Obtain an expression for the time period T of a simple pendulum. [The time period T depend upon (i) mass l of the bob (ii) length m of the pendulum and (iii) acceleration due to gravity g at the place where pendulum is suspended.
Assume the constant `k=2pi`]

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 equation with dimensions.
`[T^1] = [M^a][L^b][LT^(-2)]^c`
`[M^0L^0T^1] = [M^aL^(b+c)T^(-2c)]`
Comparing the powers of M,Land 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 k(l/g)^(1/2)= ksqrt((l)/(g))`
Experimentally `k=2pi " hence "T= 2pi sqrt((l)/(g))`.