Obtain in expression for the time period T of a simple pendulun. The time period depend upon (i) mass 'm' of the bob (ii) length 'l' of the pendulum and (iii) acceieration due to gravity g at the place where the pendulum is suspended. (Constant k = `2pi` ) i.e.

Explanation

`T alpha m^(a) l^(b) g^(c) , T =k. m^(a) l^(b) g^(c)`
Here k is the dimensionless constant. Rewriting the above equatio with dimensions
`[T^(1)] = [M^(a)][L^(b)][LT^(-2)]^(c)`
`[M^(0)L^(0)T^(1)] = [M^(a)L^(b+c)T^(-2c)]`
Comparing the powers of M,L and T on both sides, a = 0, b + c = 0, -2c = 1
Solving for a, b and c a = 0, b = `1//2`, and c = `-1//2`
From the above equation
`T = k. M^(0) l^(1//2) g^(-1//2)`
`T = k (l/g)^(1//2) = ksqrt(l/g)`
Experimentally k = `2_(p)` , hence
T = `2_(p) sqrt(1//g)`