Solving Dimensional Analysis: Find V in Terms of P & Density

In summary: Thank youIn summary, the speed of sound waves in air, v, can be expressed in terms of atmospheric pressure, P, and density by using the method of dimensional analysis. This involves balancing the units (dimensions) on both sides of the equation. The final expression for v is √(P/ρ), and the method can be applied to other cases where the dimensions of the variables do not match.
  • #1
agentnerdo
13
0

Homework Statement



The speed v of sound waves in air depends on the atmospheric presure P and density. By using the method of dimensional analysis, find an expression for v in terms of P and density.



Homework Equations



What I have so far;

V= density x P



I found a textbook which said that M can be ignored, as such...I end up getting

[L][T^-2]=[M^x+y] [T^-2x]

M---> 0 = x +y
y = x

L---> 1= -1x-3y

T---> T-1= -1
T = 0


It must be obvious by now that I am missing a key concept here...any help would be appricated!
 
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  • #2
What units are you using for v?
What units are you using for P?
What units are you using for ρ?
 
  • #3
Hey lwando,

V= [L^3]
Pressure= F/A [M][L^-1][T-^2]
Density= [M][L^-3]
 
  • #4
try:

V (velocity) = [L][T^-1]
 
  • #5
Key concept: put V on the left hand side of the equation, and put different arrangements of P and ρ on the right hand side and, based on what you know about the units (dimensions), see if the units/dimensions are equivalent. It becomes a puzzle-solving problem.
 
  • #6
I think I have a bit of a hang on this now...

[L][T^-1] = [M][L^-1][T-^2] [M][L^-3]

We ignore [M] as it is not on the left side thus;

[L][T^-1] = [L^-1x][T-^2x] [L^-3x]


-1 = -2x -------> x=1/2

1= -1x - 3y sub x= 1/2

y= -2/3


V= sqt pressure / density ^2/3


does that seem like the proper way to do? This might be a very simplistic q...but how do I take care of that 2/3?

Thanks allot! This is a second year phys course everyone seems to be flying through except for the older guy that came with a biology degree (me!)
 
  • #7
You are sort of on the right track . A coulple of points:

1) Cannot simply "ignore" M on the other side. M has to go away naturally (hint: M/M or using our notation, [M][M^-1] needs to appear somehow on the RHS. Noting your earlier post:

Pressure= F/A [M][L^-1][T-^2]
Density= [M][L^-3]

it would seem that some form of division needs to happen between P and ρ to make M go away.

2) You came up with V= sqt pressure / density ^2/3 as a result of your process that I am struggling to understand (also "sqrt" of just pressure? or the whole RHS? Not clear.) Still, the units (dimensions) should balance on both sides. You should check this result. I don't think they balance.

Have another go at it.
 
  • #8
One more point: I would start with an hypothesis expression involving V, P, and ρ, then substitute the units in place of the variables, simplify and see if the units balance. If not, tweak your V, P, ρ expression and repeat the process. Insight will eventually occurr. You started with:

[L][T^-1] = [M][L^-1][T-^2] [M][L^-3]

which is V = Pρ

Backwards from how I suggest (but who am I to force a method)

Clearly they don't balance because of M on RHS, so V = Pρ can't be the expression.
 
  • #9
I think I got it!

Pressure= F/A [M^x][L^-1x][T-^2x]
Density= [M^y][L^-3y]

For M M^y plus M^x= 0

For T -1 = -2x
x= 1/2

For L 1= -1x - 3y sub in x=1/2
x= 1/2

As such,

Velocity = sqt (pressure/density)

I hope this it? yes?

I did not ignore or 'take care' of M, I just applied the rules and realized x=-y


Could you verify if this method of taking care of M would also work in cases inwhich x does not equal y?
 
  • #10
That's correct.

As far as verifying the method, I understand what you did and it seems reasonable. In this case, x does not equal y. I don't see why it wouldn't work for other cases.
 
  • #11
You been a great help lewando.
 

FAQ: Solving Dimensional Analysis: Find V in Terms of P & Density

How do you solve for V in terms of P and density using dimensional analysis?

To solve for V in terms of P and density using dimensional analysis, you need to set up a conversion factor that relates volume to pressure and density. This can be done by using the formula V = P/(density), where P is pressure and density is the mass per unit volume.

What are the units for pressure and density in dimensional analysis?

The units for pressure are typically measured in pascals (Pa) or newtons per square meter (N/m^2). The units for density are typically measured in kilograms per cubic meter (kg/m^3) or grams per cubic centimeter (g/cm^3).

Can dimensional analysis be used to solve for other variables besides volume?

Yes, dimensional analysis can be used to solve for other variables such as mass, length, time, and temperature. As long as the equation has consistent units, dimensional analysis can be used to solve for any variable.

What are some common mistakes to avoid when using dimensional analysis to solve for V in terms of P and density?

One common mistake to avoid is not paying attention to the units and making sure they cancel out properly in the conversion factor. Another mistake is not using the correct values for P and density, which can result in an incorrect answer.

Can dimensional analysis be used in real-world applications?

Yes, dimensional analysis is commonly used in real-world applications, especially in the fields of physics, chemistry, and engineering. It is a useful tool for converting between different units and solving complex equations involving multiple variables.

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