Mathematical representation of a pulse on a rope

[zenterix]

Homework Statement

Consider a very long rope at rest on a frictionless surface and placed parallel to the $x$-axis. If its left end is quickly shaken up and down once, a perturbation of the shape of a pulse propagates along the rope from the left to the right end. If there is no energy losses, the pulse will propagate along the rope without changing its original shape.

In the figure below, the displacement in centimeters of an ideal rope with respect to its equilibrium position defined by $y=0$ is shown. The travelling pulse is shown at different times. The speed of the pulse is $2\text{m/s}$.

Relevant Equations

If the shape of the pulse at time $t=0$ can be modelled by the function $y(x)=Ae^{-Bx^2}$, where $A$ and $B$ are positive constants, what is $y(x,t)$, the mathematical representation of the shape of the pulse at any time $t>0$?

My initial thought was to model the wave as

$$y(x,t)=Ae^{-B(x-t)^2}$$

This question is part of an automated grading system and the above entry is considered incorrect.

I think I need to incorporate the information that the speed of the wave is $v$ somehow.

 

[Orodruin]

First off, the expression $x-t$ is dimensionally inconsistent. Think about how this can be corrected.

Then yes, the speed must enter into it. How fast does the wave you wrote down move. Note that the argument = 0 is the peak of $y(x)$. How fast does this peak move?

 

[zenterix]

The answer seems to be just

$$y(x,t)=Ae^{-B(x-vt)^2}$$

$x-vt$ has dimensions of length.

For any fixed $t$, the maximum of the pulse occurs at $x=vt$.

Now, how fast does this peak move?

At $t=0$ the peak is at $x=0$.

At $t=1$ the peak is at $x=v$.

I am not satisfied with this demonstration of "how fast", however. Seems there should be a more rigorous way.

$x=vt$ itself gives us the correct derivative of $v$, and I guess that is the more rigorous way.

 

[kuruman]

At $t=1$ the peak is at $x=v$.

This is meaningless. At $t=1$ doesn't say much. t = 1 what? Seconds, weeks, centuries? You could do it more rigorously if you were to note that $x$ and $vt$ have dimensions of length and the argument of the exponential must be dimensionless. What does this say about the dimensions of $B$?

 

[zenterix]

This is meaningless. At $t=1$ doesn't say much. t = 1 what? Seconds, weeks, centuries? You could do it more rigorously if you were to note that $x$ and $vt$ have dimensions of length and the argument of the exponential must be dimensionless. What does this say about the dimensions of $B$?

At $t=1$ units of time, the peak is at $x=v\cdot 1$ units of length.

Since $v=2\text{m/s}$ then it is $t=1$ second and $x=v\cdot 1$ meters.

Then $(x-vt)^2$ has dimensions of length squared, ie $m^2$ and so $B$ has dimensions of $m^{-2}$.

$x=vt$ represents, for each time $t$, the $x$-coordinate of the peak of the pulse.

The velocity of this peak is then $\frac{dx}{dt}=v$ which has units $m/s$.

 

[hutchphd]

Yikes you are told in the problem statement: "The travelling pulse is shown at different times. The speed of the pulse is 2m/s. "
Relax a little. This is not an inquisition nor is it very complicated (yet).

 

[haruspex]

At $t=1$ units of time, the peak is at $x=v\cdot 1$ units of length.

No, the second picture is at t=0.5s.

The mix of variables and constants in the answer makes it a bit tricky.
Given the 2m/s speed, the correct way to write the answer is
$$y(x,t)=Ae^{-B(x-2tm/s)^2}$$

 

[zenterix]

No, the second picture is at t=0.5s.

The mix of variables and constants in the answer makes it a bit tricky.
Given the 2m/s speed, the correct way to write the answer is
$$y(x,t)=Ae^{-B(x-2tm/s)^2}$$

I don't know why you are writing "No".

The pictures are just snapshots in time, which is a continuous variable. I made a statement about a particular time, $t=1$.

I also said the answer seems to be

$$y(x,t)=Ae^{-B(x-vt)^2}$$

I was the one who typed up the problem statement in the OP, so I know that $v$ is a known constant.

 

[Orodruin]

No, the second picture is at t=0.5s.

Nobody said anything about that plot though.

The mix of variables and constants in the answer makes it a bit tricky.
Given the 2m/s speed, the correct way to write the answer is
$$y(x,t)=Ae^{-B(x-2tm/s)^2}$$

Mixing variables and units in an expression is indeed tricky. In particular when typeset the same way. When necessary the clearest way is
$$y(x,t)=Ae^{-B(x-(2\,{\rm m/s})t)^2}$$
but there is notjing wrong with writing ”
$$y(x,t)=Ae^{-B(x-vt)^2}$$
where $v=2$ m/s”

 

[haruspex]

I don't know why you are writing "No".

Because I thought you were trying to deduce the speed from the two snapshots, which I took to have been provided with the question. But maybe you generated those.

I also said the answer seems to be

$$y(x,t)=Ae^{-B(x-vt)^2}$$

I was the one who typed up the problem statement in the OP, so I know that $v$ is a known constant.

If the speed is a given constant then I would expect that to feature in the answer, not the symbol "v".
But what does the automated grading system expect? Does it expect "(x-2t)”? My point was that that would be technically incorrect; the correct form would include "m/s".