Question

Prove that t= 2 pai underroot l by g by dimensionally

Answer 1

So the equation is,

$T=2\pi\sqrt {\dfrac {l}{g}}$

We have to check it dimensionally, whether it's true.

Here the LHS is the time period, $T.$

Well, it's time. So,

$[LHS]=[T]=M^0L^0T^1\quad\longrightarrow\quad (1)$

In the RHS, l is just the length. Hence,

$[l]=M^0L^1T^0$

And, g, we know, is the acceleration due to gravity. Well, it's merely an acceleration. So,

$[g]=M^0L^1T^{-2}$

Then,

$[RHS]\ =\ \sqrt {\dfrac {[l]}{[g]}}\\\\\\\[\][RHS]\ =\ \sqrt{\dfrac{M^0L^1T^0}{M^0L^1T^{-2}}}\\\\\\\[\][RHS]\ =\ M^0L^0T^1\quad\longrightarrow\quad (2)$

Note that constants are not taken in dimensional analysis.

From (1) and (2),

$[LHS]=[RHS]$

So the equation is dimensionally correct.

Remember, equations can't be proved by dimensional analysis, but can be verified or checked whether it's true. This is one of the limitations of the dimensional method.

Well, the given equation is theoretically true, and the expression is for calculating the time period of a simple pendulum.

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