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Units and Measurements — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsUnits and Measurements3 Marks
Question
Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period
of oscillation of the simple pendulum depends on its length (l), mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its time period using method of dimensions.
✓

Answer

The dependence of time period $T$ on the quantities $l, g$ and $m$ as a product may be written as :
$
T=k P^r g^y m^z
$
where $k$ is dimensionless constant and $x, y$ and $z$ are the exponents.

By considering dimensions on both sides, we have
$
\begin{array}{l}
{\left[ L ^0 M ^0 T ^1\right]=\left[ L ^1\right]^x\left[ L ^1 T ^{-2}\right]^y\left[ M ^1\right]^z} \\
= L ^{x \neq y} T ^{2 y} M ^z
\end{array}
$

On equating the dimensions on both sides, we have
$
x+y=0 ;-2 y=1 \text {; and } z=0
$

So that $x=\frac{1}{2}, y=-\frac{1}{2}, z=0$
Then, $T=k\ l^\frac{1}{2} g^\frac{1}{2}$
or, $T=k \sqrt{\frac{l}{g}}$
Note that value of constant $k$ can not be obtained by the method of dimensions. Here it does not matter if some number multiplies the right side of this formula, because that does not affect its dimensions.
Actually, $k=2 \pi$ so that $T=2 \pi \sqrt{\frac{l}{g}}$

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