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4) waves, the classical equation of motion / SurprizingFacts

Let's return to the equation of oscillations of a sphere on a spring

In one of the first articles of the cycle, we first derived a formula for the vibrational motion of a sphere

 $ z (t) = z_0 + A cos [ 2 π ν t ] $ "data-tex =" display

And then we found the equation of motion for which this formula was a solution

 $ d ^ 2z / dt ^ 2 = - K / M (z-z_0) $ "data-tex =" display

Here
• d 2 z / dt 2 denotes the time evolution of the change in time z (t).
• K – the force of the spring, M – the mass of the ball, z 0 – the equilibrium position.
• ν = √ K / M / 2π

The key step for obtaining the last frequency equation, expressed in terms of K and M, was the calculation of d 2 z / dt 2 for the vibrational motion of the ball z (T) = z 0 + A cos [ 2 π ν t ]. We found that

 $ d ^ 2z / dt ^ 2 = - (2 π ν) ^ 2 (z-z_0) $ "data-tex =" display

The equation of motion of waves

Now we want to do the same for waves. We found a formula for the shape and motion of a wave that oscillates both in space and in time.

$$ display $$ Z (x, t) = Z_0 + A cos (2π [ν t – x/λ]) = Z_0 + A cos (2π [t/T – x/λ]) $$ display $$

Among the solutions of which equation of motion is such a formula? You can imagine the answer. Obviously, it includes:

1. D 2 Z / dt 2 change in time of change in time Z (x, t).
2. 2 Z / dx 2 a change in the space of change in the space Z (x, t).

In a natural way, we can guess that the equation should look like Like this:

 $ C_t d ^ 2Z / dt ^ 2 + C_x d ^ 2Z / dx ^ 2 = C_0 (Z - Z_0) $ "data-tex =" display

Where C t C x and C 0 are constants. I would like to note that if C t = 1, C x = 0, and C 0 = -K / M, you return to the equation of the ball oscillation on a spring . What are the constants in this case?

We can always put C t = 1. If you wanted, say, put C t = 5, I would just ask you To divide the whole equation by 5, which would give you the equivalent of the variant in which C t = 1, just with other values ​​of the remaining constants.

After that it turns out that the values ​​of C X and C 0 are different in different physical systems. We will study two different classes of waves differing by different constants.

Both classes C x will be negative,

 $ C_x = -c_w ^ 2 $ "data-tex = <sub> w </sub> denotes the velocity of high-frequency waves). </p>
<p> These classes will differ in that the first class, Class 1, C <sub> 0 </sub> will be negative, and will equal – (2 π μ) 2, and for the second, Class 0, C <sub> 0 </sub> will be zero. </p>
<p> Let us now investigate the properties of the waves of these two classes of equations . But before that, we need to perform another calculation, which we already did before. </p>
<h2> Quick Count </h2>
<p>
For our infinite wave </p>
<p><math> $$ display $$ Z (x, t) = Z_0 + A cos (2π [ν t – x/λ]) = Z_0 + A cos (2π [t/T – x/λ]) $$ display $$ </math>
</p>
<p>
We need to know d <sup> 2 </sup> Z / dt <sup> 2 </sup> and d <sup> 2 </sup> Z / dx <sup> 2 </sup>. In the previous article we have already shown that for a ball on a spring moving in accordance with z (t) = z <sub> 0 </sub> + A cos [ 2 π ν t ]it turns out that </p>
<math> <img src=

. The change in time gives us a factor of 2 π ν, and the time variation of the time variation gives us two factors. In addition, there is a common minus sign. Therefore, you will not be surprised that:

 $ d ^ 2Z / dt ^ 2 = - (2 π ν) ^ 2 (Z-Z_0) $ "data-tex =" inline

 $ d ^ 2Z / dx ^ 2 = - (2 π / λ) ^ 2 (Z-Z_0) $ "data-tex =" inline

Each change in time gives us a factor ν = 1 / T (the longer the period, the slower the time variation), and each change in space gives us a factor of 1 / λ (the longer the wave, the slower the change in space).

The proof

For an infinite wave, we have the basic equation

$$ display $$ Z (x, t) = Z_0 + A cos (2π [ν t – x/λ]) = Z_0 + A cos (2π [t/T – x/λ]) $$ display $$

And we want to show that

 $ d ^ 2Z / dt ^ 2 = - (2 π ν) ^ 2 (Z-Z_0) $ "data-tex =" display

 $ d ^ 2Z / dx ^ 2 = - (2 π / λ) ^ 2 (Z-Z_0) $ "data-tex =" display

Several facts:

• Z-Z 0 = A cos (2π [ν t – x/λ]) (in the basic equation they transferred Z 0 to the left side)
• Since Z 0 is a constant independent of time and space, dZ 0 / dt = 0 and dZ 0 / dx = 0.
• d (cos t) / dt = – sin t, and d (sin t) / dt = + cos t
• d (F [a t +b x]) / dt = a d (F [a t +b x]) / d (a t + b x), where a and b are constants, and F is any function of (a t + b x).
• d [A f(t)] / dt = A d [f(t)] / dt, where f (t) is any function of t, and A is a constant

All together this means that:

$$ display $$ dZ / dt = d [Acos(2π[ν t – x/λ])] / dt = A d [cos(2π[ν t – x/λ])] / dt \ = A (2π ν) d [cos(1945)] and

and

(1945)

$$ display $$ d ^ 2Z / dt ^ 2 = d [-(2πν)Asin(2π[ν t – x/λ])] / dt = \ – (2π ν) A d [sin(2π[ν t – x/λ])] / Dt = \ – (2π ν) ^ 2 A cos (2π [ν t – x/λ]) = – (2π ν) ^ 2 (Z-Z_0) $$ display $$

Since the basic formula for the wave does not change when (ν t) is replaced by (-x / λ), the calculation of d 2 Z / dx 2 does not differ from the calculation of d 2 Z / dt 2 simply instead of d / dt, which gives the multiplier (2π ν), we will have d / dx, giving the factor (- 2π / λ). But since there are two such multipliers in the answer, we simply replace (2π ν) 2 by (-2π / λ) 2 = (+ 2π / λ) 2 ; Minus the value does not have (the common minus in the addition remains). As we were required to prove, [1945][1945][1945] ]

Small font: all the above derivatives are in fact partial derivatives.

Class 0: waves of any frequency and equal velocities

In this class of waves, the equation of motion will be as follows:

 $ d ^ 2Z / dt ^ 2 - c_w ^ 2 d ^ 2Z / dx ^ 2 = 0 $ "data-tex =" display

Having connected the formula Z (x, t) for an infinite wave and using the calculations just made by us, we find that:

$$ display $$ – (2 π ν) ^ 2 (Z – Z_0) – ( -c_w ^ 2) (2 π / λ) ^ 2 (Z-Z_0) = 0 $$ display $$

We divide the equation by

 $ - (2 π) ^ 2 (Z-Z_0) $ "data-tex =" inline

we get

 $ ν ^ 2 - c_w ^ 2 / λ ^ 2 = 0 $ "data-tex =" display

Since the frequencies, velocities and wavelengths are positive, one can extract the root and obtain

ν = c w / λ, or, if you want, λ = c w / ν = c w T

From this formula we learn that:

• Initially, our wave, as we recorded it, could have any frequency and any wavelength. But the equation of motion makes them depend on each other. For waves of class 0, any frequency can be selected, but after that the wavelength is determined by λ = c w / ν.
• All waves of class 0, regardless of frequency, move with speed c w . This follows from the formula λ = c w T and Fig. 3 of the previous article. Observe how the wave passes one cycle of oscillations during one period of T. What happens? The wave looks exactly the same after T, but each comb is moved to where its neighbor was, at a distance of λ. This means that the crest moves a distance λ in a time T – one wavelength in one oscillation period – and this means that the crests move at a speed λ / T = c w . This is true for all frequencies and their periods, and for all wavelengths!
• As in the case of a ball on a spring, the amplitude A of these waves can be any, arbitrarily large or small. And this is so for all frequencies.

Class 1: waves with a frequency greater than the minimum, with different speeds

For this class of waves, the equation of motion is as follows:

 $ d ^ 2Z / dt ^ 2 - cw ^ 2 d ^ 2Z / dx ^ 2 = - (2 π μ) ^ 2 (Z - Z_0) $ "data-tex =" display

Substituting the formula Z (x, t) for an infinite wave and using the fast calculation indicated above, we find that

$$ display $$ – (2 π ν) ^ 2 (Z – Z_0) – (- Cw ^ 2) (2 π / λ) ^ 2 (Z-Z_0) = – (2 π μ) ^ 2 (Z-Z_0) $$ display $$

By dividing the equation by

 $ - (2 π) ^ 2 (Z - Z_0) $ "data-tex =" inline

we get

 $ ν ^ 2 - cw ^ 2 / λ ^ 2 = μ ^ 2 $ "data-tex =" display

Since the frequencies, velocities and wavelengths are positive, we can extract the square root and obtain

 $ ν = [ (cw / λ)^2 + μ^2 ] ^ {1/2} $ "data-tex =" display

I recall that y 1/2 is the same as √y.

This formula is very different from the formula for waves of class 0, as well as the consequences of its application.

Vo First, the equation of motion indicates the presence of a minimum acceptable frequency. Since (cw / λ) 2 is always positive,

$$ display $$ ν = [ (cw / λ)^2 + μ^2 ] ^ {1/2} ≥ μ $$ display $$

]

In order to approach ν = μ, it is necessary to increase λ. For very large wavelengths, the frequency approaches μ, but it can not become smaller. For waves of class 0 this was not so. They had ν = cw / λ, so for them, the more you make λ, the stronger ν approaches zero. For waves of class 1, any value of ν greater than μ is possible.

Secondly, we found the proof that for all waves of class 0 the velocity is the same, but it does not work for waves of class 1. The only option in which it can work , If ν is taken to be very much larger than μ; For this we need to make λ very small (and, accordingly, 1 / λ is very large). In this case,

$$ display $$ ν = [ (cw / λ)^2 + μ^2 ] ^ {1/2} ≈ cw / λ $$ display $$

That is, at very high frequencies and small wavelengths for Class 1 waves, there will be approximately the same ratio between frequency and wavelength as for Class 0 waves, therefore for the same reasons as Class 0 waves, such waves will move with velocity , (Approximately) equal to cw.

What is true for waves of both classes is that the amplitude A can be arbitrary, arbitrarily small or large, and does not depend on frequency.

 image
Fig. 1. For waves of class 0 and 1, the equation of motion gives the relationship between frequency, or period, and wavelength, or 1 / wavelength. Each of the graphs shows the relationship of these quantities as a function of the equation of motion. Three graphics show the same thing, but they are built on different variables. The blue lines refer to waves of class 0. Reds denote waves of class 1 whose speed is the same at very high frequencies and small wavelengths when they coincide with the blue lines. But at the minimum frequency μ (and with a maximum period of 1 / μ), indicated by a green one, the two curves diverge with increasing wavelengths.

Small font: you may have noticed that I was a little cunning . I did not count the speed of waves of class 1. The fact is that there is a very cunning trick here. For waves of class 0, I counted their velocity, following the movements of the crests. This works because in class 0, the waves of all frequencies move at the same speed. But in class 1, or in any other, where waves of different frequencies move at different speeds, the speed of a real wave is not set by the speed of its crests! It turns out that the crests move faster than cw, but the wave velocity is less than cw. To understand this, it is necessary to use a very unobvious logic and the difference between "group" and "phase" speed. I will bypass this trick; Just wanted to draw your attention to its existence, so that you do not get the wrong idea.

Final comments about the classical waves

You can find many familiar examples of waves of class 0, including sound in air, water or metal (where cw is the speed of sound waves in the material), light, and other electromagnetic waves (where cw = c in vacuum) and waves on ropes or strings, As in Fig. 2 in the previous article. Therefore, waves of class 0 are taught in elementary physics courses. I can not give an example of class 1 waves in ordinary life, but we will soon see that these waves are as important for the universe.

We have the convenient formula E = 2 π 2 ν 2 A 2 M for the energy of the ball of mass M on the spring. Formulas for other oscillators depend on their nature, but their shape is about the same. But in the case of waves, we did not mention their energy. In particular, because we have studied waves with an infinite number of ridges to simplify mathematics. It is intuitively clear that some energy must be stored in the motion and shape of each crest and hollow, and with an infinite number of crests and depressions the amount of energy in the wave will be infinite. This can be circumvented in two ways. Exact formulas depend on the type of wave, but let's consider waves of class 0 on a rope.

• The amount of energy per wavelength (stored between the point x and the point x + λ) is, of course, 2 π 2 A 2 M λ where M λ is the mass of a piece of rope of length λ.
• In reality, waves are not infinite. As an impulse from several crests and depressions, shown in Fig. 2 in the last article, any wave will be finite, it will have a finite number of ridges and depressions. If it extends by a length L, that is, it has L / λ ridges, then the energy transferred to it will be 2 π 2 ν 2 A 2 M L where M L is the mass of a length of a string of length L. This is simply L / λ, multiplied by energy by one wavelength.

For waves propagating not along The details of the equations will differ, but the energy per wavelength of a simple oscillatory system will always be proportional to ν 2 A 2 .

In class 1 there is a very interesting wave, Which was not is 0. This class in the case where ν = μ, the minimum value, and λ = infinity. In this case, the wave takes the form

 $ Z (x, t) = Z_0 + A cos (2πμt) $ "data-tex =" display

This wave does not depend on x at any time, that is, Z (x, t) is a constant throughout the space, and Z oscillates in time exactly as a ball on a spring with frequency μ. Such a stationary wave, shown in Fig. 2, will prove to be very important in subsequent discussions.

 image
Fig. 2

Quantum waves

For a ball on a spring, the difference between a classical and a quantum system was that in the first case the amplitude could take arbitrary values, like the energy, and in the quantum case the amplitude and energy were quantized. For any similar oscillatory system, this works in the same way. Perhaps we can guess that this is also true for waves …