A horizontal string tied at both ends is vibrating in its fundame...

Question

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. (a) Calculate the maximum transverse velocity and maximum transverse acceleration of points located at (i) x = λ/2, (ii) x = λ/4, and (iii) x = λ/8, from the left-hand end of the string.

Accepted Answer

Hey everyone in this problem we have a wire stretched horizontally between two vertical stems. When it's strummed, the wire vibrates in its fundamental mode at a frequency of 95 hertz. The transverse waves on the structure wire of a wavelength of 31 centimeters. A speed of 30 m per second and an amplitude of five millimeters. Were asked what are the maximum transverse velocity, a maximum transverse acceleration of the wire particle at X equals 15.5 centimeters. X equals 7.75 centimeters and X equals 3. centimeters from the left stem. Alright, so let's just go ahead and write out what we were given. Were given the frequency f is equal to Hz. The wavelength lambda is equal to 31 cm Converting this two m we get 0.31 m. The speed V. is equal to 30 meters per second And the amplitude A. is equal to five mm or 0. 0.005. There we go m. Okay, Alright, so that's everything we've been given. Now we're asked to find the velocity and the acceleration. Well we know that those are related to the displacement. So let's go ahead and write out the general equation for the displacement of this transverse wave of Y. Of X. T. Is equal to to a sign K. X. Sign. Oh my God, T. Okay. Now in terms of our problem we have A but we don't have K. Or omega. What we can do though is we can find Kay and omega using the values were given. Okay so recall that K Is equal to two pi Over Lambda which in our case is two pi Over 0. m. And omega Similarly is related to the frequency through two pi f. Which is going to be equal to two pi hertz Which is equal to pi or pi per second. Alright, so that means we can write our equation as why of X. T. Is equal to two times a. We're just going to be 0.01 m Sine of KX two pi Radiant over 0.31 m X Times sine of Omega which is pi radiant per second times time. T. So that is our equation for the displacement of our transverse waves. We want to find the maximum transverse velocity, maximum transverse acceleration. So let's start with the velocity. How is the velocity related to the displacement? Well, it's going to be the derivative. Okay so the velocity V at X T is going to be equal to the derivative. Okay, the change in the displacement over time Which is going to be 0.01 m. And here we're taking the derivative with respect to t. So the sine term with the X is going to be treated like a constant. So we leave it alone. Now this term we're taking the derivative of sine case. We're gonna get the derivative of what's inside 190 pi radium per second many times the derivative Sine which is cosine of everything inside 190 pie radian per second. T. Alright, so this is our V This is a velocity in our case. We want to find the maximum velocity at a particular X value. Okay, so the maximum V at a particular ax is going to be given what? Well, if we're at a particular X value, the maximum is going to be given when this cosine is equal to one. Ok. That's the maximum value of that Kassian. And so that's gonna be the maximum. So it's going to be when co sign of 190 pi T is equal to one which tells us that our equation for V max Is going to be 0.01 m. Okay, I'm gonna move this constant. 190 pi to the front And 190 pie radium per second time sign two pi radiant. There is 0.31 m X chris. All right, So this is our equation for V max and you'll notice that now the equation only has the variable X. So we can go ahead and plug in our value for X that we're interested in and we're going to get the maximum velocity that we're trying to find. Okay, let's do the same for the acceleration, acceleration is related to velocity just like velocity is related to displacement. Okay, the acceleration, it's going to be the derivative DV D T. The velocity. Okay, so the change in velocity over the change in time We do this we get 0.01 m. Okay, sign the term with the X stays the same because we're taking the derivative with respect to T. So this is treated like a constant. Now when we take the derivative of coast we're gonna have to take the derivative of what's inside. So we're gonna get another 190 pi radian per second multiplied. Okay so we have 190 pi radian per second squared. Now the derivative of this coast is going to be negative sign pie radiant per second times time T. So that is our acceleration and the same thing. We want to find the maximum transverse acceleration at a particular X value. So at a particular X value, the maximum a well what's this gonna look like? Okay, we have a negative here. So we want the maximum to be when this sign term is at its lowest. Okay, so negative one. So we want sign 190 pi T to be equal to negative one. Okay and that's gonna give us our maximum a that's negative is gonna cancel with this negative and we're gonna be left with a max Is equal to 0.01 m Times pi radian per second squared time sign. two pi over 0.31 radiant meter X. Okay. And again, this is an equation with the just X. So we can substitute in the X values we're interested in and find our max acceleration. Okay? Alright, so let's give ourselves some more room to work and we're gonna start with the first X value. Okay, So X is equal to 15.5 cm is the first x value were asked to investigate. So 0.155 m. Now, we're gonna drop our units here just for the sake of space because there is a lot of writing in this problem. Okay, But let's go look first, we're gonna end up with a meter per second squared units here are going to go away. Okay, So we're gonna add meter per second squared on acceleration and similarly meter per second on velocity. Okay, So the units do check out but we're just going to omit them from writing here to save some space. But it is a good idea to always write the units out fully in your question. Okay. All right, So at 15.5 cm we're gonna have v max is equal to 0. Times 190 pi time sign. It's two pi over 0. Times 0.155. Now 0.155 divided by 0.31 is going to give us a half. So we get two pi divided by two. That's going to give us pi 0.1 times 190 pi times Sine of pi. Okay, well what sine of pi zero? So this whole thing goes to zero And we get zero m/s for maximum velocity there. Okay, similarly for our max acceleration We have 0.01 times 190 pi squared times sine of two pi over 0.31 time 0.155. And what you'll notice is that the sine term that we use in the V max equation is the same as the a max equation. Okay, so here we're going to get sine of pi as well, which is going to give us zero So we have zero m per second squared. Alright, so for the first x value that we're interested in 15.5 cm we have a speed or a velocity and acceleration of zero for the maximum. Okay, what you'll notice is because we have this sine of pi. Okay, this is actually a note. All right, let's move over here and look at our next x value which is going to be 7.75 cm Or 0.0775 m. And what we have is we have V max Is equal to 0.01 Times 190 Pi Times signed coupon 0. Times 0.0775. Now this time when we do 0.775 divided by 0.31, we're gonna get a quarter. So two pi divided by four is gonna give us pi over 20.1 times pi times Sine of pi over two and a pie over to sign is at its maximum. So this is actually an anti note And the value of Sine Pi over two is 1. So we get 0.01 times 190 pi Which is going to give us a value of five 0.969 m per second. Okay, We're gonna have a max a similarly 0.1 times 190 pie and sine of two pi over 0. times zero point whoops. And I missed the square 190 pi squared. 0.775. Now again, this sign term is going to work at the same. We're gonna get sine of pi over two which is going to be one. Okay? And this is going to give us and acceleration of 3563 m per second squared. And if we write this in scientific notation 3. times 10 to the three m per second squared. Okay, so here is our velocity and acceleration maximum max transverse velocity and acceleration at 7.75 centimeters. And now we're going to so a tiny bit more room to do the last X value. Okay. And that's going to be at X equals 3.875 cm. Which is 0. m. All right, no V max is going to be equal to 0.01 times 190 pi times signed of two pi over 0. Times 0.03875. Okay, now in this case we are going to get 1/8. Okay. And then two pi over eight is gonna be pi over four. So we get 0.1 times 190 pie time sign of pi over four which gives us a v max value Of 4.22 m/s. Now we're going to continue over here so I'm just gonna draw a line. So it's clear That we are still talking about X equals 3.875. And we'll just divide all these calculations up. Alright, so we're a max here is going to be the same equations before 0.1 is 190 pi squared times sine of two pi over 0. times 0. 875. Okay. And again we're going to get pi over four there. And when we work out the calculation we get 2.52 times 10 to the three m per second squared. Okay, so that is our max transverse acceleration at 3.875 centimeters from the left hand side. Okay, so now we've got all our values. Let's go back to the top. And what we will see is that we have answer a okay. At 15.5 centimeters, the maximum transverse velocity and maximum transverse acceleration are both zero at 7.75 centimeters we have the maximum velocity of 5.96 m per second. Maximum acceleration of 3.56 times 10 to the three m per second squared. And at 3.875 centimeters we have the maximum velocity of 4.22 m per second and a maximum acceleration of 2.52 times 10 to the three m per second square. Thanks everyone for watching. I hope this video helped see you in the next one.

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