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In my textbook, it says that a stationary wave is generated upon interference with the incident and the reflected wave. Well, if there is a reflected wave, I suppose that there would be a phase change of 180 degrees. When that happens won't we express the reflected wave as the following?

$$y = A\sin (kx + \omega t + \pi)$$

Assuming that the reflected wave is coming from the opposite end and having a negative amplitude.

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  • $\begingroup$ Of course you can. What is the problem? $\endgroup$
    – Abhijeet Melkani
    Commented Feb 20, 2017 at 14:32
  • $\begingroup$ @A.Melkani No, you can't. $\endgroup$
    – Yashas
    Commented Feb 20, 2017 at 14:32
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    $\begingroup$ Just add $y_1 = A\sin (kx + \omega t)$ to $y_2 = A\sin (kx - \omega t)$. You can add a phase angle to one of the functions to make a particular value of $x$ a node or an anti node. $\endgroup$
    – Farcher
    Commented Feb 20, 2017 at 14:39
  • $\begingroup$ Keep in mind that $\sin (x + \pi) = -\sin(x)$. See: those goofy trig identities they made you (try to) learn are good for something. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Feb 20, 2017 at 16:39

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A standing wave is formed when you have a continuous source of disturbance on one end. Consider a string whose one end is fixed and the other end is attached to a source of disturbance.

Let the equation of the wave created by the source of disturbance be:

$$y = A\sin (kx + \omega t)$$

This wave travels through the string and reflects from the fixed end of the string. The reflected wave is traveling in the opposite direction and is also 180 degrees out of phase (reflections from fixed ends of a string are 180 degrees out of phase). Therefore, the equation of the wave is given by:

$$y = A\sin (kx - \omega t + \pi)$$

The reflected wave is NOT the equation of the standing wave. As you might already know, two waves can interfere to form a new wave. The new wave is given by the sum of the two waves.

The equation of the standing wave is given by (for a general case where the phase difference is $\phi$):

$$y = A\sin (kx - \omega t - \phi) + A\sin (kx + \omega t)$$ $$ = 2A \cos(kx + \frac{\phi}{2})\sin(\omega t - \frac{\phi}{2})$$

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  • $\begingroup$ Won't the terms in your final expression cancel out? $\endgroup$
    – Karthik
    Commented Feb 20, 2017 at 14:44
  • $\begingroup$ I find it, but is not it 2Asin(kx)cos(wt)? $\endgroup$
    – Karthik
    Commented Feb 20, 2017 at 14:55
  • $\begingroup$ I have used $A_o$ instead of $2A$. $\endgroup$
    – Yashas
    Commented Feb 20, 2017 at 14:56
  • $\begingroup$ I still have a conceptual doubt though. If my incident wave was from the left to the right in the plane of the +y axis and it returned reflected in the -y axis, won't there be a phase change for the reflected wave? $\endgroup$
    – Karthik
    Commented Feb 20, 2017 at 15:03
  • $\begingroup$ I am truly sorry again. I pretty much messed up the answer. The wave which interferes with the original wave needn't necessarily be exactly 180 degrees out of phase. It can be any amount out of phase. The phase difference being 0 and 180 is a special case. I will edit the answer and add the solution for the general case. $\endgroup$
    – Yashas
    Commented Feb 20, 2017 at 15:07

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