A horizontal string tied at both ends is vibrating in its fundamental mode. The traveling waves have speed v, frequency f, amplitude A, and wavelength λ. (c) How much time does it take the string to go from its largest upward displacement to its largest downward displacement at the points located at (i) x = λ/2, (ii) x = λ/4, and (iii) x = λ/8, from the left-hand end of the string.

Video transcript

Hey everyone in this problem we have a violin with a horizontal cord. Okay and it's held in place by two fixed ends. We're told that the fundamental frequency of the stationary waste produces 440 Hz. And were asked, how long does it take? The oscillating cord to go from the max positive displacement to the max negative displacement at a point X. Equals lambda over four from the left end. Alright so we're talking about stationary waves and we're looking for the displacement at a particular point. Okay so let's go ahead and write an equation for the stationary wave. Okay now we're called the general equation for stationary wave can be written Y. Is equal Y. Of X. T. Story is equal to to a sine of K. X. Sign of omega T. Okay now we know that we can write K. As two pi over lambda. Okay so we get to a sine of two pi over lambda X. Times sine of omega. Alright now let's plug in the X. Value we're looking for and see what we can do from there. Okay so we want to look at a point X. Equals lambda over four. Okay so why lambda over four T. Is equal to to a sign of two pi Over lambda K. Times lambda over four sign of omega T. Okay so simplifying lambda pi over four T. Is equal to two A. Okay the lambdas will cancel. And we're just left with sine of pi over two time sign. I'm omega T. And sine of pi over two. Well that's just gonna be one. Okay so we're just left with two A sign of omega t. Alright now we're asked to find the time. Okay so we're looking for tea. No one has to find the time that it takes to go from max positive displacement to max negative displacement. Okay. So let's start on the left hand side here with the maximum positive displacement. All right, so let's think about this. I'm just gonna give her us some more room to work. Okay so we have this equation. When is a max positive displacement gonna occur? Well, we know that the maximum value of sine omega T. Is one. Okay. Two times a that's a positive number. Okay so the maximum positive displacement is gonna occur when sine of omega T. We're gonna call this omega one. Okay. This will be the omega where we have the max positive displacement or sorry not omega one T. One. The time that we have the max positive displacement. Okay so sine omega T. One is equal to one. Well this means that Omega T. one. It's gonna equal pi over two. OK. It could be some multiple of power. Super. When you take the first instance. Okay. Alright So T. one is going to be equal to pi over two. Omega pi over two omega. We don't know any information about omega but we do know information about the frequency. Okay so let's relate omega to our frequency and we recall that. That can be done. Omega is going to be too high times of frequency. Alright so the pie will cancel we're gonna be left with 1/4 f. Recalling from the question. Our frequency is 440 Hz. Which is going to give us T. one Equals 5.682 times 10 to the negative four seconds. Okay. All right, we found our time. But let's remember what we're looking for. This isn't our answer. Okay this is the time where we have the max positive displacement but we want to know how much time it takes to go from the max positive displacement to the max negative displacement. Okay so in order to do that, we need to find the time when the max negative displacement occurs. Okay so let's do that over here on the right max negative this way smith. Alright, so that's gonna occur in this case instead of sine omega and we're going to call it t to being equal to one. We're gonna have an equal to negative one. Okay that's the most negative value we can get. So that's when we are going to have our max negative displacement. Well in order to have sign equal to negative one, we need whatever's inside of the sign Omega T. 2 to equal to three pi over two. Okay and again there could be multiples. There are other values that would give you a negative one but we're gonna choose that first instance. Okay. Alright so this gives us T. Two is equal to three pi over two. Oh my God. Omega is equal to two pi F. Same as when we were calculating the T. One instance. Okay so we get three pi over two times two pi F. The pie will cancel. And we're going to be left in this case with 3/4 f. Which is equal to 0. seconds. Okay that's T. two. Okay. So we have our T. one. We have our tea too. And what we want to find is we want to find the change in time. Okay. How much time does it take to go from one to the other? And so what we're looking for is delta T. Which is T. Two minus T. One. Okay. And this is gonna give us 1.14 times 10 to the -3 seconds. Okay. And that's it. That's the time that it takes to go from the maximum positive displacement to the maximum negative displacement at that particular point on the violence strength. Okay so that's going to correspond with the answer. A thanks everyone for watching see you in the next video