The period of a simple pendulum is given by T = 2π(g/l)1/2, where l i

1.	The period of a simple pendulum is given by T = 2π(g/l)1/2, where l is length of the pendulum and g is acceleration due to gravity. Show that this equation is dimensionally correct.
Answer» Given \(T = 2\pi \sqrt{\frac lg}\) L.H.S. \([T] = [M^0L^0T]\) R.H.S. \([l] = [M^0LT^0]\) \([g] = [M^0LT^{-2}]\) Then \([T] = \sqrt{\frac l g}\) \([M^0L^0T] = \cfrac{[M^0LT^0]^\frac12}{[M^0Lt^{-2}]^\frac12}\) \([M^0L^0T^{-1}] =[M^0L^0T^{-1}]\) R.H.S = L.H.S

The period of a simple pendulum is given by T = 2π(g/l)1/2, where l is length of the pendulum and g is acceleration due to gravity. Show that this equation is dimensionally correct.

Answer»

Given \(T = 2\pi \sqrt{\frac lg}\)

L.H.S.

\([T] = [M^0L^0T]\)

R.H.S.

\([l] = [M^0LT^0]\)

\([g] = [M^0LT^{-2}]\)

Then

\([T] = \sqrt{\frac l g}\)

\([M^0L^0T] = \cfrac{[M^0LT^0]^\frac12}{[M^0Lt^{-2}]^\frac12}\)

\([M^0L^0T^{-1}] =[M^0L^0T^{-1}]\)

R.H.S = L.H.S