Relationship between the frequency and amplitude of wave

I wonder is there any relationship between frequency and amplitude of wave. Generally, most of the people say that frequency and amplitude are independent to each other. But in this case, where it is a sound wave, the intensity, $I$ can be expressed in this form: $$I=\frac12\rho v\omega^2x^2$$

Where:

• $I$ is intensity of the sound wave
• $\rho$ is the density of the medium
• $v$ is the speed of sound being transmitted
• $\omega$ is the angular frequency
• $x$ is the amplitude of sound

So, judging from this formula, I suppose that the frequency should be inversely proportional to the amplitude, but why it is not like what I think? I get the formula from Halliday and Resnick physics book. • I'm voting to close this question as off-topic because it's based on invalid assumptions – Carl Witthoft Jul 16 '15 at 12:35
• It is not an assumption and i cant see how it is off topic – Jet Herng Chion Jul 16 '15 at 12:38
• I wonder if there is any relationship between the amount of water in my bathtub and the temperature of my coffee. Generally, most people say these have nothing to do with each other. But if I define the variable Y to be the product of the amount of water in my bathtub with the temperature of my coffee, then I have $Y=QT$, where $Q$ is the amount of water in my bathtub and $T$ is the temperature of my coffee. So judging from this formula, I think that the temperature of my coffee should be inversely proportional to the amount of water in my bathtub. – WillO Jul 16 '15 at 13:05
• Please take note of @WillO's comment. – DanielSank Jul 16 '15 at 14:27

Let me say what others are trying to say, hopefully in a clearer fashion:

Just because you can relate two variables in an equation does not mean that they are dependant. In this case, you have to constrain intensity $I$ in order to get the relationship. At that point, it is not a general relationship, but only true when $I$ is constrained.

An example that might be easier to see intuitively would be: $$KE=\frac{1}{2}mv^2$$ If you constrain kinetic energy you can get a relationship between mass and velocity. For example: $$m=2\frac{KE}{v^2}$$ But intuitively, you know that mass and velocity are independent of one another. Why would changing the mass of an object inherently change the velocity? But, if the kinetic energy is held constant, then it would force a relationship between them. A relationship that is not generally meaningful.

So, to bring this back to your case, $x$ sound amplitude and $w$ angular frequency are independent of each other, but you can force a relationship between them by constraining $I$, but it is not a meaningful or general relationship.

I found a good answer elsewhere that explains it much better than I did here. I would recommend checking it out.

I don't know where you got this formula but I think it's wrong. See https://en.wikipedia.org/wiki/Sound_intensity

Sound Intensity is given by $I = p \cdot v$ where

p is the sound pressure;
v is the particle velocity.

Sufficiently far away from any source or diffraction object the relation ship between particle velocity and pressure is $v=\rho \cdot c \cdot p$ where $\rho$ is the density and c the speed of sound. So we simply get $$I = \rho \cdot c \cdot p^{2}$$

For any given pressure, the intensity is independent of frequency.

• I got the formula from halliday and resnick physics book.... I have attached a picture to my question, please have a look – Jet Herng Chion Jul 16 '15 at 12:37

You have the intensity:$$I=\frac{1}{2}\rho\omega^2s_m^2$$ which is a relationship between displacement amplitude $s_m$, angular frequency $\omega$ and intensity $I$. What this is telling you is how the intensity is related to the angular frequency and displacement amplitude.

There is nothing remarkable about this, all it is saying is the intensity is proportional to the square of the displacement amplitude and of the square of the angular frequency. But if you do constrain the intensity then it does give you a relation between angular frequency and displacement amplitude, but this is rather artificial.

protected by ACuriousMind♦Jun 9 '17 at 9:29

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