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(that question may sound like my last question What makes a higher frequency sound wave more energetic? but I wouldn't consider it a duplicate, the focus is very different.)

Comparing two mechanical waves with same amplitude, the one with higher frequency delivers more energy but I don't really understand why.

enter image description here

The amplitude is associated with the potential energy - ok, higher amplitude equals more energy, that makes sense. But if I just shorten the wavelength (elevating the frequency) why does that one period now contains more energy? I mean the kinetic energy is higher because the amplitude is reached with higher velocity but at that moment it also falls again and over the whole period averaging to zero.

So even if the higher frequency causes that the wave reaches the amplitude faster, that extra kinetic energy getting lost somewhere because the amplitude is fixed. I can move a mass in vacuum with the force of 1 newton over 1 meter resulting in an energy lost of 1 Joule, but I can also move that mass with 10 newton 1 meter resulting in an energy lost 10 times higher. The only difference is that I moved that block faster in the second scenario because I gave it a higher acceleration and thus giving it a higher kinetic energy, which is ultimately entirely pointless because my task was simply to move the block 1 meter in a vacuum, an infinitely small force would have been sufficient without considering time.

The additional kinetic energy is either lost as heat at the end of the 1 meter, or the block is rebounded and retains the additional kinetic energy as a result.

Now, I wonder what happens to the additional kinetic energy of a wave at its extreme points; they don't shoot beyond the amplitude, after all.

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  • $\begingroup$ Why do the strings of a string bass move more than violin strings? $\endgroup$
    – John Doty
    Sep 15, 2023 at 21:17
  • $\begingroup$ @JohnDoty with "more" you mean with higher amplitude? I think it's because a bass string has less tension, thus the kinetic energy can move it farther from equilibrium position than in the case of a violin. $\endgroup$
    – iwab
    Sep 15, 2023 at 21:21
  • $\begingroup$ @JohnDoty oh, so you want to make clear that by lifting up the frequency it's like setting up a greater restoring force? But how would that work? I mean we can't change the gravitational force causing waves in water (?) Or what about sound waves and the back and forth of the molecules? That string in a violin for example definitely have way more energy at its amplitude because the tension is much stronger and thus it's much hotter but how to generalize that special example? $\endgroup$
    – iwab
    Sep 15, 2023 at 21:36
  • $\begingroup$ Stringed instrument tensions aren't so different. Bass tunes lower than violin because strings are longer and heavier. $\endgroup$
    – John Doty
    Sep 15, 2023 at 21:51
  • $\begingroup$ Energy versus displacement depends not only on string tension, but string length. $\endgroup$
    – John Doty
    Sep 15, 2023 at 21:52

8 Answers 8

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Let's take a string of length $L$ fixed on both ends with some tension $T$. The string can be made slightly longer by stretching it, and we will only consider "small" stretchings so the tension does not change much. Since $T$ is the force resisting stretching the string, if the string length is increased by $l$, it should have taken work $Tl$ (there should be potential energy $Tl$ in the stretched string). This is the sense in which waves on the string increase the potential energy of the string. In other words, the energy of the wave is proportional to how much it lengthens the string at maximum.

Not let me focus on standing waves. So at certain times, the strings are not moving, but physically resemble a sine wave (blue). At other times, the string is straight, but there is kinetic energy as it is moving between one peak and another (green, halfway between blue and orange in time). enter image description here

In the apex of the wave we can easily calculate the total energy added to the string by the wave, because kinetic energy is zero. Say the curve has a shape $y=A\sin(2\pi n x/L)$. Then letting $a=A\pi n/L$ the curve length is: $$ \frac{2L}{\pi}\sqrt{1+a^2} E[a^2/(a^2+1)]\approx L(1+a^2/4) $$ Where $E$ is an elliptic integral. The $\approx$ in the equation above is a small $a$ approximation, and the equation is only valid for small $a$ anyway because only then is the shape of the string sinusoidal.

enter image description here Since $a$ increases with $n$ and $A$, higher amplitude and higher frequency with fixed amplitude waves have more energy. Indeed the string length is quadratically increasing with $a$, and therefore quadratically increasing with both $n$ and $A$. All of this to explain the visually apparent fact that higher frequency sine waves have more curve length... enter image description here

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It maybe helpful to think of a mass attached to a spring but the following statements are general. At the same amplitude, a higher frequency implies a higher velocity and hence a higher kinetic energy for the moving mass. To achieve this higher frequency a larger return force is needed and this means a larger potential energy, consistent with the higher kinetic energy.

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Wave equation in classical mechanics. Sometimes it's possible to recast the governing equation of an elastic medium as a wave equation,

$\dfrac{1}{c^2} \partial_{tt} u - \nabla^2 u = f$,

being $u$ the displacement, $c$ the speed of sound in the medium, $f$ the volume force. As an example, the equation governing the lateral displacement of the points of a wire with pre-stress and no bending stiffness reads with no volume forcing,

$m \partial_{tt} u - \sigma_0 \partial_{xx} u = 0$,

with $m$ the linear mass density and $T_0$ the axial preload, so that $c = \sqrt{\sigma_0 / m}$.

General expression of the solution of the wave equation. It's possible to write the solution of this equation as a superposition of standing waves as,

$u(x,t) = \sum_n \left\{ A_n e^{j(k_n x + \omega_n t)} + B_n e^{j(k_n x - \omega_n t)} + c.c \right\} = \sum_n f_n(x) e^{j \omega_n t} + c.c.$

Energy. The kinetic energy density $k$ is proportional to $\frac{|\partial_t u|^2}{2}$ so that,

\begin{align} k \sim \partial_t u^* \partial_t u & = \sum_n \left( -j \omega_n f^*_n (x) e^{-j\omega_n t} + c.c \right) \sum_m \left( j \omega_m f_m e^{j\omega_m t} + c.c \right) = \\ & = \sum_{m,n} \left( \omega_n \omega_m f_n f_m e^{j(-\omega_m + \omega_n)t} - \omega_n \omega_m f_n f_m e^{j(\omega_m + \omega_n)t} + c.c. \right) \end{align}

and averaging over a period (or a long time),

$k \sim \sum_m \omega_m^2 f^2_m(x)$,

that shows you how the contribution of each standing wave to the kinetic energy is weighted by the square of its frequency.

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Your basic error is that you are treating the wave as though it was a system of independent particles and this is not correct.
The particles are communicating with one another in some way, eg via chemical bonds between the particles which make up a string.

So even if the higher frequency causes that the wave reaches the amplitude faster, that extra kinetic energy getting lost somewhere because the amplitude is fixed.

The kinetic energy does not get lost it is stored as potential energy amongst the system of particles.
I agree that if you choose one particle at its maximum displacement (amplitude $A$) then the "spring" (spring constant $k$) it is "attached to" might appear to have potential energy $\frac 12 kA^2$ but that is ignoring the potential energy of all the other particles which make up the system.

Consider a small element of string from position $(x,0)$ to position $(x+\delta x ,0)$ now being stretched from position $(x,y)$ to position $(x+\delta x ,y+\delta y)$.

The new length of that element is now $\sqrt{\delta x^2+\delta y^2}$.
Thus, the element has been stretched by $\sqrt{\delta x^2+\delta y^2} - \delta x\approx \delta x\left(1 + \frac 12\left (\frac{\partial y}{\partial x}\right)^2 \right)$ on apply the binomial expansion assuming that $\delta y\ll \delta x$.

Assuming that the tension $T$ stays constant then the work done in stretching the element (equal to the potential energy stored in the element) is $T\left [\delta x\left(1 + \frac 12\left (\frac{\partial y}{\partial x}\right)^2 \right)-\delta x \right]=\frac T2 \left(\frac{\partial y}{\partial x}\right)^2\delta x$.

If $y=A\sin (kx-\omega t)$ then the potential energy stored in the element is $\frac T2 A^2k^2\cos^2(kx - \omega t)$

Noting that $k = \omega /\,c$, where $c$ is the speed of the wave, this means that the potential energy of the element is proportional to $\omega^2$.

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The easy way to think of this is to imagine a hand saw that is cutting a board. For the sake of our analogy the saw has teeth that cut both on the push and the pull.

We work the saw back and forth through the same amplitude, and start slowly: One back-and-forth cycle in 10 seconds. Our rate of performing work on the board is pretty slow and it takes 5 minutes to cut through it. Now we speed up the back-and-forth rate to one cycle in 5 seconds (twice the frequency) and are applying work to the board at twice the rate, and we cut the board in half the time. One complete cycle per second increases the work rate (i.e., power) ten times and the saw cuts the board in 1/10th the time.

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    $\begingroup$ Now you observed the power but I am especially interesting in the quality behind the energy difference between two waves with same amplitude but different frequency AND we just watch 1 period of each wave. Your analogy compares 1 period with multiple periods. $\endgroup$
    – iwab
    Sep 16, 2023 at 6:41
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    $\begingroup$ I mean, we just speed up the cutting. If I release the energy faster it doesn't mean I have more energy. I could cut it slowly and it needs more time or cutting it faster (higher frequency) which takes less time but in both scenarios I ended up using the same amount of energy so I lack to see how the frequency changes things that the wave got more energy. $\endgroup$
    – iwab
    Sep 16, 2023 at 13:50
  • $\begingroup$ If we stay in that analogy, maybe we can say higher frequency means more saw tooths? So with amplitude A one Periode with high frequency cutting a bit deeper than one periode with A and lower frequency. $\endgroup$
    – iwab
    Sep 16, 2023 at 13:54
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I’m not really familiar with the mathematics of this, so I’ll propose a simple answer:

First we need to look at the definition of a mechanical wave. A mechanical wave is a vertical transfer of energy through a medium. Imagine tying a rope to a pole and generating waves by swinging the other end up and down. The energy of a single wave is defined as its frequency. When you use more force and hence, more kinetic energy, to swing the rope, it produces higher waves that, in turn, have a higher frequency.

For more proof, you can refer to the general relationship that the average rate of energy transfer (i.e., the speed that the waves are traveling) is proportional to the square of the amplitude and frequency. This means that a high frequency wave of 4 would have 4 times the kinetic energy than a low frequency wave of 2. You can refer to https://shorturl.at/AHSVZ for an algebraic explanation.

Hope this helps!

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In the case of stationary waves, as the vibrating strings in a guitar, the situation is similar to an oscillator. If one fret is pressed, so that the vibrating string has half of its length, the elastic potential is bigger (keeping the same amplitude). So the energy increases with the pitch (frequency). It is like to increase the spring constant in the oscillator.

For traveller waves (in a very long string) we can start from the differential equation:

$$\frac{\partial^2 y}{\partial t^2} = \frac{|T|}{\rho}\frac{\partial^2 y}{\partial x^2}$$ where $y$ is the vibration direction, $x$ is the propagation direction, $T$ is the tension and $\rho$ is the linear density.

The solution for a traveller wave is: $$y = f(x - vt)$$ where $$v = \sqrt{\frac{T}{\rho}}$$

The kinetic energy in a small region of the string is: $$E_k = \frac{1}{2}\rho \left(\frac{\partial y}{\partial t}\right)^2 = \frac{1}{2} \rho v^2\left (\frac{\partial f(x - vt)}{\partial (x - vt)}\right )^2$$ Using $$v = \sqrt{\frac{T}{\rho}}$$ and noting that $$\frac{\partial f(x - vt)}{\partial (x - vt)} = \frac{\partial y}{\partial x}$$ $$E_k = \frac{1}{2}T\left (\frac{\partial y}{\partial x}\right )^2$$ The slope of the curve $y(x)$ for a given $t$ increases with a higher frequency (and smaller wave length), so $E_k$ increases.

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But if I just shorten the wavelength (elevating the frequency) why does that one period now contains more energy?

In a mechanical wave, the oscillation corresponds to some periodic bending or stress on the material medium. Squeezing more bends into the same length of material, with a constant amplitude, increases the energy required to reach this state. Moreover, each individual bend is tighter, which takes more energy (imagine bending the ends of a stiff spring toward each other). These two effects both increase energy inversely proportional to wavelength (so $E\propto\omega^2$).

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