Waves & Sound: Combine Pressure & Density to Produce Dimensions of Speed

Homework Statement

Two important quantities associated with air are the pressure and density. How can the pressure and density be combined to produce dimensions of speed? Look up the pressure and density of air at sea level and use your equation to estimate the speed of sound. Is your answer reasonable?

Homework Equations

pressure (P)= force/area
area of sphere= 4$$\Pi$$d$$^{2}$$
density (p)= mass/volume

The Attempt at a Solution

somehow i'm assuming you're supposed to use the equation for the speed of a wave that includes force, which is V=$$\sqrt{T/\mu}$$
but T=tension of the string. Since sound is not a wave on a string.. i am thoroughly confused.

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cepheid
Staff Emeritus
Gold Member
Welcome to PF sammmy!

I think the first thing would be to write out the fundamental dimensions of each quantity:

I'll use [L] to mean "length", [M] for mass, and [T] for time:

$$[P] = \frac{[F]}{[A]} = \left([M] \cdot \frac{[L]}{[T]^2}\right)\left(\frac{1}{[L]^2}\right)$$

in the last part on the right hand side, the quantity in the first set of parentheses is force (mass*acceleration), and the quantity in the second set of parentheses is 1/area.

So, pressure reduces to dimensions of:

$$[P] = \frac{[M]}{[L][T]^2}$$

similarly for density:

$$[\rho] = \frac{[M]}{[V]} = \frac{[M]}{[L]^3}$$

Now, does anything jump out at you? Hint: you want to get rid of [M] entirely.

Well, since dimensions of speed are meters/second- It seems as if the density equation would need to be flipped to 1/$$\rho$$ which is equal to L$$^{3}$$/M so the masses cancel out

but I'm not sure how you can equate pressure and density.. ?

cepheid
Staff Emeritus
Gold Member
Well, since dimensions of speed are meters/second- It seems as if the density equation would need to be flipped to 1/$$\rho$$ which is equal to L$$^{3}$$/M so the masses cancel out

but I'm not sure how you can equate pressure and density.. ?
You can't "equate" them. But you can combine them together algebraically in an expression, in such a way that you are left with something that has dimensions of [L]/[T] (speed).

$$\sqrt{P/\rho}$$ so that $$\sqrt{(M/LT^{2})/(M/L^{3})}$$ which equals L/T

cepheid
Staff Emeritus
Gold Member
$$\sqrt{P/\rho}$$ so that $$\sqrt{(M/LT^{2})/(M/L^{3})}$$ which equals L/T
Yes. Good work.

EDIT: Sound is indeed a wave, and you could have derived this result from the appropriate wave equation. I bring this up because your attempt in your original post to make an analogy with the wave speed for waves propagating in a string was a good insight. P is sort of analogous to T in its role of "forcing" or driving the oscillations (I think). The mass density appears in both equations and is a sort of "inertial" factor.

yes. Good work.

Edit: Sound is indeed a wave, and you could have derived this result from the appropriate wave equation. I bring this up because your attempt in your original post to make an analogy with the wave speed for waves propagating in a string was a good insight. P is sort of analogous to t in its role of "forcing" or driving the oscillations (i think). The mass density appears in both equations and is a sort of "inertial" factor.
thank you!!!