Velocity of Propagation of a wave

[jumbogala]

Homework Statement

A wave is sent down a cable. When it hits the end of the cable a reflected wave is sent back. You are given the:
- time between the initial and reflected pulse
- the amplitude of the initial and reflected pulse.

You also know the length of the cable (you may not need this measurement).

Find the 1) the velocity of propagation, and 2) the attenuation of the wave medium.

Homework Equations

None are given, but:
velocity of propagation is the velocity at which a wave propagates along a rope.

Attenuation is when is a wave traveling on a rope loses a constant fraction of its amplitude per meter of travel along the rope.

The Attempt at a Solution

I really have no idea what I'm doing here, but I think the time between the initial and reflected pulse would be double the time it takes the wave to travel along the rope...

The distance the wave travels would be the length of the rope, correct? So would the velocity of propagation be (length of rope)/(0.5*time between reflected and initial)?

For the attenuation, is it the difference between the initial wave's velocity and the reflected wave's velocity, divided by the length of the rope?

Those are just guesses... please help!

 

[inhere]

The velocity of propogation is simply the distance traveled over the time of propogation. or v = t(sent to reception)/2L. The attenuation is generally defined as final amplitude over initial amplitude so A/Ao.

 

[lewando]

A/Ao would be the attenuation of the system. The media (cable) attenuation should have units of attenuation per unit length.

 

[arildno]

The distance the wave travels would be the length of the rope, correct? So would the velocity of propagation be (length of rope)/(0.5*time between reflected and initial)?

Your formula is certainly correct.

However, the distance ACTUALLY traveled is 2L, L being the length of the rope, so that rewriting your expression to 2L/time might be a more transparent way of putting it.

Attenuation is when is a wave traveling on a rope loses a constant fraction of its amplitude per meter of travel along the rope

For the attenuation, is it the difference between the initial wave's velocity and the reflected wave's velocity

Difference is not the same as "fraction"!

The fractional loss is (A0-A)/A0, where A0 is intial amplitude, A final.

Since you are to find out how much is the fractional loss is per unit distance, simply divide this with 2L, gaining:
(A0-A)/(2LA0)