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.

 

[Gaussian97]

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.

 

[Orodruin]

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

 

[Gaussian97]

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?

 

[Orodruin]

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

 

[Gaussian97]

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?

 

[Gaussian97]

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.

 

[Orodruin]

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?

 

[Gaussian97]

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

 

[Gaussian97]

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

 

[Orodruin]

And the solution to this linear system of equations is?

 

[Gaussian97]

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?

 

[Gaussian97]

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?

 

[Orodruin]

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##.

 

[berkeman]

The OP @kaydis has re-posted the algebra portion of this question in a separate thread. Please see her solution there. I'll lock this thread since the dimensional analysis part is done.

https://www.physicsforums.com/threa...knowns-given-3-equations.975151/#post-6210360