Calculating Mass on a Wave Pulse Along a Wire

[996gt2]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

L=8.0m

m_{string}=60g=0.06kg

<br /> \mu=m_{string}/L=0.0075 kg/m<br />

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!

 

[LowlyPion]

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

 

[996gt2]

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

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?

 

[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.

 

[996gt2]

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.

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

 

[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°?

 

[996gt2]

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

 

[LowlyPion]

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

 

[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 :)

 

[chengenlee]

996gt2 is right. The answer is 17.8.