Question

The time period of simple pendulum depends on length and acceleration due to gravity .Using dimensions derive the formula of time period

Answer 1

T= 2π√l/g

T^2= 4 π2 l/g
T^2=L/ LT^-2
T^2= T^2
T = T

Answer 2

Explanation:

Let,

$T \propto l^ag^b$

$[L^0T^1] = [L^1T^0]^a[L^1T^{-2}]^b$

$[L^0T^1] = [L^{a+b}T^{-2b}]$

On comparing,

$a+b=0$

$-2b=1$

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

So,

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

Put the value of a and b,

$T \propto l^ag^b$

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

$T \propto \sqrt{\dfrac{l}{g}}$

Add constant of proportionality,

$T =k\sqrt{\dfrac{l}{g}}$