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$$
psi_i(x,t) ~=~ sinleft(omega t - k x right)
,.$$ 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$$
psi(x,t) ~=~ 2 cos{left(omega t - k Lright)} sin{left(-k x - k Lright)}
,,$$using the trigonometric identity$$
sin{left(aright)} - sin{left(bright)}
~=~ 2 cos{left( frac{a+b}{2} right)} sin{left(frac{a-b}{2}right)}
,.$$

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

Where is the mistake in the derivation?

Farcher · Answer

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.

$$psi_i(x,t)+psi_r(x,t)=sin(omega,t - k ,x) + sin(omega,t + k ,x+ phi)=2cosleft( k,x + dfrac phi 2right)sinleft(omega,t + frac phi 2 right)$$

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.

$$2cosleft( k,
L + dfrac phi 2right)sinleft(omega,t + frac phi 2 right)=0$$

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

$$psi_i(x,t)+psi_r(x,t)=2cosleft( k,x -k, L + frac pi 2 right)sinleft(omega,t -k,L + frac pi 2 right)$$

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

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

$$psi_i(0,t)+psi_r(0,t)=2cosleft(-k, L + dfrac pi 2 right)sinleft(omega,t -k,L + dfrac pi 2 right)$$

At this position the amplitude of the combined waves is $2cosleft(-k, L + dfrac pi 2 right)$ 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!

Non-standing waves on string

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