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In my text book equation of travelling wave is given as $y=A\sin(\omega t-kx)$ and they deduced standing wave equation using the same wave equation and arrived at $Y=2A\cos(kx)\sin(\omega t)$ but suddenly they changed to $y=A\sin(kx-\omega t)$ to derive standing wave equations on a string and arrived at $y=2A\sin(kx)\cos(\omega t)$

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closed as unclear what you're asking by Aaron Stevens, Buzz, Jon Custer, ZeroTheHero, user191954 Feb 6 at 4:22

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ We have MathJax running on the site, so by writing y(x,t) = A \sin ( \omega t - k x ) between dollar signs (or double dollar signs) you get a neatly typeset inline (block-set) equation (and with the right symbol for angular frequency, too!): $y(k,x) = A \sin ( \omega t - k x )$. $\endgroup$ – dmckee Feb 4 at 3:21
  • $\begingroup$ Sines and cosines are just off by a phase shift $\endgroup$ – Aaron Stevens Feb 4 at 3:24
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While $\sin ( k x - \omega t )$ and $\sin (\omega t - k x )$ are mathematically distinct the difference between them can be absorbed by the physicist in their choice of coordinate systems (where you put the origin, and what direction you point the $x$ and $y$ axes).

And the difference in the results formula for the standing wave amplitude $Y$ can likewise be absorbed in a choice of coordinate system (though now you may also have to decide when to start the clock.

There is no difference in the physics described by your result and that you found in the book.


As a side note, one thing that students sometimes forget when working this problem is that waves are inverted when reflected from a fixed point (i.e. a very stiff interface), so the the starting form should be something like \begin{align} Y(x,t) &= &y_+(x,t) &\color{red}{-} y_-(x,t) \\ &= &A \sin( k x - \omega t) &- A \sin( k x + \omega t) \;. \end{align} The handling of that minus sign between the traveling wave terms may be where you and the book are diverging.

It is not entirely clear to me from your answer, but if you are complaining that your book/instructor formed $Y$ as $\sin(k x - \omega t) + \sin(\omega t - k x)$, then the issue is exactly one of taking care of the inversion of the reflected wave (though I find the $\sin (\omega t - k x)$ form more opaque than $-\sin ( k x + \omega t)$ ).

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  • $\begingroup$ positions of nodes and antinoeds are changed due to different final results $\endgroup$ – teja Feb 4 at 3:43
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    $\begingroup$ Again, you can fix that by deciding where you put your origin and when you start your clock. Keep in mind that I'm not arguing that there isn't a mathematical difference in the results you found, I'm arguing that the results represent the same physics in different coordinate systems. Without seeing exactly the treatment you are looking at I can only guess at exactly what the author had in mind, but this particular subject is very forgiving of differences of treatment, notation and assumptions, so people are sometimes sloppy with it. $\endgroup$ – dmckee Feb 4 at 3:46
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If you wish to use the sine function for a wave travelling in the positive x-direction then you would write $y=A\sin(\omega t-kx+\phi)$ and then your choice of the phase angle $\phi$ will enable you to decide as to where the origin of position, $x=0$, is and at which part of the oscillation of the wave you started the clock, $t=0$.

If $\phi = 2n\pi$ where $n$ is an integer then $y=A\sin(\omega t-kx)$ could be the form of the wave equation with the displacement $y=A\sin(\omega t)$ at position $x=0$.

If $\phi = (2n+1) \pi$ where $n$ is an integer then $y=A\sin(kx-\omega t)$ could be the form of the wave equation with the displacement $y=-A\sin(\omega t)$ at position $x=0$.

However you have even more control as to the properties of the wave $y=A\sin(\omega t-kx+\phi)$.

As an example let $\phi = \pi$

You could then write the wave equation as $$y= A\sin\left ((kx+\frac \pi 2)-(\omega t-\frac \pi 2)\right)= A\sin\left (k(x+\frac {\pi}{ 2k})-\omega (t-\frac {\pi} {2\omega})\right) =A\sin (kx'-\omega t')$$
where $x' = x + \frac{\pi}{2k}$ and $t'=t-\frac{\pi}{2\omega}$ ie you have shifted the origin of position and at which part of the oscillation of the wave you started the clock - a shift of position and time coordinates.

Then to form the standing wave you need to add a wave travelling in the positive x-direction $y=A\sin(\omega t-kx+\phi)$ to a wave travelling in the negative x-direction $y=A\sin(\omega t+kx+\psi)$ and you can decide on $\phi$ and $\psi$ to satisfy where you would like the nodes and anti-nodes to be.

In the main one does not perform such a general analysis and sticks to $y=A\sin(\omega t-kx)$ or $y=A\sin(kx-\omega t)$ or $y=A\cos(\omega t-kx)$ or $y=A\cos(kx-\omega t)$ for the wave travelling in the positive x-direction with a similar four choices for the wave travelling in the negative x-direction.

Addition of two such waves will produce a wave where the amplitude of the displacement of a particle depends on the equilibrium position of the particle from the origin $x$ with the particles all undergoing simple harmonic motion in phase between two nodes and in antiphase between adjacent inter-nodal positions.

Note that your two standing wave equations

$$y=2A\cos(kx)\sin(\omega t)=2A\sin \left[k\left(x+\frac{\pi}{2k}\right)\right]\cos\left[\omega\left(t-\frac{\pi}{2\omega}\right)\right]=2A\sin kx'\cos \omega t'$$

are really referring the same standing wave looked at with different position and time coordinates.

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