Question

Obtain the expression for the period of oscillation of a pendulum assuming that it may depends on mass of the bob , length of the pendulum and acceleration due to gravity at the place using dimensional analysis .​

Answer 1

Solution

Let t∝malbgc where a, b, c are the dimensions. Or t=kmalbgc…(4)

where k is dimensionless constant of proportionality.

Writing the dimensions in terms of M, L, T on either side of (4), we get

[M0L0T1]=MaLb(LT−2)c=MaLb+cT−2c

Applying the principle of homogeneity of dimensions, we get

a=0b+c=0....(5)

−2c=1orc=−12

From (5),b=−c=−(−12)=12

Putting the value of a,b,c, in (4), we get

t=km0I1/2g−1/2

t=kI–√g

Using other methoud, we calculate the value of dimensionless constant,  k=2π:,t=2πI/g