# Homework Help: Speed of a Transverse Wave

1. May 30, 2010

### jlsoccer4

1. The problem statement, all variables and given/known data
A wire that is 10 m long and has a uniform density(u) of 7.75 g/cm^3 is pulled to a tension of F=80 N. The wire, however, does not have a uniform thickness; rather, it varies uniformly from an initial radius of 1mm to a radius of 3mm where it is attached to a wall. If you send a wave pulse down the length of the string, how long does it take to reach the wall?

2. Relevant equations
v = sqrt(F/u)
u = m/L (m= mass, L= length)

3. The attempt at a solution
If I could find the volume of the wire, then I could determine the mass of the string, and the velocity of the pulse using the equations above. From there, once I had the velocity, since the units were m/s, I would divide by the distance to find the time it would take for the pulse to reach the wall. (Does this logic make sense?)

But how would I determine the volume?

2. May 30, 2010

### Fightfish

That would be by integration.

3. May 30, 2010

### jlsoccer4

What would the integration look like? Would the limits be from 1mm (.001m) to 3mm (.003m) and the integrand be pi*(r^2)*10m (volume of a cylinder)?

4. May 31, 2010

### Fightfish

No. One method of how to obtain the integrand would be the classic "divide-and-conquer". We will divide the wire into infinitely many thin slices. We then consider the volume of each thin slice.

The volume of each thin slice would be $$\pi r^{2} dx$$. However, we realise also that the radius of each slice varies, and can be expressed as $$0.01 + 0.002x$$, where x represents the distance of the slice from the starting point (taken to be the free end). So, we have the volume of each slice being $$\pi (0.01 + 0.002x)^{2} dx$$.

The total volume of the wire would thus simply be a summation of the volume of each slice across the entire length of the wire: $$\int^{10}_{0}\pi (0.01 + 0.002x)^{2} dx$$

5. May 31, 2010

### jlsoccer4

Wouldn't r be equal to .001+.0002x (in terms of millimeters) and (1*10^-6) + (2*10^-7)x (in meters)? Since my limits are in meters, wouldn't my integrand need to be in the same units?

6. Jun 1, 2010

### Fightfish

Oops, it should be r=0.001+0.0002x in terms of meters. Accidentally left off a factor of 10.
Why do you think that would be in milimeters?

7. Jun 1, 2010

### jlsoccer4

I think that you are correct. I had my units mixed up. Thank you for the help on this problem, I truly appreciate it.

8. Jun 2, 2010

### bballman1441

would this happen to be for a phys 132 class

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