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

`Tproptom^(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^(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, we get `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(1/g)^(1//2)=ksqrt(1//g)`
Experimentally `k=2pi`, hence `T=2pisqrt(l//g)`