(b) A metal bar with a length of 1.50 m has density 6400 kg/m3 . ...

Question

(b) A metal bar with a length of 1.50 m has density 6400 kg/m3 . Longitudinal sound waves take 3.90 * 10-4 s to travel from one end of the bar to the other. What is Young's modulus for this metal?

Accepted Answer

Hey everyone, welcome back in this problem. We have a longitudinal sound wave. Okay, It's generated by an ultrasonic emitter and it's going to travel along the 3.5 m copper rod in one m in one millisecond. Sorry? Okay, we're told the volume of the rod and its mass. And we're asked to find the young man Angeles for copper. Okay. Okay, so let's think we want to find the young's module asse And we're told that we're in a copper rod. Okay. And a copper rod, it's gonna be a thin solid. So, when we're thinking about an equation with the youngs modulates in a thin solid, we have the following recall, we can write the velocity is equal to or the speed, sorry, is equal to the square root of why? Over row. Okay, why is our young's module asse? So, that's what we're looking for. V. Is the speed and rho is the density. All right, so we're looking for y so we need to find V and we need to find row. Now, v speed while we're told the distance. Okay, 3.5 m and we're told the time one millisecond. Well, with the distance and time we can find speed. Okay, so let's find v. We know it's going to be the change in distance over the change in time And in this case that is 3.5 m. Okay? And then we have one millisecond. Well, putting this into our standard unit, this is going to be 0.001 seconds And this is going to give us a speed of m/s. Alright, so now we have 3500 m/s. V. We can use that in our equation. All we need now is row. Okay, so we need to find the density rho Well, we do know the volume of the rod and we know the mass of the rod. And let's recall density is just mass per unit volume. Okay, so in order to find the density, we need to divide the mass by the volume. This is gonna give us one kg divided by 1.116 times 10 to the minus four m cubed. And this is going to give us a density rho Of 8960.5735. Okay. And the unit here is kilogram per meter cubed. Okay, great. So we have the we have row. Okay, so we can sub those in here. V. Andro. And what we will get is we will get meters per second is equal to the square root of Why the young's module is that we're looking for divided by 8960. kg per meter. Cute mm. If we square both sides, we're gonna have 12 million. 250,000. Okay, big number. We could put this into scientific notation. Um if you wanted to make it a little cleaner here, Y divided by 8960.5735. Okay. And finally, multiplying, we're gonna get Y is equal to 1. times 10 to the 11 in the unit here is pascal, and so this is our youngest modulates that we were looking for that's going to correspond with answer D. We have Y is equal to 1.9 times 10 to the 11 pascal's. Thanks everyone for watching. See you in the next video.

(b) A metal bar with a length of 1.50 m has density 6400 kg/m3 . Longitudinal sound waves take 3.90 * 10-4 s to travel from one end of the bar to the other. What is Young's modulus for this metal?

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