Question

The period of a pendulum depends upon the length of the pendulum (l) and the acceleration due to gravity (g). Obtain the expressions for the time period of the pendulum

Answer 1

Let the time period of the pendulum depends upon length of the string $\sf{l}$ and acceleration due to gravity $\sf{g}$ as,

$\longrightarrow\sf{T\propto l^a\,g^b\quad\quad\dots(1)}$

Taking dimensions of each term,

$\longrightarrow\sf{[T]=[l]^a\,[g]^b}$

$\longrightarrow\sf{T=L^a\,(LT^{-2})^b}$

$\longrightarrow\sf{L^0\,T^1=L^{a+b}\,T^{-2b}}$

On equating corresponding powers,

$\longrightarrow\sf{-2b=1}$

$\longrightarrow\sf{b=-\dfrac{1}{2}}$

And,

$\longrightarrow\sf{a+b=0}$

$\longrightarrow\sf{a-\dfrac{1}{2}=0}$

$\longrightarrow\sf{a=\dfrac{1}{2}}$

Hence (1) becomes,

$\longrightarrow\sf{T\propto l^{\frac{1}{2}}\,g^{-\frac{1}{2}}}$

$\longrightarrow\sf{\underline{\underline{T\propto\sqrt{\dfrac{l}{g}}}}}$

Let $\sf{k}$ be the proportionality constant. Then,

$\longrightarrow\sf{\underline{\underline{T=k\sqrt{\dfrac{l}{g}}}}}$

Practically, $\sf{k=2\pi.}$ Hence,

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

Answer 2

Answer:

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