplied by Y one divided by X. So let's plug in our known variables, the sulfur theta one. So why one is 0.40 m divided by 4.0 m? So when you plug that into a calculator, you should get that theta one equals 5.71 degrees. OK. So going back to equation one, we can use equation one and we can solve for the slit width or the, which is our final answer to me is we're trying to find the slit width. So when we rearrange equation one to solve for a, the slit width, we get that A is equal to LAMBDA, which is the wavelength divided by sine theta. So let's plug in or known variables to solve for the slit width. So let's remember that the value that we found the wavelength was 0.133 m divided by sign of which is the theta one value we just found which was 5.71 degrees. So when we plug that into a calculator, we should get 1.34 m, which is our final answer. So this is the width of the wall opening. Fantastic. OK. So that means our final answer has to be D 1.34 m. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

Verified Solution

Key Concepts

Wave Speed and Frequency

Diffraction of Waves

Interference Patterns