Dimension Analysis: α,β in x=α+(2/3)βt^(3/2)

In summary, we are discussing the physical dimensions and corresponding SI units of the constants α and β in the expression x=α+(2/3)βt^(3/2). We determine that the units of β are meters per root second cubed and that this unit is valid for any real value of a and b in the formula y = x^a t^b. We also note that while there is no specific SI unit for LT^(-3/2), it is a valid unit for a quantity.
  • #1

Homework Statement



In the following expression x is a position and t represents time. What are the physical dimensions of each of the constants α and β? Also, for each what are the corresponding SI units?

Homework Equations



x=α+(2/3)βt^(3/2)

The Attempt at a Solution



I know that since x is a position, it must have the dimension of L and the unit of m. That makes the right have a dimension of L as well. That said [α]=L, but the β is throwing me off because of the t^(3/2). I can't for the life of me figure out how to manipulate that exponent to end of with a dimension of L or L/T or L/T^2.
 
Physics news on Phys.org
  • #2
Basically, the strategy for finding the units of β is to make sure all other units cancel besides one power of [x].

Since the factor of 2/3 doesn't matter (it is unitless or "dimensionless"), and you already have figured out α, we can write:

[x]=[βt3/2] = [β] [t]3/2
where the square brackets mean "the units of" whatever is enclosed. One way to solve for the units of [β] is to use algebra (really some sort of pseudo algebra, but everything works out just fine) like this:

[x]/[t]3/2 = [β]
or
[β]=[x][t]-3/2

So this means the units of β are units of length times time to the power -3/2, or meters per root second cubed.

If you plug this in you can verify the result:
[β t3/2] = [β] [t]3/2 = ([x][t]-3/2) [t]3/2 = [x]
as desired.
 
  • #3
Try to figure out what must be the dimensions of β so that the dimensions of βt[itex]^{3/2}[/itex] ends up as L.
 
  • #4
I tried this approach and it made sense algebraically. However, I don't know of an SI unit associated with [x][t]^-3/2. Am I missing something?
 
  • #5
studentofphy said:
I tried this approach and it made sense algebraically. However, I don't know of an SI unit associated with [x][t]^-3/2. Am I missing something?

Yes; the SI unit is [itex]\mathrm{m}\,\mathrm{s}^{-3/2}[/itex].

The formula [itex]y = x^a t^b[/itex] makes dimensional sense for any real [itex]a[/itex] and [itex]b[/itex], and if [itex]x[/itex] and [itex]t[/itex] are measured in metres and seconds respectively then [itex]y[/itex] will be measured in units of [itex]\mathrm{m}^a\,\mathrm{s}^b[/itex].
 
  • #6
It's a tragedy that there isn't an SI unit with dimensions of LT^(-3/2), but it's still OK if a quantity has units which are a mixture of only basic units. There simply aren't enough people to provide names for every combination of basic units.
 
  • #7
Thank you so much. Being a new physics students I made a rookie mistake in assuming the only dimensions were the ones listed on a chart with quantities.
 
  • #8
I mean, I already told you the name of the unit:
meters per root second cubed
Is it a problem that the units of acceleration are meters per second squared? That is an extremely commonly used one, as in g=9.8 m/s2
 

What is dimension analysis?

Dimension analysis is a mathematical method used to analyze the relationships between physical quantities. It involves using the dimensions of the quantities involved in an equation to determine the relationship between them.

What does α represent in the equation x=α+(2/3)βt^(3/2)?

α represents the initial position or starting point in the equation. It is the value of x when t=0.

What does β represent in the equation x=α+(2/3)βt^(3/2)?

β represents the coefficient of the time variable in the equation. It determines the rate of change of x with respect to time.

What are the units of α and β in the equation x=α+(2/3)βt^(3/2)?

The units of α are the same as the units of x, and the units of β are the units of x divided by the units of t^(3/2). For example, if x is measured in meters and t is measured in seconds, the units of β would be meters per second cubed.

How can dimension analysis be applied to solve problems?

Dimension analysis can be used to check the correctness of equations, to derive new equations, and to convert between different unit systems. It can also be used to identify relationships between different physical quantities and to determine the units of unknown quantities.

Suggested for: Dimension Analysis: α,β in x=α+(2/3)βt^(3/2)

Back
Top