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I Amplitude vs Frequency

  1. Nov 11, 2016 #1
    Hello all,

    First of all, let me say that I am aware that a higher amplitude doesn't translate into a higher frequency.
    However, my understanding of how for example a string vibrates makes me draw other conclusions.

    From what I understand, hitting a string with a particular force makes the string vibrate over a certain width distance x. The time duration in which that string goes up and down (2x = 1 period) once during the vibration determines the frequency of the waves it generates. Thus it depends on the speed in which the string travels over 2x.

    Now, if I hit the string harder, it would have a higher amplitude and thus it would vibrate over a wider distance (> x). However, to keep the frequency, the string has to go up and down in the same time interval as previous. But since the width distance of string vibration has increased, the string has to travel faster over that width distance to keep the same time interval for 1 period.

    If the string has to travel faster, doesn't it give more kinetic energy to the air molecules which results in the high pressure column of airmolecules in soundwaves moving faster as well? If so, how can faster moving high pressure columns of air not result in a higher pitch since pitch also depends on the speed of the waves?
    Last edited: Nov 11, 2016
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  3. Nov 11, 2016 #2
    Everything in your description is correct except the conclusion. Higher amplitude is by definition more motion. It does require the string to move farther and it will move faster in the middle of the motion. It does cause higher variations in the pressure of the air with higher compressions and lower rarefactions and faster instantaneous changes in pressure and faster motion of the air molecules. All these things you describe are true, and they in no way require the frequency to change. (For completeness I'll mention the nonlinearity of over driven oscillators later)

    I can describe the pressure wave in the air mathematically in one dimension:

    P(t,x) = A * sin [2 pi ( k x + f t)]

    Where k is 1 / (wavelength * speed of sound)

    You are concerned with how the rate of change of pressure relates to frequency, so I take the derivative

    d P(t,x) / dt = - A * 2 pi f cos [ 2 pi ( k x + f t)]

    Which reaches its maximum as the cosine equals 1. The maximum rate of change of pressure is given by the pre factor

    A * 2 pi * f

    So, yes, a more rapid change in pressure is associated with a higher frequency, but it is also independently associated with a higher amplitude. There is no reason to associate the fact that a higher amplitude gives a more rapid instantaneous change in pressure and faster motion of the molecules with some necessity of higher frequency.

    That being said, I should mention that frequency can be coupled to amplitude, but in a way and for reasons completely unrelated to what you were asking. If the object doing the oscillating (the string in this case) is driven to too high of an amplitude the restoring force may become nonharmonic. The force holding the molecules together are not infinitely deep parabolic potentials and at large enough amplitude the restoring force will be less than proportional to the displacement. In that case the resonance will change and the frequency will change. However, once again I only mention that because it seems vaguely related, but it really has nothing to do with the question being discussed.
  4. Nov 11, 2016 #3

    I thank you so much for the extensive explanation. Sorry but I have quite some questions regarding your post if you don't mind.

    1. Does d P(t,x) / dt show how fast the pressure changes over time? So that the larger the value of this derivative, the larger the change of P in a fixed time interval?

    2. I think my problem is that I (wrongly) think that the speed of sound would travel faster at a higher amplitude. The formula you gave shows how fast a change in pressure is created by the string vibration but not how fast that high pressure travels. Because the string gives a higher kinetic energy to the air molecules, I'd think that the high pressure column of a wave would travel faster (contains higher kinetic energy). Imagine for example me pushing air at a lower speed with my hand compared to me pushing air at a higher speed of my hand. Isn't it better to show the mathematical relationship of amplitude and the speed of sound instead?

    3. Maybe I'm asking too much here but is there a possible non-mathematical explanation for this? How it physically shows that it doesn't change the frequency?

    With "molecules", do you mean the molecules of the string itself? So that, if the string has a very high amplitude, its restoring force decreases and thus the string would have a smaller vibration velocity than you'd expect with the proportionality of amplitude and vibration velocity?
  5. Nov 12, 2016 #4
    RE 1: yes, exactly

    RE 2: until the amplitude is large enough to exceed the linear response (something akin to the elastic limit) the speed of waves does not depend on the amplitude or how fast the particles are moving as they oscillate. This is easy to show empirically. However, your fundamental question is why not. I think to best explain that you want to look in detail at the forces and motions in a ball and spring model. I'm going to cop out here and ask you to take a shot at it. I'll work on it too.

    RE 3: yes, I'm sure there is. In looking at the ball and spring model and deriving the wave velocity, I think you'll find that things cancel. A larger displacement gives larger forces. That gives larger accelerations, and the time remains the same. I'll see if I can work it out in detail.

    RE molecules, yes exactly. In the spring analogy when you get to a high enough force that the spring permanently bends (which is a materials failure and depends on the bonding at the molecular level) all bets are off.
  6. Nov 12, 2016 #5


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    The frequency of a sound does not change on the way from transmitter (loudspeaker etc.) to receiver (ear etc.) when both are stationary relative to each other. It would be impossible for your eardrum to vibrate more times in a second than the loudspeaker cone source or extra cycles would have to be introduced on the way. As the sound gets absorbed on the way, it just gets quieter.
    In a musical instrument (simplest example) there is something (e.g. a string) that vibrates with approximately Simple Harmonic Motion. That, by definition, has a waveform that's a sine wave of a particular frequency. Any practical oscillator will behave itself OK and follow this rule up to a limited amplitude, for which the 'restoring force' is directly proportional to the displacement. There are many mechanical oscillators that don't actually follow that rule but they are not always much good for making harmonious music. In fact many instruments have a different frequency during the attack phase.
    If you take a wooden ruler and hold it against a desk top and 'twang it', the frequency of the sound gradually rises as it gets quieter. That's because the mechanism is not SHM ; the part of the ruler that's in contact with the desk will only vibrate for large amplitudes and the active part of the oscillator gets shorter as the amplitude drops (= rising frequency). The frequency rises, just before the vibrations stop.
  7. Nov 12, 2016 #6
    ?? While interesting I'm having a hard time understanding how this is connected to anything else on this thread??
  8. Nov 12, 2016 #7


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    Well. Sounds do sometimes change frequency as they die down. The reason has to be the source and not the transmission medium. (I could have just written that, I guess! But never use one word when ten will suffice.)
  9. Nov 12, 2016 #8


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    the OP is specifically referring to a string being struck/plucked. Tho he didn't specifically state a guitar or other musical instrument, the info supplied infers a string fixed at each end rather than your single fixed end ruler example. It would be difficult to get a string fixed at one end only to do anything but droop like a wet noodle :wink:
    Last edited: Nov 12, 2016
  10. Nov 12, 2016 #9


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    A string is not perfect. It has an effective length, which involves the method of terminating and the radius of the string. Large displacements will shorten the length of the string and the tension may well not be proportional to the excursion of the centre. Very much like the simpler arrangement of a twanging ruler. Not SHM at all. Open or short circuited ends produce different end effects.
    PS You can hang a string vertically with only one end fixed.
  11. Nov 12, 2016 #10
    Thanks again for your very clear answers!

    I think you've hit the nail on my question when you said that my fundamental question is actually "why not". Why shouldn't the amplitude influence the frequency if it's within the proportionality of vibration velocity?

    I think, the fact that a very hard force causes a disproportionality between the amplitude and the vibration velocity which then results in a different frequency, makes me think that frequency actually DOES depend on the vibration velocity but that there is something cancelling that influence when the amplitude is within the proportionality limit with vibration velocity. The question is what exactly that cancelling factor is.

    I'll try to look up and understand it from the ball and spring model but I'm also looking forward to your explanation on that ;).
  12. Nov 12, 2016 #11
    So here it is. N balls with mass m attached to each other in the x dimension by springs. Each ball is constrained to move only in the y dimension to make the math easy. The force and therefore the acceleration of the ith mass is determined only by the relative displacement of his two neighbors which stretches the springs between them.

    Fi = k [(yi-1-yi) - (yi-yi+1)]

    Where k is the spring constant. This implies the acceleration is

    d2y / dt2 = (k /m)* [(yi-1-yi) - (yi-yi+1)]

    Look at the right side of the equation. This is the discrete version of a second derivative in x. In the continuos limit this becomes

    d2y / dt2 = (k/m)* d2y / dx2

    It is this relation between the temporal and spatial derivatives that fixes the speed of propagation. It was arrived at without reference to the amplitude or frequency and is true so long as the interaction between elements is harmonic (linearly proportional to displacement).

    To see how this enforced the speed regardless of frequency or amplitude, I write a general wave equation

    Y(x,t) = A ei (w t + C x)

    d2y / dt2 = -w2 A ei (w t + C x)

    d2y / dx2 = -C2 A ei (w t + C x)

    Relating these as noted above requires

    w2 = (k/m)*C2

    which gives

    C = sqrt(m/k) w

    Looking at the wave equation C is the wavenumber (k was taken!) and is 2 pi/ wavelength. w is the angular frequency and is 2 pi f. We know wavelength times frequency is speed so

    v = f wl = sqrt(k/m)

    The speed is independent of the wavelength or frequency so long as the restoring force is harmonic. This is because the force and acceleration are tied to the curvature which is a direct consequence of the restoring force being proportional to the displacement.
  13. Nov 14, 2016 #12
    Thanks. I'm trying to visualize here what you're saying. I assume the balls are the air molecules. Because in my head, I pictured a horizontal model where the string hits an air molecule and that air molecule hitting the second, etc. like this:
    And then using this model to calculate the speed of propagation of the wave copmression. Since a higher amplitude gives a larger vibration velocity, this would give a larger Fstring. However, I'd have to prove that this doesn't influence the speed of propagation, I assume using the mass of the air molecules and the spring constant and the resulting acceleration of the molecules.

    Your descricption however made me picture it like this.
    Here the middle air molecule got already moved the force of the string, and I guess you're trying to calculate the deceleration caused by the increased spring tension that wants to pull the middle air molecule down, which compensates for the acceleration caused by the string force?
  14. Nov 14, 2016 #13
    I started to do the longitudinal version, but it made the notation messy. For each ball I needed a displacement from initial position: delta xi and I needed the spatial dimension x for the derivative. I wanted to call them both x so it was clear they were on the same axis, but I needed to be clear the were different things. In the end I chose clarity. The math is just the same for the transverse wave and the notation is easier, so that is what I did.

    Yes, I am calculating the restoring force, and it is proportional to the second derivative. Make any straight line with your three balls, and regardless of displacement the net force on the middle one is zero. His restoring force is proportional to the difference of what is happening to his left and right, and that is the second derivative in space. The second derivative in time (force, acceleration) is tied to the second derivative in space (curvature) and the speed of the wave becomes fixed regardless of amplitude or frequency.
  15. Nov 14, 2016 #14


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    From what you are writing, I can't be sure whether you are trying to model the oscillations of the string or the transmission of sound through the air. Sound in air is strictly a longitudinal wave (molecules can't make each other move transverse to the wave).
    A string, otoh, has a transverse wave. But the fact that they are different forms of wave is not necessarily relevant.
    Which are you interested in - sound or vibrating strings? Remember, a simple vibrating string doesn't deliver much energy into the air and it is necessary to couple the string to a diaphragm / soundboard , as in a banjo or guitar, to match the string to the air and to get any significant sound out of the string. Sound propagation would usually be described in terms of a vibrating 'piston' as in a loudspeaker, where the energy is supplied over a relatively large area and produces a wave front that is plane over a big area (the simplest way to look at it).

    I have to say that you are more likely to get a proper introduction into this stuff by reading some reasonable text book or on-line equivalent. The Q and A method is not a good way to learn Science because 'the beginner' doesn't actually know the relevant questions to ask and the 'helper' doesn't always understand the question he's being asked.
  16. Nov 14, 2016 #15
    I'm interested in the speed of sound and I want to prove that, regardless of how hard you hit an air molecule, the speed of the compression propagation caused by that hit would stay the same. So that the speed of wave propagation is independent of how hard you hit the air molecules, just like a higher amplitude of a vibrating string would give a harder hit on air molecules but it wouldn't make the sound propagation any faster (thus frequency stays the same).

    Thanks, this explanation makes sense to me.

    I might be a bit vague when I ask this, but is it possible to hit an air molecule so hard, that the spring tension wouldn't be able to decrease the speed of that air molecule sufficiently before it hits the second air molecule? Thus, it hits the second air molecule in a shorter time period than normal? Is that when the proportionality of restoring force with spacial displacement breaks??
  17. Nov 14, 2016 #16
    Yes, absolutely. If you think of the positions of the molecules as a potential well, the equilibrium position is always the bottom of the well. Expanding the potential as a Taylor series, near the bottom where the derivative goes to zero the first non-zero term in the Taylor series and the only one which isn't negligible at small displacements is the x^2 term. That is to say every potential looks like a parabola at sufficiently low energy. Every potential obeys Hooke's law for sufficiently small displacements. Every system follows a set of linear differential equations for weak enough excitations.

    However no potential goes to infinity. No potential can keep following a parabolic curve at all displacements. No system will continue to have a linear response at all excitation amplitudes. If you drive any system hard enough the forces will stop looking like a parabola, and the governing differential equations will become nonlinear. All sorts of fun things happen.

    So this is the regime people are talking about when they discuss nonlinear optics, or nonlinear dynamics, or of particular interest here nonlinear acoustics. Look up "nonlinear acoustics" or "nonlinear system"
  18. Nov 14, 2016 #17


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    Didn't the post nearer the top give you a perfectly good explanation about that? It applies to gases over a massive range of pressures but falls down if you exceed the range of the simple gas laws. You get into the realms of shock wave production where the impressed displacement is faster than the speed of sound in the medium. Only when you are fully happy with the theory (not just arm waving level) that describes normal sound propagation can you expect to get into non linear behaviour.
  19. Nov 15, 2016 #18
    Which post do you mean? In any case, I wanted to have a more detailed physical explanation for it to which mike.albert99 responded with the ball-spring model.
    Like you just explained, I now indeed understand why a stronger force on an air molecule (i.e. higher amplitude) doesn't change the speed propagation of sound. Thanks!

    I'll definitely look that up!

    I have one request if you don't mind again. Out of interest, I've tried to approach my problem mathematically from a different perspective, using longitudinal waves, and I concluded that sound propagation doesn't depend on the kinetic energy that an airmolecule receives from the force of a hit (by a string for example). I was wondering if my math statements are correct and if you could verify these. Here it goes:

    The way I did it is an analogue to kinetic energy being transferred to potential energy, as if you're shooting a ball up in the air until the ball stops at a certain height.

    So, let's say that an air molecule receives kinetic energy by a hit from a string. The air molecule would then posess a kinetic energy of:

    1/2 ⋅ mair ⋅ v2 = Eair

    This energy of the air molecule is equal to the force that it exerts on the spring, that it's connected to, times the amount of spring compression Δx that the spring will undergo.

    1/2 ⋅ mair ⋅ v2 = Eair = Fair ⋅ Δx

    The question is, when is the spring compressed enough until it exerts a large enough force to stop the air molecule from moving any further? This could be considered as when the kinetic energy of the air molecule is all transferred to the spring (analogue to a ball being shot in the sky and stopping at a certain height). The restoring force of the spring must then equal the force the air molecule exerts on the spring. Since the restoring force Fspring is also equal to the spring constant k times the amount of length Δx the spring is compressed by, this means that:

    1/2 ⋅ mair ⋅ v2 = (Δx ⋅ k) ⋅ Δx = Δx2 ⋅ k

    Now, what is the time duration for the air molecule to cover that distance Δx while it gets decelerated by the spring (just like a shooting ball is getting decelerated in the air until it reaches a certain height)? This can be calculated by:

    Δx = 1/2 ⋅ a ⋅ t2

    So , the starting velocity of the air molecule is equal to the v in its kinetic energy formula, and that velocity gets decelerated by a until distance Δx is reached. Substituting this equation of Δx and putting it in the previous formula gives:

    1/2 ⋅ mair ⋅ v2 = 1/4 ⋅a2 ⋅ t4 ⋅ k

    Since v is the starting velocity and the time t is the time until that velocity is 0, this means that a ⋅ t = v and substituting that in the formula gives:

    1/2 ⋅ mair ⋅ v2 = 1/4 ⋅ v2 ⋅ t2 ⋅ k

    Here's when the v's on both sides cancel out and thus the time it needs for an air molecule to transfer its energy to the spring is independent from its velocity. So, according to this formula, the time duration is equal to:


    So the time of propagation of sound is fixed and dependent on the mass of the air molecule and the spring constant (which is probably the Young's modulus?)

    Does this math make any sense?
    Last edited: Nov 15, 2016
  20. Nov 15, 2016 #19
    Almost. The problem is that F is not constant over the delta x, so instead of just multiplying you need to integrate F over delta x. The good news is that the integral of - k x dx is -1/2 k x2 so only a factor of 2 different from what you use below. Just lose the 1/2 out front, and the equation will be correct.

    Continuing on

    Here again the acceleration is not constant so you have to integrate. I'm afraid the answer doesn't work quite as well in the rest of your argument, so that's where I have to leave your argument.

    Better to get the equation of motion from the diff eq

    F = ma = - k x


    d2x/dt2 = - k x/m

    So x(t) = C*sin(sqrt(k/m) t)

    And determine C from setting the first derivative equal to the initial velocity

    x(t) = v0 * sqrt(m/k) * sin(sqrt(k/m) t)

    You can see in the argument of the sine that the time is independent of v0.
  21. Nov 16, 2016 #20
    Ah, of course. This seriously came to my mind a minute before reading your post.

    I have one last question:
    I now understand that the full transfer of the kinetic energy of the first air molecule occurs in a fixed time duration regardless of how large the compression Δx is (when it's within the proportionality limit).

    This made me think about the second air molecule though. Because that second molecule doesn't necessarily have to wait that fixed time duration until the kinetic energy of the first air molecule is fully transferred to start moving as well. That second molecule could already start moving at a smaller certain compression length (< than Δx) of the spring (between the 1st and 2nd) when that spring reaches enough force to push that 2nd molecule, thus before the kinetic energy of the first molecule has been fully transferred (so before Δx has been reached). Let's call that certain length in which the 2nd air molecule starts moving towards the 3rd, length Δd (which is < Δx)

    A higher kinetic energy of the first air molecule would have a larger Δx compared to the Δx of a lower kinetic energy molecule, but since both Δx's are reached at a fixed time duration, that means that a higher kinetic energy air molecule would reach that length Δd in a shorter time, thus provoking the 2nd molecule to start moving earlier towards the 3rd one. So it's more when the 2nd molecule starts to move than the time duration until the 1st molecule reaches Δx. I would therefore think that all the consecutive air molecules would start moving earlier compared to a lower kinetic energy and thus the propagation of the sound would go faster.

    And yet, the propagation still happens at a fixed speed regardless of that. I might have a theory why. Because a higher kinetic energy of the 1st molecule compresses the spring more, that means that the spring has a larger restoring force, and thus pulling the 2nd molecule back stronger which compensates for its early movement.

    Not sure if this is a plausible explanation though.
    Last edited: Nov 16, 2016
  22. Nov 16, 2016 #21
    You are working towards the reason that it is the second derivative that matters. Don't think if the first molecule. Think of a molecule in the middle. He's getting pushed by the compression on one side, but he's having to push to compress the molecules on the other side. The only time he experiences any net force is when the compression on the left has a different magnitude than the compression on the right. That is the definition of the second derivative of displacement. And as you point out there is no waiting. As soon as the compression is at all unbalanced the molecule experiences a force and starts to accelerate.
  23. Nov 22, 2016 #22
    Oh ok. So my described problem is already taken into consideration by the wave equation in your post #11?

    Looking at your post #11, it starts to make sense now. What you've done is;
    1. Calculating the deceleration by using (k / m) ⋅ Δx = a, but since a changes over Δx, you've written a down as a function of dx2 / dt2 which should be the same as k / m.
    2. You then showed 2 formulae of actual wave functions that show the relationship of 1) y with the time t, and 2) y with the position x for an air molecule.
    3. This means that the second derivations of those 2 wave functions, combined in a way that it is equal to
    dx2 / dt2, must show the acceleration of that air molecule over time. After simplifying I got: w2/C2 = dx2 / dt2
    4. Since dx2 / dt2 is the acceleration just like (k / m) is, this means that w2/C2 = k / m and thus w2 = (k/m)*C2

    Is my understanding of your actions correct?

    One question though regarding steps 1) and 2); what are exactly the y's in those functions? Because, I see you've given the formula Y(x,t) that shows the relationship of Y with x and t but what info does that Y give when you fill in the wave equation with known x and t values? (do you get a speed, for example?)
  24. Nov 22, 2016 #23
    For clarity I've done a transverse wave. The Ys are the displacement from the equilibrium position. The math is exactly the same for the longitudinal wave. Take the yi's to mean the displacement of each molecule from its equilibrium position. It doesn't really matter whether that is longitudinal or transverse displacement. All that matters is that we be able to distinguish these displacements from the variable x which means position along the wave.
  25. Nov 22, 2016 #24
    Yes, I was indeed aware that the maths are the same for transverse and longitudinal waves. That's why I used dx instead of dy in my previous post. I got confused because there's also an x variable in your formulas along with the displacement y and that made me wonder which one is actually the displacement. I thought you don't need both to know the function of acceleration. Isn't the acceleration function dx2/dt2 the same as the acceleration function dy2/dt2 since they both give how fast the air molecule is changing in space?

    Sorry if I'm missing something obvious here.
  26. Nov 22, 2016 #25
    The subtlety comes from the two different meanings of the coordinate x. to keep from carrying around a lot of deltas you can give each and every molecule it's own coordinate system where xi = 0 for every molecule at equilibrium. However when you write the wave equation or take a derivative to see how things vary along the length you need a global x coordinate. Rather than try to explain that I called the individual displacements "y" and said it's transverse.
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