# Where is the energy stored in destructively-interfering waves?

Let's say we have two waves moving along a string. One of them is represented by the function: $$f_1(t)=\sin(\omega t)$$

The other one is represented by a function:

$$f_2(t)=-\sin(\omega (\tau-t))$$

Both of these functions are defined over one period.

At time $t=\tau/2$, the waves are overlapping perfectly and destructively interfere. This means we have: $$y(\tau/2)=f_1(\tau/2)+f_2(\tau/2)=\sin\left(\omega\left(\frac \tau 2\right)\right)-\sin\left(\omega\left(\frac \tau 2\right)\right)=0$$

This is fine and good; it shows that the waves have destructively interfered. But there's a weird part to this. But obviously, not only does $y(t)=0$, but $y^{(n)}(t)$ must also be zero (as $f_1(t)+f_2(t)=0$). However, we know full well that, because they're waves on a string, if we advance to time $\tau$ the waves will pass each other and head in the opposite directions.

How can this be? There's an instant at $\tau/2$ where the wave is not only flat, but there is no velocity, acceleration, jerk, snap, or anything that would cause a change in motion.

Where did the energy go? How does the wave start moving again?

• Whoops. I just realized I made a very silly error in calculation. Obviously taking $\frac d{dt} y(\tau/2)$ is going to be zero because $y(\tau/2)$ doesn't vary with $t$. – Aza Mar 6 '15 at 3:01

Let us consider this with two symmetrical pulses traveling in opposite directions; they are exactly alike except that one is positive & other negative. As they pass through each other, there comes a moment at which the whole string is straight. Where did the energy go during the annihilation?

After some time, the pulses reappear. But "what is it that preserves the memory of them through the stage of zero displacement, so that they are recovered again in their original form"? It is the velocity of the different parts of the system. The string at the instant of zero deformation has a distribution of transverse deformation has a distribution of transverse velocities characteristic of the two superposed pulses. Transverse displacements cancel, but transverse velocities add, and for this one instant the whole energy of the system resides in the kinetic energy associated with these velocities.

$y(\tau/2)$ doesn't vary with time, so obviously $\frac d{dt} y(\tau/2)=0$, since $y(\tau/2)$ is a constant value with respect to $t$.

• This should be an edit or a comment. – Jimmy360 Jun 18 '15 at 1:17
• @Jimmy I mean, it is the answer to the question. I can expand on why, this actually answers it, though. – Aza Jun 18 '15 at 2:13

You made a mistake. Let me explain it simply by taking two waves that begin simultaneously from opposite sides of the chord. The waves won't look as you say, but

$$f_{+}(x, t) = Asin(kx - \omega t) \tag{i}$$

and

$$f\_(x, t) = Asin(-kx - \omega t) \tag{ii}$$

i.e. they advance toward one another from opposite direction. Their superposition produces

$$f_{+}(x, t) + f_{+}(x, t) = 2Asin(kx)cos(\omega t), \tag{iii}$$.

or, calculating the intensity,

$$I(x, t) = (|A|^2/2)[1 - cos(2kx)][1 + cos(2\omega t)]. \tag{iv}$$

As you see, there are positions where the $cos(2kx) = 1$, (minima), and positions where $cos(2kx) = -1$ (maxima). Note also that whenever $cos(2\omega t) = -1$ we have zero amplitude on all the string, and when $cos(2\omega t) = 1$ we have respectively maximum amplitude. So, we get static waves.

Now, if you ask where goes the energy of the minima, it is redistributed in place. From the places of minima it goes to the places of maxima. As to the behavior in time, you know that an oscillator passes from kinetic to potential energy. Through the region with amplitude zero the chord passes with maximal velocity, at maximal amplitude the velocity vanishes and then changes direction, i.e. when the cord is at maximal amplitude of oscillation, the energy is only potential. So, the total energy in time and space, is the same. No energy is lost, none appears out of nowhere.