How to Determine the Units of a Homogeneous Equation in Physics

[MemoNick]

1. So for today's exam, I was given this equation, and I was required to get the value and units of K:

T⁴ = (4K∏⁴)/g² - (8d∏⁴)/g²​

I had to follow an experiment, plot the results, get the gradient. After getting the value of g, I was required to find the value of K, and its units.

T⁴ has units s⁴, d has units m, g is gravitational acceleration, hence I believe it is m.s^-2



2. The problem lies in the fact that I don't believe I was given a homogeneous equation, hence I couldn't provide the units. Any help? Note that the units are enough for me, I managed to solve for K (correctly, I believe).

Thanks in advance :)

 

[Enigman]

You can only add or subtract things with same units.

 

[MemoNick]

So you confirm the equation is not homogeneous?

 

[Enigman]

I neither confirm nor deny, I merely point out the obvious. What then is ∏? Or did you mean \pi?

 

[MemoNick]

Well, after working it out, the left side is this: T⁴ has units s⁴.

For the K part, I can't work it out without knowing whether it's homogeneous or not. So I go where there's d.

(8d∏⁴)/g² would then become:
Numerator: m
Denominator: m².s^-4

Therefore this part would be s⁴.m^-1

For homogeneity with addition/subtraction, all parts have to be the same. These two parts aren't the same, evidently. Any confirmation would be very much appreciated, so I could e-mail my invigilators.

 

[Enigman]

What is the experiment?

 

[MemoNick]

That should be a pi :) I couldn't find the symbol :P

 

[Enigman]

Nope, it doesn't look homogenous. What's the Experiment?
You sure d is in meters and not m^2 ?

 

[MemoNick]

The experiment isn't very important per-se, but for the record, there are two stand clamps, with a wire attached to both, forming a triangle. At the bottom, there's a pendulum, and d is the distance between the two stands. So yeah, definitely in meters. It's also given, but because of copyright issues, I'm afraid I'm not allowed to take a picture of it and upload it.

 

[gneill]

Could be a typo in the given equation; I suspect that the 'd' in the second term should have been $d^2$.

Even if the second term has a typo, the terms should individually have the same units as the LHS of the equation. So you can still determine what k should be and even repair the typo!

 

[MemoNick]

While I did as you said, gneill, I couldn't ascertain which was the right one - the 'd' part or the LHS.

 

[gneill]

While I did as you said, gneill, I couldn't ascertain which was the right one - the 'd' part or the LHS.

I think you can be pretty certain the LHS would consist of a single variable, and that as missed typos go, dropping a square on a variable in the midst of a term is more likely than dropping a variable on the LHS.

 

[Enigman]

Well, I haven't got the derivation of the formula yet but with off-hand information the formula's probably wrong. When you put d=0 the bifilar pendulum becomes a simple pendulum the formula should be reduced to a simple one.
Making $$k= 4 l^2$$ from the standard formula of time period. Formula's probably missing a square factor on d, will confirm it and let you know.
Whoops gneil got there before me...

 

[MemoNick]

Thanks a lot for your help :) I really appreciate it!