My problem is the following: Assuming, we have a string (homogeneous, no energy loss), with a given propagation speed: $c$.

Let the origin (the source) of wave at $x=0$, thus the incident (direct) wave will be in a form of$$ \psi_i(x,t) ~=~ \sin\left(\omega t - k x \right) \,.$$ The source is sinusoidal.

At the other end of the string (at $x=L$), there is a fixed end, no energy loss. At this point the phase difference is $k*L$. The reflected wave's source is in the $x=L$ point, the distance from the new source is $L - x$, thus the reflected wave will be in a form of$$ \psi_r(x,t) ~=~ -\sin{\left( \omega t - k L - k L + k x \right)} ~=~ -\sin{\left( \omega t + k x - 2 k L \right)} \,.$$

Adding the incident and the reflected wave, we have$$ \psi(x,t) ~=~ 2 \cos{\left(\omega t - k L\right)} \sin{\left(-k x - k L\right)} \,,$$using the trigonometric identity$$ \sin{\left(a\right)} - \sin{\left(b\right)} ~=~ 2 \cos{\left( \frac{a+b}{2} \right)} \sin{\left(\frac{a-b}{2}\right)} \,.$$

The problem: this wave function $\psi$ will always be a standing wave, independently of the wavelength, however, experience shows that only $\lambda=n*2*L$ wavelengths generate standing waves.

The $\psi(x,t)=2 \cos{\left(\omega t - k L \right)} \sin{\left(-k x - k L\right)}$ is similar to $\cos{\left(\omega t\right)} \sin{\left(k x\right)},$ only difference is the phase displacement.

Where is the mistake in the derivation?

  • $\begingroup$ But have you considered what happens to the reflected wave when it reaches the 'source end' of the string, and so on? There will be interference between an infinite number of waves. It's not that hard to handle it mathematically, but easier if you lose a small fraction of the amplitude on each reflection. $\endgroup$ – Philip Wood Jul 31 '18 at 20:13
  • $\begingroup$ If your source is vibrating one end of the string, then that end is no longer fixed, and the normal standing-wave conditions (which assume two fixed ends) don't apply. $\endgroup$ – probably_someone Jul 31 '18 at 20:44

You have started with an incident right travelling wave.

$$\psi_i(x,t)=\sin(\omega\,t - k \,x)$$
and added to it a left travelling wave which represents the reflected wave

$$\psi_r(x,t)=\sin(\omega\,t + k \,x+ \phi)$$

noting that you do not yet know the phase relationship between those two waves.

You then added the two waves to represent their superposition.

$$\psi_i(x,t)+\psi_r(x,t)=\sin(\omega\,t - k \,x) + \sin(\omega\,t + k \,x+ \phi)=2\cos\left( k\,x + \dfrac \phi 2\right)\sin\left(\omega\,t + \frac \phi 2 \right)$$

and then stated a constraint which was that at $x=L$ the sum of the incident wave and the reflected wave was zero for all time.

$$2\cos\left( k\, L + \dfrac \phi 2\right)\sin\left(\omega\,t + \frac \phi 2 \right)=0$$

A solution is that $k \, L + \dfrac \phi 2 = \dfrac \pi 2 \Rightarrow \phi = \pi - 2\, k \, L$

Putting this value of $\phi$ into the summation of the two waves gives

$$\psi_i(x,t)+\psi_r(x,t)=2\cos\left( k\,x -k\, L + \frac \pi 2 \right)\sin\left(\omega\,t -k\,L + \frac \pi 2 \right)$$

which certainly has all the characteristics of a standing wave for all values of $k$ and wavelength $\lambda = \dfrac {2\pi}{k}$

Let's see what is happening at $x=0$

$$\psi_i(0,t)+\psi_r(0,t)=2\cos\left(-k\, L + \dfrac \pi 2 \right)\sin\left(\omega\,t -k\,L + \dfrac \pi 2 \right)$$

At this position the amplitude of the combined waves is $2\cos\left(-k\, L + \dfrac \pi 2 \right)$ which is not zero.

If you want the sum to be zero at $x=0$ then you have to include a second constraint which for example could be

$$-k\, L + \dfrac \pi 2 = - \dfrac \pi 2 \Rightarrow k\, L = \pi \Rightarrow L = \dfrac \lambda 2$$

So a perfect reflection will always produce a standing wave but if you then require that there is a node at a certain position then only certain wavelengths can satisfy that condition.

PS - Please check the Mathematics as there is ample room for error on my part!


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