Energy of a Point at different times in a Standing Wave

Question

Two waves traveling in opposite directions meet to Produce Standing Waves. At Nodes The displacement of one wave is +x and for the other wave, it is -x. Both of which cancel to produce a Node. At this point, there is Kinetic Energy & Potential Energy in Both waves.

I am confused with the Part that will the Nodes have Energy. Because adding Kinetic Energy of Both waves at this position will give Kinetic Energy on the Node but we say the Energy of a Node is Zero.

How will the energy at the Nodes Change in different positions of Standing Wave a) When all the points are at their Mean Position. b) When all the Points are at their extreme Position.

I have the understanding that In a standing wave a) When all the points are at their Mean Position all have Varying Kinetic Energies and Zero Potential Energies.

b) When all the points are at their Amplitude Position all have Varying Potential Energies and Zero Kinetic Energies.

Does this Hold for the Nodes also?

Answer 1

The quantity you plot isn't defined, but let's assume it corresponds to displacement and that the square of displacement corresponds to potential energy. In a travelling wave there are always two energy storage mechanisms. Let's say the other one is kinetic energy which corresponds to velocity-squared. In a traveling wave the displacement and velocity are 90 degrees out of phase. This means that the potential and kinetic energy sum to be a constant at any point in space or time.

Two waves traveling in opposite directions create a standing wave. In a standing wave, there are points where the displacements cancel (nodes) where there is zero potential energy. At these points the velocities add and there is a maximum of kinetic energy.

A quarter-wavelength away from the nodes, there are points where the displacements add and the potential energy reaches a maximum. At these points the velocities cancel and there is zero kinetic energy.

In a standing wave, the total energy is constant but the proportion of potential and kinetic energies changes with position.

An electromagnetic wave would provide a different example. In a standing wave the total energy would be constant but the proportion of electric and magnetic energies would change with position.

Answer 2

A standing wave stores potential and kinetic energy as shown in the series of displacement-time graphs below.

At times $t=0,\,\frac T2,\,T, . . . . .$, where $T$ is the period, the kinetic energy of all the particles is zero and the displacement of all the particles is a maximum with the standing wave storing a maximum amount of potential energy.
At times $\frac T4,\,\frac {3T}{4},......$, when the displacement of all particles is zero the potential energy is a minimum and the kinetic energy is a maximum.
At all times the sum of the potential energy and kinetic energy is constant.

It is certainly true that a standing wave can be thought of as being produced by the superposition of two travelling waves of equal amplitude, frequency and velocity travelling in opposite directions but all you can observe is the resultant standing wave.

So it just happens that at a node the phases of the two travelling waves differ by $\pi$ and so the displacement of particles at such positions is zero for all time.
At anti-nodes the amplitude of motion of the particles is a maximum as the two travelling waves are in phase at that point.

Thus it is not a matter of, at a particular position, adding the kinetic energy of wave $i$ to the kinetic energy of wave two, eg $\frac 12 mv_1^2 + \frac 12 mv_2^2$, but rather finding the resultant velocity, $\vec v_1+\vec v_2$ of a particle which is under the influence of the two travelling waves and then using that value to find the kinetic energy at that position, $\frac 12 m|\vec v_1+\vec v_2|^2$.
And one can also do the same in terms of the potential energy by first adding the displacements due to the two individual travelling waves, $\vec x_1+\vec x_2$, and then finding the potential energy at a given position using the net displacement.