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Dimensional analysis (Speed of sound)

  • Thread starter kaydis
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  • #1
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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.
 

Answers and Replies

  • #2
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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.
 
  • #3
Orodruin
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Please show your efforts so far explicitly, ie, show us exactly what you have done.
 
  • #4
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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
 
  • #5
23
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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
 
  • #6
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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?
 
  • #7
Orodruin
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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##.
 
  • #8
23
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Ok, be careful with the ##L## exponent. But you didn't answer my question, what are the dimensions of speed?
okay, and i havent been given that in the question paper but im guessing it would be LT^-1
 
  • #9
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okay, and i havent been given that in the question paper but im 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?
 
  • #10
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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
 
  • #11
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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?
 
  • #12
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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?
 
  • #13
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You have something like ##L^n##, the number ##n## is the exponent of L



No, why they should cancel each other?
I dont know, I'm dont understand this at all. I havent done any maths qualifications other than GCSE's and i have been thrown in the deep end.
 
  • #14
Orodruin
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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?
 
  • #15
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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##?
 
  • #16
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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?
 
  • #17
23
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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
 
  • #18
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equate the powers?
i.e. a+b = 0
Exact, this is one equation, now you can find another two.
 
  • #19
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-2a-2c = -1
-a-3b+c= 1
 
  • #20
Orodruin
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And the solution to this linear system of equations is?
 
  • #21
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Exact!, Now you only have 3 equation that depends on 3 unknowns, you just have to solve it!
 
  • #22
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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?
 
  • #23
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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##.
 
  • #24
23
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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 equasions it doesn't work?
 
  • #25
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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##.
 

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