# Homework Help: Wave pulse along a wire

1. Dec 5, 2008

### 996gt2

1. The problem statement, all variables and given/known data

The figure shows two masses hanging from a steel wire. The mass of the wire is 60.0 g. A wave pulse travels along the wire from point 1 to point 2 in 24.0 ms.

What is mass m?

2. Relevant equations
3. The attempt at a solution

L=8.0m

$$m_{string}=60g=0.06kg$$

$$\mu=m_{string}/L=0.0075 kg/m$$

$$v=4.0m/24ms=166.667m/s$$

$$v=\sqrt{T/\mu}=\sqrt{T/0.0075kg/m}$$

$$T=\sqrt{(2mgsin\theta)^2+(2mgcos\theta)^2)$$

Using $$\theta=40 degrees$$, I got m to be 10.6 kg. However, this answer is wrong. Can anyone tell me where I made the mistake? Thanks!

2. Dec 5, 2008

### LowlyPion

Aren't you only interested in the horizontal tension in the wire from 1 to 2?

3. Dec 5, 2008

### 996gt2

So you mean I should keep using 8.0m for L but use 2mg cos 40 for T?

I tried doing that-->setting 2mg cos 40 for T and then solving for T.

I got 13.9 kg, which is still not right...

Am I supposed to use 4m for L too?

Last edited: Dec 5, 2008
4. Dec 5, 2008

### LowlyPion

Why is it 2mgCos40 ?

But as to the μ, you simply calculate the density as you did. The density can be considered not to change between the various segments.

5. Dec 5, 2008

### 996gt2

Well, I thought that it was 2 mg cos 40 since each of the weights exerts a force of mg cos 40 on the string...if not, I'm confused as to how I should find T

6. Dec 5, 2008

### LowlyPion

Consider the free body diagram of 1 of the masses in isolation.

Isn't the force of the tension required to hold that mass in equilibrium equal to just mgCos40°?

7. Dec 5, 2008

### 996gt2

Ah I figured it out. T=mg/tan(40)...

8. Dec 5, 2008

### LowlyPion

As I suggested before I think you are only interested in the horizontal Tension between points 1 and 2.

9. Dec 5, 2008

### 996gt2

I figured out the correct answer using T=mg/tan(40) so I am pretty sure that is the correct value for the tension in that part :)

10. May 11, 2011

### chengenlee

996gt2 is right. The answer is 17.8.