Transverse mechanical waves in a steel wire

[Brian_D]

Homework Statement

A steel wire can tolerate a maximum tension per unit cross-sectional area of $\frac{ 2.7 \mathit{GN}}{m^{2}}$ before it undergoes permanent distortion. What is the maximum possible speed for transverse waves in a steel wire if it is to remain undistorted? Steel has a density of $\frac{ 7.9g}{\mathit{cm}^{3}}$.

Relevant Equations

$V = sqrt(F/mu); where*V*is*wave*velocity, F*is*tension*force*on*wire, mu*is*mass*per*unit*length*of*wire;$

I tried various manipulations of the variables to get the formula to work. For example, given that density is mass per cubic units, I tried representing this as mass/(area x length) so that I could relate this to mu as mass per unit length. Similarly, I reasoned that force per unit cross sectional area multiplied by length should give the tension force on the mass of the whole wire. But no matter what I tried, I could not come up with an expression for velocity that produced a numerical answer. Your thoughts?

P.S. I tried to preview the homework statement and relevant equations, but didn't see how to do it.

 

[renormalize]

P.S. I tried to preview the homework statement and relevant equations, but didn't see how to do it.

 

[Brian_D]

Thank you, but the preview button only seems to work for text in the text box, not for the homework statement and relevant equations boxes. I didn't see a preview option for the latter two boxes.

 

[Brian_D]

P.S. The answer that the textbook gives for this problem is 585 m/s

 

[renormalize]

Thank you, but the preview button only seems to work for text in the text box, not for the homework statement and relevant equations boxes. I didn't see a preview option for the latter two boxes.

Sorry, my mistake. You're right, I see no way to preview LaTeX in the title, problem statement or relevant equations. But you can edit your post:
"$V=\sqrt{F/\mu}$; where $V$ is wave velocity, $F$ is tension force on wire, $\mu$ is mass per unit length of wire"

 

[Brian_D]

Yes, and the solution is straightforward (it just occurred to me). We can preview the problem statement and the relevant equations simply by putting them into the text box, then copy and paste them into the other boxes.

 

[fanofakirakurosawa]

$$ V = \sqrt{\frac{F}{\mu}} $$
$$ F = \sigma A $$
$$ \delta = \frac{m}{A L} $$

edit: deleted some part of the reply due I provided the entire solution

 

[Brian_D]

Thank you, fanofakirakurosawa. However, I am not familiar with the second and third formulas. For the second, I assume F is the tension force on the wire and A is the amplitude. What is sigma? For the third formula, I assume that m is the mass of the wire, A is the amplitude, and L is the length of the wire. What is delta?

 

[fanofakirakurosawa]

Thank you, fanofakirakurosawa. However, I am not familiar with the second and third formulas. For the second, I assume F is the tension force on the wire and A is the amplitude. What is sigma? For the third formula, I assume that m is the mass of the wire, A is the amplitude, and L is the length of the wire. What is delta?

I'm sorry for not giving context, my fault.
Sigma is stress tension, F is the tension force and A is the area of the cross-section.
Then the third equation is the density. It's mass over volumen and volumen is A (area) times L (large of the wire). Now you need to determine what is μ. And then you'll be able to calculate the wave speed.

 

[Brian_D]

Ah, that explains everything! Now I was able to manipulate all three formulas and discovered that $\frac{F}{\mu}$ is really the same as $\frac{\sigma}{\delta}$. Plugging the given values into this fraction and taking the square root, I get 585 m/s, which is the same answer as given by the textbook. Thank you, fanofakirakurosawa.

 

[fanofakirakurosawa]

Ah, that explains everything! Now I was able to manipulate all three formulas and discovered that $\frac{F}{\mu}$ is really the same as $\frac{\sigma}{\delta}$. Plugging the given values into this fraction and taking the square root, I get 585 m/s, which is the same answer as given by the textbook. Thank you, fanofakirakurosawa.

Your welcome Brian!