A transverse wave pulse travels to the right along a string wi...

Question

A transverse wave pulse travels to the right along a string with a speed v = 2.0 m/s. At the shape of the pulse is given by the function D = 0.45 cos (2.6x + 1.2), where D and x are in meters. Determine a formula for the wave pulse at any time t assuming there are no frictional losses.

Accepted Answer

Hey, everyone in this problem. A physics student is experimenting with a rope, one end is fixed to the wall and the other end is in his hand, the rope is horizontal and he's gonna generate a wave pulse on the rope by giving it a slight shake afterward. The pulse is gonna travel along the rope to the right with a speed V of 1.5 m per second at time T equals zero. The student notices that the shape of the pulse is described by the function P is equal to 0.25 cosine of two X plus one where P and X are in meters and works to find the formula for the wave pulse at any time. T assuming there is no friction at play. We're given four answer choices, options A B and C all contain a different equation for this wave pulse. And option D is that it is none of the above. We're gonna come back to these equations. Um Once we're done working through this problem, so the first thing we wanna no notice is that we're given this equation for the shape of the wave at time zero and we have a cosine wave. OK. So we recall that we can represent waves in terms of our cosine function or in terms of our sine function like either is OK. And they can be equivalent if we take into consideration some phase shift. OK. So if we shift our cosine curve, it's gonna become the sine curve and vice versa. OK. So in this case, we have a cosine curve. When we think about the equation, we're gonna be thinking about how to represent this in terms of cosine. And that kind of matches with those answer choices. The answer choices give us equations with cosine. OK. So if we think about the general equation in terms of cosine, we have that the displacement D can be written as the amplitude of the wave a multiplied by cosine KX minus omega T plus the face of five. That's our general equation to describe a wave. When we're talking about describing it with a cosine curve. OK. So if we think about what happens at time, T equals zero at T equals zero, what we have is that this equation becomes D is equal to a multiplied by kung of KX plus F. Now, the way we were given P is equal to 0.25 cosign of two X plus one. So we can compare these two equations. When we substitute T equals zero into our general equation. We get the equation on the left A A cosine of KX plus five. And the equation we're told that we get is 0.25 cosine of two X plus one. So we can notice that our amplitude A is gonna be 0.25. That coefficient open front multiplying our cosa. We also noticed that K. So the coefficient inside of the cosine multiplying that X term is going to be two and our phase shift P is going to be equal to one. We were just comparing these two equations. Now, we know our A value, our K value and our five value. Now, the last thing we need to do is to calculate omega. OK. So what we have so far is that A is equal to 0.25 K is equal to two and th is equal to one. But you'll notice in our general equation for D, we need this omega term as well. So let's think about the relationship between omega and some of these values we've already figured out and we will recall that Omega can be written as two pi multiplied by the frequency F. Well, we don't have the frequency app. So how can we relate the frequency app to some of these variables that we do have? Well, we can recall that the frequency is related to the speed in the wavelength and the frequency can be written as the speed of the wave divided by its wavelength. And we know that the wavelength is related to this K value, the wavelength is equal to two pi divided by this K value. And we kind of have these three steps to relate the information we were given in the problem to this Omega value that we're trying to find. OK. So Omega is two pi divided by F which we now know we can write as two or sorry, two pi multiplied by F which we now know we can write as two pi multiplied by V divided by length. Hm We can then write that as Omega is equal to two pi multiplied by V divided by two pi divided by K. And this makes our Omega value just equal to K multiplied by B OK. So this K value we know right, this K value we found using that general equation and the equation we were given to represent the wave at times zero and the speed we were given in the problem. And so we get omega is two multiplied by 1.5 m per second, which is going to be equal to three. And so our general equation for this wave at any point in time T is going to be equal to 0.25 cosign of two X minus three T who was one. And that is the equation we were looking for. If we compare this to the answer choices we were given, what we'll see is that this corresponds with answer choice. A answer choice B and answer choice C both had the wrong values for all of these terms. So the Amplitude Omega K and Phi they all had different values there that do not match what we found. So the correct answer here is option A. Thanks everyone for watching. I hope this video helped see you in the next one.

A transverse wave pulse travels to the right along a string with a speed v = 2.0 m/s. At the shape of the pulse is given by the function D = 0.45 cos (2.6x + 1.2), where D and x are in meters. Determine a formula for the wave pulse at any time t assuming there are no frictional losses.

Key Concepts

Wave Function

Wave Speed

Time Dependence of Waves

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