CALC. A thin, taut string tied at both ends and oscillating in it...

Question

CALC. A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x, t) = (5.60 cm) sin[(0.0340 rad/cm)x] sin[(50.0 rad/s)t], where the origin is at the left end of the string, the x-axis is along the string, and the y-axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern.

Accepted Answer

Hey everyone in this problem we have a second harmonic generated in an elastic cord of length, little L stretched between two clamps, the vertical displacement Y. Of a cord element as a function of the horizontal position X. And the time T. Is given by the equation Y. Of X. T. Is equal to to a sine two pi over lambda X. Sine of omega T. And we are asked to make a sketch of the standing wave pattern. Alright, so when we look at our general equation of our standing wave pattern, now we're given this equation here and we know that the coefficient in front of the signs is going to be the amplitude. And so what that tells us is in this case the amplitude Is going to be equal to two. And so if we look at our possible answer choices, we can already eliminate B and D. Because in the diagrams we can see that they show an amplitude of A. But our amplitude is to A. Because of this coefficient in front of the signs in our equation. So we have either A R C. No we have a standing wave. Okay, so this is fixed at both ends and recall that that means that we will have notes Okay occurring at x equals zero land over to λ over two, 3 λ over two etcetera. Okay. And it also means that we have anti notes that are gonna occur between those points at the quarter intervals. Okay, so that land over four, 3 λ over four etcetera. So we have an idea of where the nodes are, Where the anti nodes are. So what is this lambda. Okay. How can we relate lambda to our problem to the length of our strength? To figure out how many nodes and anti nodes we can expect. Okay, because the question is really coming down to, do we have another node within our length L In which case we have to ante notes or do we have only one anti node and no additional note inside? Well, recall that Land N for a standing wave Is given by two L over N. So in this case it's two times little L over end because little L is the length of our strength and were asked to draw the diagram of this second harmonic. Okay, now if we're looking at the second harmonic That tells us that N is equal to two and so lambda two is going to be two L over two which is just equal to L. Alright, so we know our wavelength now, we can figure out where the nodes and anti nodes occur combining the information. So we have notes At x equals zero at lambda over two. So what L over two? At two lambda over two. So just at L and that's going to be anything else is going to be longer than the length of our string, which isn't possible. So we see that we actually do indeed have three notes. Okay, and so we are expecting that we're gonna have option C. If we look at the anti notes, Those are gonna occur at λ over four. L over four. At three lambda over four. So at three All over four and anything further is going to be longer than that stream. Which is impossible. And so let's just go ahead and draw what we were given before we choose that answer. We have this okay, we have standing way, we have node at both ends. We found that we had a node At L Over two. So we have another note here. We have anti notes at L over four And three. L. Over four. So we can draw those. And if we connect what we have drawn or where we know this wave goes. It's going to look something like this. This is going to be the wave pattern for the second harmonic. Now we already talked about the amplitude, we know the amplitude is to A And so the distance from the center up to this top is going to be to A And so when we look at our answer choices, we see that this is going to correspond with answer choice. C. We have three nodes located at X equals zero X equals L over two and X equals L. We have two anti nodes and an amplitude of to a. That is option C. Thanks everyone for watching. I hope this video helped see you in the next one

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