$$f=Ae^{i\omega t}\big(e^{-ikz}-e^{ikz}\big) = -2iAe^{i\omega t}\sin(kz)$$

This is the displacement equation I got. As it is complex, I used the real part to represent the nth normal mode of standing wave which is having both position and time dependent term as sine function. At $t=0$ , this gives me zero amplitude at any point $z$. I am following H. J. Pain, The Physics of waves and vibrations book.

  • $\begingroup$ It can be correct! Please notice that the time derivative does not vanish at $t=0$. If this does not help, please clarify what is the difficulty. $\endgroup$ Feb 4, 2022 at 16:27
  • 1
    $\begingroup$ How is this zero at t=0? $\endgroup$
    – nasu
    Feb 4, 2022 at 16:34
  • $\begingroup$ What are you stuck with? $\endgroup$
    – Cream
    Feb 4, 2022 at 16:40
  • $\begingroup$ Hi Tarun. I've replaced the picture of your equation with MathJax; if you need a reference to do this yourself in the future, you can find one here. $\endgroup$
    – J. Murray
    Feb 4, 2022 at 16:41

1 Answer 1


The displacement of the standing wave vanishes for any point $z$ twice every period. There is nothing wrong here


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