Question 12 Marks
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
View full question & answer→Let us assume that $T \propto m^a l^b g^c$
$
\text { or, } T=km^{a} l^{b} g^{c}\ldots(i)
$
where, k is a dimensionless constant.
The dimensions of various quantities are
$
\begin{aligned}
& {[T]=T,[m]=M,} \\
& {[l]=L, \text { and }[g]=LT^{-2}}
\end{aligned}
$
Substitute these values in Eq.(i), we obtain
$
\begin{aligned}
& T=[M]^{a}[L]^{b}\left[LT^{-2}\right]^{c} \\
& \text { or, } M^0 L^0 T^1=M^{a} L^{b+c} T^{-2 c}
\end{aligned}
$
Now equate the powers of $M , L$ and T on both sides, we obtain
$
a=0, b+c=0,-2 c=1
$
On solving, $a =0, b=\frac{1}{2}, \quad c=-\frac{1}{2}$
$
\therefore T=k m^0 l^{1 / 2} g^{-1 / 2}=k \sqrt{\frac{l}{g}}
$
From experiments, $k =2 \pi$
Therefore, $T =2 \pi \sqrt{\frac{l}{g}}$, which is the required expression.
$
\text { or, } T=km^{a} l^{b} g^{c}\ldots(i)
$
where, k is a dimensionless constant.
The dimensions of various quantities are
$
\begin{aligned}
& {[T]=T,[m]=M,} \\
& {[l]=L, \text { and }[g]=LT^{-2}}
\end{aligned}
$
Substitute these values in Eq.(i), we obtain
$
\begin{aligned}
& T=[M]^{a}[L]^{b}\left[LT^{-2}\right]^{c} \\
& \text { or, } M^0 L^0 T^1=M^{a} L^{b+c} T^{-2 c}
\end{aligned}
$
Now equate the powers of $M , L$ and T on both sides, we obtain
$
a=0, b+c=0,-2 c=1
$
On solving, $a =0, b=\frac{1}{2}, \quad c=-\frac{1}{2}$
$
\therefore T=k m^0 l^{1 / 2} g^{-1 / 2}=k \sqrt{\frac{l}{g}}
$
From experiments, $k =2 \pi$
Therefore, $T =2 \pi \sqrt{\frac{l}{g}}$, which is the required expression.
