can anybody explain dimensional analysis?

Question

can anybody explain dimensional analysis?

could u also answer this question?

The position of a particle moving under uniform acceleration is some function of time and the acceleration. Suppose we write this position s = k * a^m * t^n, where k is a dimensionless constant. Show by dimensional analysis that this expression is satified if m = 1, and n = 2. Can this analysis give the value of k?

thanks

Answer 1

dimensional analysis means using the dimensions in an equation to ensure that the correct operations are being performed.
for instance, the units of each are:
s = position, IE distance (m)
a = acceleration (m/s^2)
t = time (s)
k = (no dimension, negligible)

so substituting (using dimensional analysis) gives:

m = (m/s^2)^M * (s)^N

m = (m/s^2)^1 * s^2

m = m/s^28s^2

m = m

This shows that M=1 and N=2 is correct. You can gain no information about k. Note that k was assumed to be dimensionless. If this assumption were wrong, then the information about M and N would possibly be wrong as well.

Answer 2

Dimensional analysis is basically the process of following the units of your quantities as you go through your equation. It is most useful in conversion operations - e.g. if you want to turn 5 seconds into hours you do ( 5s/1 ) * ( 1min/60s ) * ( 1hr/60min ). Essentially, you want all your units in these divisions and multiplications to cancel out, except for the one unit you want to end up with (in this case, hours).

In your second question, you need to look at the units. I'll just assume some of the more common ones as an example.
m = meters, s = seconds

s = k * a^m * t^n
m = k * ( (m/s^2)^1 ) * ( s^2 )

Combining the stuff on the right side:
m = (k * m * s^2) / (s^2)

You can see that the seconds cancel out in the division, so you are left with:
m = k * m

Solving for k, you end up with 1. Since there are no units here, you can see that k must be a dimensionless constant!

Hope that made sense!

Answer 3

Dimensional analysis is basically just keeping track of your units - for instance, if you are trying to compute a volume, your answer had darn well better have units of length cubed, or else you screwed up your calculation. What needs to be shown here is that if you have acceleration multiplied by the square of a time, that you get units of length. We can take, for instance, acceleration in m/s², and time in s, and find that m/s²^1*s²=m, which is a length, and thus this expression is satisfied. For the second part of the question: changing the value of a dimensionless constant will not change the units at all, and thus dimensional analysis can say nothing aboutthe value of any such constant. As such, dimensional analysis cannot give the value of k (it happens to be 1/2, just in case you were wondering).

Answer 4

In order for your equation to be valid, it must have the same units on both sides. Since k has no units, you have a * t^2. If you assume that a is in m/s and t^2 is in s^2. This reduces to m/s, so s must be in m/s. Since k is dimensionless, you can't find its value because it was not included in the analysis, and dimensional analysis only looks at units, not actual values.

Answer 5

You need some units. The idea is that running the units through in the equation provides a double check to make sure you set up the equation properly.

In the second question, the units for the constants would have to be known. I would think 's' must stand for position, and be measured in meters; 'a' would be acceleration with m/sec^2; and 't' would be seconds. m and n must be dimensionless, since they're exponents. That would give me:

m/sec^2 * sec^2

The seconds cancel out, leaving meters, which is compatible with 's'. That means if the a^m*t^n were moved under 's' as a denominator, the meters would cancel out, leaving a dimensionless 'k'.

Answer 6

The guiding principle of diimensional analysis is that physically meaningful formulas should give quantities with physically meaningful units.

The problem you give here is a good example. We measure acceleration in units of [L]/[T]^2 (length over time squared), time in units of [T] and position in units of length [L]. The unit k has no dimensions.

Now apply the formula to the abstract units given here:

s = k * a^m * t^n

[L] = [1] * ([L]/[T]^2)^m * [T]*n
[L] = [L]^m * [T]^(-2m) * [T] *n

The exponents of [L] left and right should match:
1 = m
0 = -2m + n = -2 + n ==> n = 2

This analysis cannot give the value of k.

Answer 7

Not sure, but I have an absurdly beautifully example dimensional reduction.

Consider miles per gallon. What that is in its most basic form is liner distance per unit of volume. Reduced, that is 1 dimension per 3 dimensions or

1d / 3d

Now we have a fraction. By simple algebra a dimension can be removed from each side of the fraction by dividing each side by 1 dimension. The result is

1 / 2d

One dimension is distance, 2 dimensions is area, 3 dimensions is volume. So now we have 1 / area.

As a result, fuel economy could be represented in terms of per-area. So with some arithmetic, you could convert miles per gallons into "per Square feet" or "per acres".

Sounds crazy? Through deminisional reduction, it is just as valid as miles per gallon. So how many per-acres does your car get?

Answer 8

I think you should do your own homework