Dimensional consistency problem, need help

  • #1
Beginning student striving to learn. The problem was generated by people from the University of Winnipeg. Source: http://theory.uwinnipeg.ca/physics/ (this website was actually suggested by a user on the forums who suggested utilizing it as a learning source).


Homework Statement


Problem is, quote:

The period of a simple pendulum, defined as the time for one complete oscillation, is measured in time units and is given by:

T = 2∏√l/g

where,l is the length of the pendulum and g is the acceleration due to gravity, in units of length divided by time squared. Show that this equation is dimensionally consistent; that is, show that the right hand side of this equation gives units of time.


Homework Equations


The solution is actually given. The full equation was as follows:

[2∏√l/g]= √L/L/t2. This all equals: √t2 which = T


The Attempt at a Solution



Given the description which is italicized, I understood how they got: √L/L/t2 which ultimately equals T, but I do not understand how "2∏" disappeared so easily in the process of equating T.

My attempt: The italicized description already tells us that "g"= L/t2. Given this information, dividing "L" by L/t2 will result in t2 (all under the square root symbol). Then, the √t2 is T. I get stuck on the conversion process when it comes to 2∏ however.

I've thought about it for several minutes, and I still can't understand how to mathematically describe T and the aforementioned equation including 2∏. Can anyone help? Thank you for your time.
 
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Answers and Replies

  • #2
NascentOxygen
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Pi is a constant, it has no units. It is dimensionless. It is just a scaling factor, likewise the 2.
 
  • #3
Pi is a constant, it has no units. It is dimensionless. It is just a scaling factor, likewise the 2.


This is hard for me to understand, and I probably need to enhance my mathematical skills as it relates to physics. However, during the entire time of trying to make sense of the equivalence, I kept thinking about multiplying 2 times 3.14 or something like that, and connect it to the rest of the equation. This is why I was stuck.

I really appreciate your input, but if 2∏ was ultimately dimensionless to begin with, why was it put in the equation? Thanks for your time.
 
  • #4
SteamKing
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I really appreciate your input, but if 2∏ was ultimately dimensionless to begin with, why was it put in the equation? Thanks for your time.

When you do dimensional analysis on equations in general, you must learn to distinguish any constants which might be contained in the equation from the parts of the equation which might contain units. Not all of the problems you encounter will be 'pre-digested' for your convenience.
 
  • #5
When you do dimensional analysis on equations in general, you must learn to distinguish any constants which might be contained in the equation from the parts of the equation which might contain units. Not all of the problems you encounter will be 'pre-digested' for your convenience.

I think I understand what you're saying, but that doesn't really answer my question. Why were the constants put there in the first place if they were ultimately irrelevant in the conversion process?

I mean, another conversion problem was posed on the website related to dimensional analysis (which dealt with converting miles into meters into kilometers), and I had no problem with that whatsoever.

Yet, with this problem it was different. I don't wan't things necessarily pre-digested for me, but only an understanding of things...in the same way that someone would like an understanding as to why ∏=3.14592...; or something to that effect.
 
  • #6
NascentOxygen
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Dimensionless constants are necessary in some formulae to make the formula correct, so that it gives the right answer. They are a scaling factor.

Pi is the ratio of a circle's circumference to its diameter, so it is metres divided by metres, these cancel leaving no units. You have to recognize in formulae those constants which have no units, and dismiss them when checking dimensions.

You may recognize the formula F=Gm1 m2 / r2

here G is a constant, but it is not dimensionless, it has units.
 
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  • #7
I am in the same boat. I went to one of the physics teachers in my high school and asked. He explained it quite well. Looking at T = 2∏√l/g you are trying to make sure the equation in dimensionally consistent, meaning that both sides must end up in the same dimension. (time) Let's pick units for length and the acceleration due to gravity. Length=m (meters) and acceleration due to gravity=m/second squared. That ends up as T = 2∏√m/ m/s2. (I will just use 2 as squared) You can express m as m/1 because that doesn't change the value, so you can use m/1/ m/s2 in the equation. When dividing by a fraction, you can multiply by its inverse, so m/1/ m/s2 can be expressed as m/1 x s2/m, so you cross multiply, and the m's cancelled out, leaving s2. Now the equation is T = 2∏√s2. The square root of s2 is s, and 2 and ∏ are just values to multiply by, not separate variables or quantities, so that means the equation is dimensionally consistent, as both sides are expressed as a value of time. (T=time, 2∏s is equal to a certain amount (6.14318...) of seconds

Reference https://www.physicsforums.com/threads/dimensional-consistency-problem-need-help.745475/
 
  • #8
Ray Vickson
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This is hard for me to understand, and I probably need to enhance my mathematical skills as it relates to physics. However, during the entire time of trying to make sense of the equivalence, I kept thinking about multiplying 2 times 3.14 or something like that, and connect it to the rest of the equation. This is why I was stuck.

I really appreciate your input, but if 2∏ was ultimately dimensionless to begin with, why was it put in the equation? Thanks for your time.

It was put in to obtain a correct formula. The value ##\sqrt{l/g}## does not give the correct pendulum period, although it has the right "dimensions" in terms or units. To get a correct formula, you also need the factor ##2 \pi##. A pendulum with length ##l = 1 ## meter has a period of nearly ##T = 2.01## sec; as can be verified experimentally. Without the ##2 \pi## factor you would get about ##0.319## sec., which does not agree at all with experiment!

You will meet lots of similar examples when you continue your studies. For instance, the kinetic energy, ##KE##, of a body of mass ##m## moving at speed ##v## is ##KE = \frac{1}{2} m v^2.## Without the factor ##1/2## the result would be incorrect, even though the "1/2" is dimensionless.

Such dimensionless factors are put in because Nature demands it!
 
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