# Wave travel along a tensioned rope

• macgirl06
In summary, Oliver, weighing 90.0 kg, uses a 4.60 m long rope to pull Jordan, weighing 53.6 kg, across the floor with a friction coefficient of 0.200. Jordan signals to Oliver to stop by "plucking" the rope, sending a wave pulse forward along the rope. The pulse reaches Oliver 149.0 ms later. According to the given formula, the mass of the rope can be calculated by using the tension in the rope (which equals the friction force) and the speed of the pulse. The equation used is m_{rope} = \frac{\mu m_{person} g L}{v^2}. After some confusion and clarification, it is determined that

#### macgirl06

Oliver (m=90.0 kg) uses a 4.60 m long rope to pull Jordan (m=53.6 kg) across the floor (μk=0.200) at a constant speed of 1.14 m/s. Jordan signals to Oliver to stop by "plucking" the rope, sending a wave pulse forward along the rope. The pulse reaches Oliver 149.0 ms later. What is the mass of the rope?

I am doing m= ukmg*L\v*squared but it is not working, am I suppose to find a new velocity or something. Any help appreciated.

Why do you say it is not working? Where did you get that equation and what do the variables mean?

Work out the tension in the rope by calculating the force it takes to overcome the friction. Then use the relation between wave speed, mass, tension and length to get the mass.

I did exactly what you are telling me and I am not getting the answer. I am doing M=ukmg*L\v*squared and ukmg is my Force tension that I calculated. These are my values that I am putting in
uk= 0.2000
m= 53.6 kg
g= 9.8
v= 1.14
L=4.60

It seems like a very easy question but I don't know where I am messing up.

I'm confused by your symbols. What is M ? You must explain what your symbols mean, otherwise it's like a riddle, we have to guess. For instance I'm guessing the uk is not a country but a coefficient of friction.

its mass...

Well, that does not make sense because you've got m on both sides of your equation. I don't recognise the formula

M=uk*m*g*L\v^2

so I'm stuck also. You'll have to hope someone else does since you haven't explained it.

ok i got that from the formula: vsquared = T\u and u= m\L (mass over length) and these are equations in my physics textbook used for waves.

macgirl06 said:
I am doing m= ukmg*L\v*squared but it is not working, am I suppose to find a new velocity or something.
There's nothing wrong with that equation, which is derived from the formula for the speed of a wave (or pulse) on a stretched rope:
$$v^2 = \frac{T}{\rho}$$

Where T is tension in the rope (which equals the friction force $\mu m_{person} g$) and $\rho = m_{rope}/L$.

I'll rewrite that equation like this:

$$m_{rope} = \frac{\mu m_{person} g L}{v^2}$$

macgirl06 said:
I did exactly what you are telling me and I am not getting the answer. I am doing M=ukmg*L\v*squared and ukmg is my Force tension that I calculated. These are my values that I am putting in
uk= 0.2000
m= 53.6 kg
g= 9.8
v= 1.14
L=4.60

It seems like a very easy question but I don't know where I am messing up.
v should be the speed of the pulse, not the speed of the person. Calculate the speed of the pulse from the data given in the problem.

there is no other way to find a new velocity it just takes me back to square one...I keep on getting 0.372 grams for my mass of rope...which other way can I find velocity

Question: Would the speed of pulse to an observer watching Oliver and Jordan be the sum of the speed of the rope and the speed of the pulse if the rope where not moving?

macgirl06 said:
there is no other way to find a new velocity it just takes me back to square one...I keep on getting 0.372 grams for my mass of rope...which other way can I find velocity
To find the speed of the pulse just use the distance it travels (the length of the rope) and the time it takes (which is given).

e(ho0n3 said:
Question: Would the speed of pulse to an observer watching Oliver and Jordan be the sum of the speed of the rope and the speed of the pulse if the rope where not moving?
Yes.

k got it, just a calculational error on my part. Thanks again for the clarification...:)