# Units and Dimensions - Use of proportionality constant

In units and dimensions we learn about Establishing a Formula :

(example) : to establish a relationship between T (Time Period) , m (Mass) , l (length of the string) and g(acc. due to gravity) - Case of Pendulum :

$$T \propto m ,\quad T \propto l ,\quad T \propto G$$

So : $T\propto m^al^bg^c$. By taking proportionality constant, we get : $T = k \space m^a \space l^b \space g^c$

By using homogeneity principle , I get : $$a = 0 \quad , b = \cfrac{1}{2} , \quad c = \cfrac{-1}{2}$$

So I established the formula as : $$T = k \sqrt{\cfrac{l}{g}}$$

To complete the formula, we put k(proportionality constant) = $2\pi$ . My question is that how we got the value for proportionality constant? Also, is it fixed or it varies according to the equation?

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Using this method, I don't think there's a way to calculate $k$. But, there's another way to do this assuming the oscillation to be harmonic (of course, it already is). When you relate the acceleration with $-\omega^2x$, you'll obtain the value for $\omega$ to be $\sqrt{\frac{g}{l}}$.

Since the angular velocity is simply the velocity required to complete one rotation, it's given by $$\omega=2\pi/T$$

Relating both the equations, we obtain $$T=2\pi\sqrt{\frac{l}{g}}$$

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Thanks @Crazy Buddy , sorry I am not able to vote up your solution as I have reputation < 15 . Thanks –  Kushashwa Ravi Shrimali Mar 31 '13 at 10:11

The only way to get the constant of proportionality is to solve the equations of motion for your system, or do an experiment to get an approximate value.

The period of the pendulum is frequently used to teach dimensional analysis because it is simple to understand and works well. However in real life physicists don't often use dimensional analysis in this way. Typically you use it to check that equations you've derived are dimensionally consistent - if they aren't it means you made a mistake somewhere!

If you're interested in finding out where the $2\pi$ comes from in the example of the pendulum there are lots of easily Googlable articles about it. Wikipedia has a thorough if somewhat involved article on it.

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Thanks for your help John , I searched it on google but I expected some better and understandable solutions from this site... that I got. Thanks! –  Kushashwa Ravi Shrimali Mar 31 '13 at 10:09