A fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x, t)=2.30mm cos[(16.98 rad/m^)x+(742 rad/s)t]. Being more practical, you measure the rope to have a length of 1.35 m and a mass of 0.00338 kg. You are then asked to determine the following: (d) wave speed; (e) direction the wave is traveling;

Video transcript

Hey everyone in this problem. We have a laboratory experiment where a group of students observes a formation of a progressive wave along string using a simple harmonic oscillator and the string has a length of 1.2 m a mass of five g. Were given the wave function Y is equal to eight millimeters sine of three pi radiance per centimeter x minus 33 pi radian per second. T. Okay. And the teacher asked the students to find the speed V. Okay. And direction of propagation of the progressive wave. Okay, art. So let's recall that we can write the general way function. Why is equal to a sign K. X minus omega T. Okay. All right now we can use that to compare it to the equation. We're given to find certain values. Alright, so let's start with part one. The speed and recall that the speed of the wave be going to be equal to the frequency F. There's a wavelength lambda. So we need to find F. We need to find lambda. Alright, well how can we find those things? So let's recall that the frequency F. Is related to Omega from our equation. Right, so we know that we can write omega equals two pi f. Now, in our general equation, omega is here this minus omega And the coefficient with the variable t. So in our equation that's going to correspond to this value here. 33 pi radiance per second. So omega is going to be 30 three pi. Okay, radiance for a second, this is gonna equal to two pi. Okay, when we have two pi we have unit radiant. Okay times the frequency F. Which gives a frequency F. When we divide the unit of radiant will cancel. Well F is equal to 16.51 over second. Okay? Or hurt if you want to write that um for this one we're calculating speed, we'll just leave it as one over second. Okay so we found omega or sorry we found F. Using omega. Now we need to find lambda so that we can use that in our equation for B. Okay so here's our f. Let's go ahead and find lambda. And similarly how we can relate omega two F. We can relate the value of K. Two r wavelength lambda that we're trying to find. Okay so recall that K. Is equal to two pi divided by lambda. Well in our equation K. Is going to be the coefficient with that extra. And in the equation we're given it's going to be three pi radian per centimeter. So we get K. is equal to three pi radian per centimeter. Which is equal to two pi unit radiant. Okay, divided by the wavelength lambda. So we get λ is equal to two pi over three pi centimeters. The unit of radium will cancel. We get divided by per centimeter. So centimeter and this is just going to be two thirds centimeter. Alright so we have F. We have lambda. So now we can go ahead and find that value V. That we were looking for the speed of the wave. Okay, so the speed of the wavy is equal to the frequency F times of wavelength lambda. Just going to be 16. per second times two thirds centimeter. Which gives us a speed v. of 11 and here we have cm per second. All right. There we go. We have the speed of our wave 11 cm/s. Okay Now let's go back up. Part two wanted us to find the direction of propagation. Alright, so part to want us to find the direction of propagation since our speed V is positive, our omega is positive. We are going to the right. All right. And I say omega is positive. We have this negative so minus omega. And in our equation we have minus omega. Okay, if we had a plus here that wave would be going to the left or the propagation of the wave would be going to the left. Okay in this case omega is positive. We have this wave propagating to the right, Okay, so we're gonna have answer. C Speed of the wave is 11 m/s to the right. Thanks everyone for watching. I hope this video helped see you in the next one