# Wave speed for non-uniform density?

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1. Nov 20, 2016

### Blue_Angel

1. The problem statement, all variables and given/known data
A long rope with mass m = 10 kg is suspended from the ceiling and hangs vertically. A wave pulse is produced at the lower end of the rope and the pulse travels up the rope.
(a) Explain why the speed of the wave pulse change as it moves up the rope; does it increase or decrease?
(b) What will happen to the wave pulse if a section near the middle of the rope has been soaked in water shortly before the rope was hung from the ceiling?

2. Relevant equations
v=sqrt(T/(m/L))
3. The attempt at a solution
For part a v increases as height increases because there's greater tension....
My problem is with part C, I think soaking the string in water would increase the mass of the string near the centre section, however this means it would also increase the tension proportionally. Because wave speed is dependant on the ratio of the tension to mass the wave speed before it was soaked in water would be exactly the same as the wave speed after.

To put this into equation form v=sqrt(T/(m/L)) where T=mg so v=sqrt(Lg), The velocity has no dependence on mass.
can someone verify whether this is true or false??

2. Nov 20, 2016

### Cutter Ketch

T is not just mg. If it were you wouldn't have answered part A the way you did. I bet you can write a better expression for T as a function of position along the length.

3. Nov 20, 2016

### Cutter Ketch

Also, notice that it isn't really m in your speed equation. It is m/L, a linear density. It asumes the density is constant over the length. That isn't the case anymore when the middle is wet.

4. Nov 20, 2016

### Blue_Angel

Ok but if T were tension at a point on the string and Y was how high up the point was, can I say
T=mg*(Y/L)
therefore v=sqrt((mg(Y/L))/(m/L))=sqrt(gY)

5. Nov 20, 2016

### Cutter Ketch

Right for the case without water, and notice you now have your expression with explicit dependence on Y for answering part A.

Now can you write an expression for T vs Y including the water? You will also have to replace m/L with an expression for density vs Y. You'll get 3 results for the three regions: below the water, in the water, and above the water.

6. Nov 20, 2016

### Blue_Angel

Under the water would be exactly the same as above. Through the water density has increased a lot but tension is scarcely changed so velocity will decrease? and above the water density would be low but tension would be very high because of the added mass so it speeds up?

7. Nov 20, 2016

### Cutter Ketch

Uhmm... I wrote out the solutions on a white board and then erased them. Let me think. Yes, that sounds right.