Ratio of µ in a rope with differing thickness/Waves

[gmmstr827]

Homework Statement

For the cord in the following diagram, what is the ratio of µ for the different sections to create a Γ/2 on the right?

*** The picture depicts a rope with an upward oriented reflected pulse on the left (thinner rope) and a downward oriented transmitted pulse on the right (thicker rope). ***
http://i29.photobucket.com/albums/c261/gmmstr827/rope.jpg

Homework Equations

Useful formulas might include:
Γ = v/f
Γ_n = 2L/n where n=1 represents 1st harmony; n=2 represents 2nd harmony, etc.
v = velocity
f = frequency = 1/T
T = period = 2π√(L/g)
g = 9.8 m/s^2
L = length
µ = m/L
m = mass

The Attempt at a Solution

Perhaps if you solve Γ_n = 2L/n where n=1 for L = Γ/2 then plug that into L in µ = m/L to get µ = (mΓ)/(2L) that could be your ratio?

It seems like the answer would be that the rope on the right is twice the µ than the rope on the left, however I'm unsure how to show this. There don't seem to be many connections between the formulas for µ and Γ for this to easily be solved. If someone could guide me to the correct formulas, it would make this much easier. Thank you!

 

[anathema]

Thank you for your question. I would like to help you find the correct solution to this problem.

To start, let's define some variables:
- Γ: the amplitude of the reflected pulse (upward oriented) on the left side of the rope
- Γ/2: the amplitude of the transmitted pulse (downward oriented) on the right side of the rope
- L: the length of the rope
- µ: the linear density of the rope
- m: the mass of the rope
- n: the harmonic number

From the given formulas, we can see that Γ = v/f. This means that the amplitude of the reflected pulse is equal to the velocity divided by the frequency. We also know that the frequency is equal to 1/T, where T is the period. The period can be calculated using T = 2π√(L/g), where g is the acceleration due to gravity. Therefore, we can rewrite the formula for Γ as Γ = v/(1/T) = vT.

Next, let's look at the formula for Γ_n = 2L/n. This formula tells us that the amplitude of the nth harmonic is equal to 2L divided by the harmonic number. In this case, we are interested in the first harmonic (n=1), which corresponds to the reflected pulse on the left side of the rope. So, we can rewrite this formula as Γ = 2L/1 = 2L.

Now, let's combine these two formulas for Γ:
Γ = vT = 2L

We can also use the formula for µ = m/L to calculate the linear density of the rope. Since we are looking for the ratio of µ for the different sections, we can set up the following equation:
µ_left/µ_right = (m_left/L_left)/(m_right/L_right)

We know that the mass of the rope is equal to its linear density multiplied by its length, so we can rewrite this equation as:
µ_left/µ_right = (m_left/µ_left)/(m_right/µ_right)

We also know that the mass is directly proportional to the amplitude of the pulse, so we can rewrite this equation as:
µ_left/µ_right = (Γ_left/µ_left)/(Γ_right/µ_right)

Now, we can substitute the formulas for Γ_left and Γ_right into this equation:
µ_left/µ