# Dimensional analysis (Speed of sound)

kaydis
Homework Statement:
Use dimensional analysis to predict how the speed of sound (u) in a gas may be influenced by the gas pressure (p), gas density (r) and acceleration due to gravity (g).
Relevant Equations:
u∝p^a r^b g^c

p = ML^-1T^-2
r= ML^-3
g=LT^-2
I've been going round in circle and am stuggling as maths is not my strong suit. Any ideas?
Up to the point where i equate the powers and then confuse myself, any help would be greatly appreciated.

Homework Helper
First of all, what are the dimensions of speed, pressure, density and acceleration?
If you don't know that you will not be able to solve a problem with dimensional analysis.

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Please show your efforts so far explicitly, ie, show us exactly what you have done.

kaydis
First of all, what are the dimensions of speed, pressure, density and acceleration?
If you don't know that you will not be able to solve a problem with dimensional analysis.
p = ML^-1T^-2
r= ML^-3
g=LT^-2

kaydis
Please show your efforts so far explicitly, ie, show us exactly what you have done.
Please show your efforts so far explicitly, ie, show us exactly what you have done.
u = p^a r^b g^c
u = (ML^-1T^-2)^a (ML^-3)^b (LT^-2)^c
u = M^a+b L^-1a-3b-c T^-2a-2c
im not convinced this is correct though

Homework Helper
u = p^a r^b g^c
u = (ML^-1T^-2)^a (ML^-3)^b (LT^-2)^c
u = M^a+b L^-1a-3b-c T^-2a-2c
im not convinced this is correct though
Ok, be careful with the ##L## exponent. But you didn't answer my question, what are the dimensions of speed?

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u = p^a r^b g^c
u = (ML^-1T^-2)^a (ML^-3)^b (LT^-2)^c
First of all, note that the second expression here should be ##[u ]##, i.e., the dimension of ##u##, not ##u## itself. Second, what is ####? Also, be careful about the exponents of ##\mathsf L##.

kaydis
Ok, be careful with the ##L## exponent. But you didn't answer my question, what are the dimensions of speed?
okay, and i haven't been given that in the question paper but I am guessing it would be LT^-1

Homework Helper
okay, and i haven't been given that in the question paper but I am guessing it would be LT^-1
Exact, so now you should have the same dimensions at each side of the =, right?. What does this tell you?

kaydis
First of all, note that the second expression here should be ##[u ]##, i.e., the dimension of ##u##, not ##u## itself. Second, what is ####? Also, be careful about the exponents of ##\mathsf L##.
okay, i'll change that now. and what do you mean about the exponent of L? I'm sorry to sound stupid but I haven't ever done dimensional analsis before

kaydis
Exact, so now you should have the same dimensions at each side of the =, right?. What does this tell you?
Yes, they cancel each other out so the gas isn't influenced by the p,r or g?

Homework Helper
okay, i'll change that now. and what do you mean about the exponent of L? I'm sorry to sound stupid but I haven't ever done dimensional analsis before
You have something like ##L^n##, the number ##n## is the exponent of L

Yes, they cancel each other out so the gas isn't influenced by the p,r or g?

No, why they should cancel each other?

kaydis
You have something like ##L^n##, the number ##n## is the exponent of L

No, why they should cancel each other?
I don't know, I'm don't understand this at all. I haven't done any maths qualifications other than GCSE's and i have been thrown in the deep end.

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The physical dimensions of both sides of an equation must match. For example, a length cannot be equal to a time or an acceleration. So the dimensions of the LHS is ##\mathsf{L/T}## and the dimension of the RHS is ##\mathsf M^{a+b} \mathsf L^{c-a-3b} \mathsf T^{-2(a+c)}##. For these two to be equal, the exponents of each independent physical dimension (##\mathsf{MLT}##) must be equal on both sides. What does this tell you?

Homework Helper
u = p^a r^b g^c
u = (ML^-1T^-2)^a (ML^-3)^b (LT^-2)^c
u = M^a+b L^-1a-3b-c T^-2a-2c
im not convinced this is correct though

Until now you have found that the dimensions of speed must be $$\left[u\right]=M^{a+b}L^{-a-3b-c}T^{-2a-2c}.$$ (Although we have told you that you have an error in the exponent of ##L##, so you should fix it first).

Then you have said that the dimensions of speed are ##LT^{-1}##, or, equivalently ##M^0L^1T^{-1}##. What can you do with this to find the numbers ##a, b## and ##c##?

kaydis
okay,
The physical dimensions of both sides of an equation must match. For example, a length cannot be equal to a time or an acceleration. So the dimensions of the LHS is ##\mathsf{L/T}## and the dimension of the RHS is ##\mathsf M^{a+b} \mathsf L^{c-a-3b} \mathsf T^{-2(a+c)}##. For these two to be equal, the exponents of each independent physical dimension (##\mathsf{MLT}##) must be equal on both sides. What does this tell you?
that the dimension of M must be 0 as its not present on the LHS?

kaydis
Until now you have found that the dimensions of speed must be $$\left[u\right]=M^{a+b}L^{-a-3b-c}T^{-2a-2c}.$$ (Although we have told you that you have an error in the exponent of ##L##, so you should fix it first).

Then you have said that the dimensions of speed are ##LT^{-1}##, or, equivalently ##M^0L^1T^{-1}##. What can you do with this to find the numbers ##a, b## and ##c##?
equate the powers?
i.e. a+b = 0

Homework Helper
equate the powers?
i.e. a+b = 0
Exact, this is one equation, now you can find another two.

kaydis
-2a-2c = -1
-a-3b+c= 1

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And the solution to this linear system of equations is?

Homework Helper
Exact!, Now you only have 3 equation that depends on 3 unknowns, you just have to solve it!

kaydis
Exact!, Now you only have 3 equation that depends on 3 unknowns, you just have to solve it!
Yes! awesome, what is the easiest way to work this out? or is it just trial and error?

Homework Helper
Well, it's a linear system of equations so, if you have studied matrices you can solve it easily using Gaussian elimination.

If not you can do the following: use the first equation to find ##a## as a function of ##b## (it's trivial). Then you can substitute in the other two equations so you will have a system of two equations with two variables.
Then you can repeat the same, write ##b## as a function of ##c## and substitute, you will get finally an equation that depends only on c. Solve it and then compute ##b## and ##a##.

kaydis
So I've tried and I'm stuck...

a+b=0
-2a-2c = -1
-a-3b+c= 1

1 1 0 ¦ 0
-2 0 -2 ¦ -1 (R2 - 2*R1)
-1 -3 1 ¦ 1 (R3 + R1)

1 1 0 ¦ 0
0 2 2 ¦ -1 (R2 - 1/2)
0 4 1 ¦ 1

1 1 0 ¦ 0
0 1 1 ¦ -1/2
0 4 1 ¦ 1 (R3 - 4R2)

1 1 0 ¦ 0
0 1 1 ¦ -1/2
0 0 -3 ¦ 3 (-1/3*R3)

1 1 0 ¦ 0
0 1 1 ¦ -1/2
0 0 1 ¦ -1

so this means that:
a= -1/2
b= 1/2
c= -1

This must be wrong because when I substitute it back into the original equations it doesn't work?

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1 1 0 ¦ 0
-2 0 -2 ¦ -1 (R2 - 2*R1)
-1 -3 1 ¦ 1 (R3 + R1)

1 1 0 ¦ 0
0 2 2 ¦ -1 (R2 - 1/2)
0 4 1 ¦ 1
In the first expression, R2 - 2*R1 is (-2 0 -2) - 2*(1 1 0) = (-4 -2 -2), not (0 2 2). Also R3+R1 is (-1 -3 1) + (1 1 0) = (0 -2 1), not (0 4 1). I have not checked the rest, but it seems you need to work on your arithmetic.

Honestly, doing this for a 3x3 matrix is overkill anyway. Just solve one equation at a time and insert the solution into the others. As already stated in #23, the first equation is trivially ##a = -b##.