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]

Obtain an expression for the time period T of a simple pendulum. [The

1.	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]
Answer» Solution :`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))`.