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Consider a travelling wave produced by vibrating one end of a rope while the other end is made to freely move along a vertical line. Mathematically, the equation of the traveling wave that also represents the equation of motion of each point on the rope is given as follows.

$$y(x,t)=A\sin(kx-\omega t)$$

In the book I read, a certain particle on the rope moves up and down perpendicular to the propagation vector as wave does not transfer material but energy.

I don't agree with the statement that "a certain particle on the rope moves up and down perpendicular to the propagation vector". It is because I think to maintain the same point on the rope to move up and down, the arc length of $y(x,t)$ must be equal to that of $y(x,t+\Delta t)$ for $0\leq x \leq a$ and any $\Delta t$. However, according to the Notes below, their arc length are not generally equal.

enter image description here

From the figure above, how can $P$ and $Q$ be the same point of the rope while the red and blue curves have different length?

So my question are: Does each point of the rope move up and down?

Notes

From $x=0$ to $x=a$, generally the arc length of $y=\sin x$ is not equal to that of $y=\sin(x-x_0)$.

Let $S(x_0)$ be the arc length of $y=\sin(x-x_0)$ for the given interval.

\begin{align} S(x_0) &= \int_0^a\sqrt{1+\left(\frac{dy}{dx}\right)^2}\, \textrm{d}x\\ &= \int_0^a\sqrt{1+\cos^2(x-x_0)}\, \textrm{d}x\\ \end{align}

If $S(0)=S(x_0)=\text{const}$ for any $x_0$ then $\frac{\textrm{d}S(x_0)}{\textrm{d}x_0}=0$. Thus I have to check whether or not $\frac{\textrm{d}S(x_0)}{\textrm{d}x_0}=0$.

\begin{align} \frac{\textrm{d}S(x_0)}{\textrm{d}x_0} &= \frac{\textrm{d}}{\textrm{d}x_0}\int_0^a\sqrt{1+\cos^2(x-x_0)}\, \textrm{d}x\\ &= \int_0^a\frac{\textrm{d}\left(\sqrt{1+\cos^2(x-x_0)}\right)}{\textrm{d}x_0}\, \textrm{d}x\\ &= \int_0^a\frac{\cos(x-x_0)\sin(x-x_0)}{\sqrt{1+\cos^2(x-x_0)}}\, \textrm{d}x\\ &= \int_{-x_0}^{a-x_0}\frac{\cos y\sin y}{\sqrt{1+\cos^2y}}\, \textrm{d}y\\ &= \sqrt{1+\cos^2 x_0}-\sqrt{1+\cos^2(a-x_0)}\\ &\not = 0 \end{align}

It implies that generally they are not equal.

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You seem to be focusiing on the change in length as a problem, but really it is not.

The force which causes the rope to return is an elastic force. Just as with a spring it tries to return to it's original shape because it has been deformed.

Yes, it is generally discussed in term of tension in the rope, but what is tension if not the elastic forces between the molecules of the fibers. Keep in mind that all materials stretch a little under tension. It's not much in a natural fiber rope, but it is some---ask anyone who has ever tried to walk a taunt-line.

It is also worth being aware that

  • Diagrams in texts generally use a very exaggerated transverse scale.
  • The claim of purely transverse motion depends on small displacements
  • Demos where a real rope is driven by a wheel, do introduce some longitudinal motion
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  • $\begingroup$ So a certain point on the rope will oscillate in both x and y directions, doesn't it? $\endgroup$ – kiss my armpit Jan 24 '14 at 3:29
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I think it can be answered with an animated gif:

enter image description here

Clearly the nodes (red dots) are stationary, as are each point if you drop a perpendicular line (like a ruler) onto your screen.

Now obviously this isn't exactly what you have because there is a fixed end on the right. But if you put your hand over the right end, you'll see the same thing as a the free end. Or you can click this link and watch the wave go (I suggest going to the 5th or 7th harmonic).

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  • $\begingroup$ It is a stationary wave (rather than a single travelling wave) and how can the rope changes its length if the rope is not stretchable? $\endgroup$ – kiss my armpit Jan 24 '14 at 2:37
  • $\begingroup$ Can you make a wave with an unstretchable rope? $\endgroup$ – Kyle Kanos Jan 24 '14 at 3:04
  • $\begingroup$ Then ropes should not be used to illustrate the travelling wave. $\endgroup$ – kiss my armpit Jan 24 '14 at 3:16
  • $\begingroup$ That is one persons opinion. I think it's a fine example because you can clearly see the waves in it. It's a lot more difficult to show waves by using pistons to compress a gas... $\endgroup$ – Kyle Kanos Jan 24 '14 at 3:18

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