Question

Show that the expression of the time period T of a simple pendulum of length l given by T = 2pi sqrt((l)/(g)) is dimensionally currect

Answer 1

The given formula of time period is dimentionally correct (proved)

Step by step explanation:

Time period (T) is the time taken by a pendulm in completing one oscillation.

In this question the relation between time period of a pendulum with the length (l) and gravity or gravitational acceleration (g) is given:

$T = 2\pi \sqrt\frac{l}{g}$                                                             (equation 1)

Here LHS = T

        RHS =$2\pi \sqrt\frac{l}{g}$

The dimention of Time period (T) =  $[M^0 L^0 T^1]$

So, the dimension of LHS = $[M^0 L^0 T^1]$           (equation 2)

The dimension of Length of pendulum (l) = $[M^0 L^1 T^0]$

The dimention of gravity (g) = $[M^0 L^1 T^-^2]$

2π is constant so it is dimension less = $[M^0 L^0 T^0 ]$

So the dimention of RHS

$=[M^0 L^0 T^0 ]\times \sqrt \frac{[M^0 L^1 T^0]}{[M^0 L^1 T^-^2]}\\\\=[M^0 L^0 T^0 ]\times \sqrt \frac{1}{[M^0 L^0 T^-^2]}\\\\=[M^0 L^0 T^0 ]\times \frac{1}{[M^0 L^0 T^-^1]}\\\\=[M^0 L^0 T^1 ]$                         (equation 3)

By comparing equation 2 and 3.

LHS = RHS

$\textbf{\Large So the formula of time period is dimentionally correct }$