Dimensional analysis (Speed of sound)

In summary: That the dimension of M must be 0 as its not present on the...what does this have to do with anything?
  • #1
kaydis
23
1
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.
 
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  • #2
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
Please show your efforts so far explicitly, ie, show us exactly what you have done.
 
  • #4
Gaussian97 said:
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
Orodruin said:
Please show your efforts so far explicitly, ie, show us exactly what you have done.
Orodruin said:
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
kaydis said:
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
kaydis said:
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
Gaussian97 said:
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
 
  • #9
kaydis said:
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?
 
  • #10
Orodruin said:
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
Gaussian97 said:
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
kaydis said:
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

kaydis said:
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
Gaussian97 said:
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.
 
  • #14
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
kaydis said:
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
okay,
Orodruin said:
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
Gaussian97 said:
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
kaydis said:
equate the powers?
i.e. a+b = 0
Exact, this is one equation, now you can find another two.
 
  • #19
-2a-2c = -1
-a-3b+c= 1
 
  • #20
And the solution to this linear system of equations is?
 
  • #21
Exact!, Now you only have 3 equation that depends on 3 unknowns, you just have to solve it!
 
  • #22
Gaussian97 said:
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
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
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?
 
  • #25
kaydis said:
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##.
 

Related to Dimensional analysis (Speed of sound)

1. What is dimensional analysis?

Dimensional analysis is a mathematical tool used to convert between different units of measurement. It involves using the dimensions (such as length, time, mass, etc.) of physical quantities to create equations that can be used to convert between units.

2. How is dimensional analysis used to calculate the speed of sound?

By using dimensional analysis, we can create an equation that relates the speed of sound to other physical quantities, such as density and elasticity. By plugging in the correct units for these quantities, we can calculate the speed of sound in various mediums.

3. What are the units of measurement for the speed of sound?

The most commonly used unit for the speed of sound is meters per second (m/s). However, it can also be expressed in other units such as feet per second, kilometers per hour, and miles per hour.

4. How does the speed of sound vary in different mediums?

The speed of sound varies depending on the medium it is traveling through. In general, it travels faster in denser and more elastic mediums, such as solids, and slower in less dense and less elastic mediums, such as gases. It also varies with temperature and pressure.

5. What is the significance of the speed of sound in scientific research?

The speed of sound is an important measurement in various fields of science, including physics, acoustics, and meteorology. It is used to study the properties of different materials, determine the composition of the atmosphere, and predict weather patterns. It also has practical applications, such as in the development of supersonic aircraft and medical imaging techniques.

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