I would like to refer to the answer by John Rennie to the following (previously asked) question: How to derive the phase difference of a standing wave?

It's a short one so please skim thorugh that.

Ok, I understand that $\Delta \phi = k(x_2-x_1)$, and that was what I expected.

But please see the problem below with which I am struggling:

enter image description here

I thought that obviously the answer is D, because the separation of 0.60 m is 3/4 of the wavelength and $$\Delta \phi = k\Delta x = \frac{2\pi}{\lambda} \times \frac{3}{4} \lambda = \frac{3}{2}\pi.$$

But the answer, surprisingly, was C. I do not understand how this answer is valid. One can roughly say that the phase difference of $\pi$ means that when one point is going up, at that instant the other point is going down. But I would like a both more physically intuitive and mathematically rigorous answer.


Here are two suggestions which, combined, make a possible explanation:

  1. If it's the string that has a length of $0.80\rm\, m$, then the length of the fundamental vibration is $\lambda_\text{fund} = 1.6\rm\,m$. Editors get tired, and tired people make typos.

  2. Discussions of string harmonics are prone to fencepost errors. Is the fundamental frequency the same as the first harmonic, so that the $n$th harmonic has wavelength $\lambda_n = \lambda_\text{fund}/n$? Or should the "first harmonic" be the lowest excited state, distinct from the fundamental, in which case the $n$th harmonic has wavelength $\lambda_n = \lambda_\text{fund}/(n+1)$? Both kinds of labels have advantages, both appear in the literature, and some of us are even guilty of using both conventions in the same paragraph, for which we had to run a lap around the physics building as penance.

In the convention where the "first harmonic" and the "fundamental" have different wavelengths, the third harmonic on a string of length $L$ has wavelength $2L/4 = L/2$, and two points separated by $3L/4$ are out of phase by a half-cycle.

I don't know whether this is the simplest interpretation of the question or not. I hope you'll let us known once you've figured it out.


the third harmonic of a wave with 80cm has wavelength 80/3, that is your fault as I read the text.

  • $\begingroup$ Pls read the question again. The wavelength is given by the problem by 80 cm. $\endgroup$
    – curious
    Aug 31 '20 at 0:31
  • $\begingroup$ I think if the wavelength is 80cm why mention the third harmonic? but I agree , the text can also sustain your interpretation, So it is not clear what the exact question is, $\endgroup$
    – trula
    Aug 31 '20 at 17:08

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