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 the method of dimensions.

(N/A) The dependence of time period $T$ on the quantities $l, g$,and $m$ as a product may be written as:
$T = k l^{x} g^{y} m^{z}$
where $k$ is a dimensionless constant and $x, y$,and $z$ are the exponents.
By considering dimensions on both sides,we have:
$[M^0 L^0 T^1] = [L]^x [L T^{-2}]^y [M]^z$
$[M^0 L^0 T^1] = M^z L^{x+y} T^{-2y}$
On equating the dimensions on both sides,we have:
$z = 0$
$x + y = 0$
$-2y = 1$
Solving these,we get $y = -1/2$,$x = 1/2$,and $z = 0$.
Substituting these values,we get $T = k l^{1/2} g^{-1/2} m^0$.
Thus,$T = k \sqrt{\frac{l}{g}}$.
Note that the value of the constant $k$ cannot be obtained by the method of dimensions. Experimentally,$k = 2\pi$,so $T = 2\pi \sqrt{\frac{l}{g}}$.