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

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The discussion focuses on determining the physical dimensions and SI units for the constants α and β in the equation x=α+(2/3)βt^(3/2). It establishes that α has dimensions of length (L) and units of meters (m). For β, the analysis reveals that its dimensions must be L·T^(-3/2), leading to the SI unit of m·s^(-3/2). The conversation highlights the challenge of naming units that combine basic dimensions, emphasizing that such combinations are valid even if they lack specific names. Ultimately, the discussion clarifies the dimensional analysis needed for understanding the relationship between position and time in this context.
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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.
 
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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.
 
Try to figure out what must be the dimensions of β so that the dimensions of βt^{3/2} ends up as L.
 
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?
 
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 \mathrm{m}\,\mathrm{s}^{-3/2}.

The formula y = x^a t^b makes dimensional sense for any real a and b, and if x and t are measured in metres and seconds respectively then y will be measured in units of \mathrm{m}^a\,\mathrm{s}^b.
 
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.
 
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.
 
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
 

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