Question 6.10: The time period τof a simple pendulum depends on its effecti...
The time period \tau of a simple pendulum depends on its effective length l and the local acceleration due to gravity g . Using both Buckingham’s Pi theorem and Rayleigh’s indicial method, find the functional relationship between the variables involved.
Application of Buckingham’s Pi theorem:
The variables of the problem are \tau, l and g and the fundamental dimensions involved in these variables are L (length) and T (time). Therefore the no. of independent \pi term = (3 – 2) = 1, since \tau is the dependent variable, the only choice left for the repeating variables to be l and g .
Hence, \pi_{1}=l^{a} g^{b} \tau
Expressing the equation in terms of the fundamental dimensions of the variables we get \mathrm{L}^{0} \mathrm{~T}^{0}=\mathrm{L}^{a}\left(\mathrm{LT}^{-2}\right)^{b} \mathrm{~T} . Equating the exponents of L and T on both sides of the equation, we have,
a+b=0, \quad \text { and } \quad-2 b+1=0which give a=-\frac{1}{2}, \quad b=\frac{1}{2} ; and hence \pi_{1}=\tau \sqrt{\frac{g}{l}}
Therefore, the required functional relationship between the variables of the problem is
f\left(\tau \sqrt{\frac{g}{l}}\right)=0 (6.48)
Application of Rayleigh’s indicial method:
Since \tau is the dependent variable, it can be expressed as
\tau=A l^{a} g^{b} (6.49)
where A is a non-dimensional constant. The Eq. (6.49) can be written in terms of the fundamental dimensions of the variables as
T=A \mathrm{~L}^{a}\left(\mathrm{LT}^{-2}\right)^{b}Equating the exponent of L and T on both sides of the equation, we get, a+b=0 and -2 b=1 which give a=\frac{1}{2} and b=-\frac{1}{2}
Hence, Eq. (6.49) becomes \tau=A \sqrt{\frac{l}{g}}
Or \tau \sqrt{\frac{g}{l}}=A
Therefore, it is concluded that the dimensionless governing parameter of the problem is \tau \sqrt{\frac{g}{l}} . From elementary physics, we know that A=2 \pi .