Question 6.7: The time period τ of a simple pendulum depends on its effect...
The time period τ of a simple pendulum depends on its effective length l and the local acceleration due to gravity g. Using both Buckinghams π theorem and Rayleighs indicial method, find the functional relationship between the variables involved.
Application of Buckinghams π theorem:
The variables of the problem are τ, l and g and the fundamental dimensions involved in these variables are L (length) and T (time). Therefore the no. of independent π term = (3 2) = 1, since τ 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 L ^{0} T ^{0}= L ^{a}\left( LT ^{-2}\right)^{b} T. Equating the exponents of L and T on both sides of the equation we have,
a + b = 0, and 2b + 1 = 0
which 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 (6.47)
Application of Rayleighs indicial method:
Since τ is the dependent variable, it can be expressed as
\tau=A l^{a} g^{b} (6.48)
where A is a non-dimensional constant. The Eq. (6.48) can be written in terms of the fundamental dimensions of the variables as
T=A L ^{a}\left( LT ^{-2}\right)^{b}Equating the exponent of L and T on both sides of the equation, we get, a + b = 0 and 2b = 1 which give a=\frac{1}{2} \quad \text { and } \quad b=-\frac{1}{2}.
Hence Eq. (6.48) 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 = 2π.