How can I solve dimensional analysis problems involving exponents and constants?

In summary, dimensional analysis involves finding the relationship between different physical quantities using their units. To find the exponents of each quantity, set up a dimensional equation and simplify the exponents on each side. For the first problem, T^(2y+1)=M^(x+y), we can solve for y and then plug in the value to solve for x. In the second problem, v=(CB^x)(p^y), we can simplify the exponents and compare them on both sides to find the relationship between x and y.
  • #1
nilesthebrave
27
0
Hi, sorry for asking this but my brain still seems to be on lockdown from the summer. I have a pretty good idea of how dimensional analysis works and only seem to be having issues on one type of problem currently. something like:

t=(Cm^x)(k^y)

where:
t-oscillations of mass
m-mass
spring constant-k(force/length)
C-dimensionless constant

to find x and y.

T=(M^x)(Force/L)^y=(M^x)(ML/L(T^2))^y
T=(M^x)(M/T^2)^y
T=(M^x)(M^y/T^2y)
T=(M^(x+y))(M^y/T^2y)
T(T^2y)=M^(x+y)
T^(2y+1)=M^(x+y)

Then you get

2y+1=0
x+y=0

solving for y at top equation:
y=-1/2
then plugging in for second equation you get
x=1/2

So I have that one.

Now where I'm having hangups is on one like say:

v=(CB^x)(p^y)

B-bulk modulus
p-density
c-dimensionless constant
v-velocity
Find x and y

So I know I start with:

L/T=(M/LT^2)^x(M/L^3)^y

But honestly, I get stuck at this point. I can't figure out how to get things to cancel or how to make things simplify down easier. Do I distribute the exponent? Do I multiply the left hand by a reciprocal of one of those? Honestly, I don't get how to do one like this even though I fully get the first one which is fairly similar. Any hints to nudge me in the right direction to solve this, its been bugging me for awhile now.
 
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  • #2
Simplify the exponents and compare them on both sides, just like you did in the last question (for example the exponent of L must be equal on both sides.)
 

What is simple dimensional analysis?

Simple dimensional analysis is a mathematical technique used to convert units of measurement from one system to another. It involves using the relationships between different units to determine the appropriate conversion factor.

Why is simple dimensional analysis important in science?

In science, it is crucial to use and understand units of measurement correctly. Simple dimensional analysis allows for consistency and accuracy when working with different units, making it an essential tool for scientists.

How do you perform simple dimensional analysis?

To perform simple dimensional analysis, you must first identify the units you are converting from and to. Then, use the conversion factor, which is the ratio of the two units, to convert from one unit to another. Finally, multiply the value by the conversion factor to get the desired unit.

What are the benefits of using simple dimensional analysis?

Simple dimensional analysis helps avoid errors and inconsistencies when working with different units. It also allows for easy comparison and understanding of measurements across different systems of measurement.

Can simple dimensional analysis be used for complex calculations?

Yes, simple dimensional analysis can be used for complex calculations as long as the units are consistent and appropriate conversion factors are used. It is a versatile and useful tool for a wide range of scientific calculations.

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