What must be the stress (F/A) in a stretched wire of a material w...

Question

What must be the stress (F/A) in a stretched wire of a material whose Young’s modulus is Y for the speed of longitudinal waves to equal 30 times the speed of transverse waves?

Accepted Answer

Hey everyone in this problem. We're adjusting the tension in a stretched copper wire. Can we do so so that the velocity ratio of longitudinal waves to transverse waves is equal to phi Okay, we're told the cross sectional area of the wire is one millimeter squared. The young modulates of copper is one times 10 to the 11 pascal's. Okay. And we are asked to find the tensile stress sigma in the wire. Alright, So let's start with the tensile stress. Okay, we are looking for sigma and we know that the tensile stress can be given by the force of tension. Okay. Divided by the cross sectional area. Alright, well we don't really know F. We don't we do know a but we don't know F. So let's see if there's something else we can do um with the information we're given. Okay, so we're given the velocity ratio of longitudinal transverse waves. So let's recall the velocity of a longitudinal wave. First. Recall that this is going to be the square root of why? Over rope? They were wise, the young module asse rho is the density. And we also know the velocity of a transverse wave is given by the square root of F. Force through detention divided by mu where mu is a linear density. Alright, so let's try to work with what we have here. So we're looking for force per area. We have a force and we have this new value. Well recall that mu the linear density can just be written as rho the density times a the cross sectional area. Okay, if we write it like this now we have this f divided by a term. Okay we have this F over a which is just equal to sigma. Alright so substituting that and we're gonna have VT transverse velocity is going to be sigma over row. Alright so now we have this sigma that we're looking for. Okay, sigma is what we're looking for. We have this in our equation. Now let's use some more information that we've been given. So we've been told the ratio of longitudinal waves to transverse waves is five. Okay, so that means the velocity of the longitudinal wave divided by the velocity of the transverse wave is equal to five. Okay in other words the velocity of the longitude noise is equal to five times the velocity of a transverse wave. Alright so let's use this and substitute in the equation that we have. Guys we have V. L equals five V. T. On the left hand side we get the square root of why divided by row. On the right hand side? We have five times the square root of sigma over rho the one over the square root of row that's in both terms. Okay, so that's going to cancel or we can divide out we're gonna be left with the square root of y is equal to five times the square root of sigma Squaring both sides. We get Y is equal to 25 sigma. Okay and we're so close again, we're solving for sigma This is what we want to find. Mhm. And we know the value of why K. We were given it in the question so we can write um sigma is going to be y divided by 25. And why we're told is one times 10 to the Pascal's divided by 25. Okay, And this gives us a sigma value of four times 10 to the nine K pascal's or newton poor per meter squared. And so this is a tensile stress that we were looking for. Okay, it's four times 10 to the nine newtons per meter squared. That's going to be answer C. Alright. Thanks everyone for watching. See you in the next video.

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