Non-standing waves on string

Question

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?

Answer 1

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!