standing wave

Standing waves (most commonly) result from a right-moving travelling wave interfering with a left-moving one.

So, the wave equation for a right-moving wave is \[y_{R}(x,t)=A\sin \left( kx-\omega t+\phi_{1} \right)\] and for a left-moving wave, it is \[y_{L}=A\sin(kx+\omega t+\phi_{2})\]. To get rid of the phase shift, \[\phi\], one can either assume both they both have no phase shift, \[\phi_{1}=\phi_{2}=0\] or \[\phi_{2}-\phi_{1}=\pi\], i.e. the original wave hits a boundary (like the wall of a pipe as shown above) and gets its phase flipped by \[\pi\] upon reflection. We will show that both assumptions will lead to the same result.

So, starting with assuming both waves have no phase shift and have equal amplitude, we have simplify both equations into \[y_{R}(x,t)=A\sin(kx-\omega t)\] and \[y_{L}(x,t)=A\sin(kx+\omega t)\], then adding both waves together (as they interfere), \[y(x,t)=y_{R}+y_{L}=A\sin(kx-\omega t)+A\sin(kx+\omega t)\]. Using the trigonometric identity \[\sin a\pm\sin b=2\sin(\frac{a\pm b}{2})\cos(\frac{a\mp b}{2})\] and \[\cos\theta=\cos(-\theta)\],

\begin{align*} y(x,t)&=A \left( 2\sin \left( \frac{(kx-\omega t)+(kx+\omega t)}{2} \right)\cos \left( \frac{(kx-\omega t)-(kx+\omega t)}{2} \right) \right)\\ &=A \left( 2\sin \left( kx \right)\cos \left( -\omega t \right) \right)\\ &=2A\sin \left( kx \right)\cos \left( \omega t \right)\\ \end{align*}

Similarly, if we were to define \[\phi_{2}-\phi_{1}=\pi\], assuming that the first wave does not have phase shift, i.e. \[\phi_{1}=0\], \[\phi_{2}=\pi\]. Thus \[y(x,t)=A\sin(kx-\omega t)+A\sin(kx+\omega t+\pi)\], similar to the process above,

\begin{align*} y(x,t)&=A\sin(kx-\omega t)+A\sin(kx+\omega t+\pi)\\ &=A \left( \sin(kx-\omega t)+\sin(kx+\omega t+\pi) \right)\\ &=A \left( 2\sin \left( \frac{(kx-\omega t)+(kx+\omega t+\pi)}{2} \right)\cos \left( \frac{(kx-\omega t)-(kx+\omega t+\pi)}{2} \right) \right)\\ &=A \left( 2\sin \left( \frac{2kx+\pi}{2} \right)\cos \left( \frac{-2\omega t-\pi}{2} \right) \right)\\ &=A \left( 2\sin \left( kx+\frac{\pi}{2} \right)\cos \left(-\omega t-\frac{\pi}{2} \right) \right)\\ &=2A \sin \left( k \left( x+\frac{\pi}{2k} \right) \right)\cos \left( \omega \left( t+\frac{\pi}{2\omega} \right) \right)\\ &=2A\sin(kx^{\prime})\cos(\omega t^{\prime})\\ \end{align*}

or an easier way one could achieve this is by utilising \[\sin(\theta+\pi)=-\sin\theta\] to get \[y(x,t)=A\sin(kx-\omega t)-A\sin(kx+\omega t)\] and start solving from there.

Either way, while this equation is mathematically distinct from the first, to a physicist it is essentially just referring to the same standing wave observed at a different position and time coordinates.

Visit here and press play to watch the equation in action. Notice that the wave goes up and down but is stationary.

We can then calculate where the positions of the nodes and antinodes are on any standing wave by substituting \[\sin(kx)=0\] for nodes, \[kx=\frac{2\pi}{\lambda}x=0,\pi,2\pi,3\pi,\dots\], which tells us that each node can be found at the positions \[x=0,\frac{\lambda}{2},\lambda,\frac{3\lambda}{2},\dots\], which is indeed consistent to our observations. Then, for antinodes, or positions where \[y\] oscillates between \[\pm A\], we substitute \[\sin(kx)=\pm1\], similar to the above, \[\sin(kx)=\pm1\implies x=\frac{\lambda}{4},\frac{3\lambda}{4},\frac{5\lambda}{4},\dots\].

When standing waves are formed due to boundaries enforced on a medium through which the waves propagate, such as a string with two ends fixed, we know that two ends have to be nodes. This means that only specific wavelengths can fit on the string of a fixed length, with nodes always at the ends, and otherwise nodes and antinodes alternating on the length of the string.