How is Wave Propagation Time Calculated for a Rope in a Mine Shaft?

[sapiental]

Hello,

I have a couple of questions about my assignment. Here are the assigned questions plus my attempts.

1) One end of a rope is tied to a stationary support at the top of a
vertical mine shaft 80.0 m deep. The rope is stretched taut by a box of mineral samples with a mass of 20.0 kg attached at the lower end. The mass of the rope is 4.00 kg. The geologist at the bottom of the mine shaft signals a colleague at the top by jerking the bottom of the rope
sideways. How long does it take for the wave pulse to arrive at the top of the 80.0 m long rope?

For transverve waves v = sqrt(F/u) u = mass per unit L

F = mass_rope + mass_box (g)
F = 24.0kg (9.8m/s^2)

v = sqrt(325N/.05kg/m)

v = 80.6m/s

so it would take about 1 second for the wave pulse to arive

2) Motion of a Wave Pulse :
If the end of a string is given a single shake, a wave pulse
propagates down the string. A particular wave pulse is described by the function:

y(x,t) = (A^3/(A^2 + (x - vt)^2))where A = 1.00 cm, and v = 20.0 m/s.
a) Sketch the pulse as a function of x at t = 0. How far along the string does the pulse
extend?

I used MATLAB for this but don't know if I did it correctly. Since t = 0 the equation becomes:

y(x, t = 0) = (A^3/(A^2 + (x - v(0))^2))

i used the command plot(x,y) I let x = 0 through 1

plot(x, (.01m^3/(.01m^2 + (x)^2)

If my plot is correct it looks like the pulse extends roughly .16m.
Should my sketch look like a wave? Because mine just looks like one curve..

b) Sketch the pulse as a function of x at t = 0.001 s.

I attached a picture of my graph but I basically used the function plot(x, (.01m^3/(.01m^2 + (x - 20m/s * .001s)^2)

c) At the point x = 4.50 cm, at what time t is the displacement maximum?

Well I think I messed up on the graphs at this point because one axis is time and my y axis is x.. If anyone could tell me how I should go about graphing this function It would be much appreciated.

d) At which two times is the displacement at x = 4.50 cm equal to half its maximum value?
e) Show that y(x, t) satisfies the wave equation.

I roughly undertand the wave equation and I can recognize that this equation posesses similar properties. But I don't know what my first step should be. Maybe partially differentiate either the wave function or this particular function?

Thanks any help is much appreciated.

 

[Chi Meson]

a) since the mass of the rope is significant, where do you think the tension of the rope is greater, at the top or bottom, and hence, will the speed of the wave be faster or slower at the bottom?

It does not appear that you are doing calculus-based physics, so you could assume an average speed for the wave.

And anyway you tranposed the number for the weight of rocks + rope. Weight is 235 N not 325 N

 

[sapiental]

thanks. that was a typo on my part with. Plugging in 235N the average v = 68.6m/s so it takes about 1.16s to travel a distance of 80m. What is the "calculus based" method for calculating the speed of the transverse wave. I believe it still depends on the same algebraic formula of v = sqrt(F/u)...

 

[Chi Meson]

It does, but since the tension (F) of the rope will increase, as it holds up its own weight, v becomes a function of the position along the length of the rope. Time would be the integral of dx/v as the pulse traveled up the rope. My calculus is rusty, but I'm pretty sure that this integral is not exactly the same as using the average of the slowest and fastest speeds.