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

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. \([l] = [M^0 LT^0]\) \([g] = [M^0 LT^{-2}]\) Then \([T] = \sqrt{\frac lg}\) \([M^0L^0T] = \frac{[M^0LT^0]^{^1/_2}}{[M^0LT^{-2}]^{^1/_2}}\) \([M^0L^0T^{1}] = [M^0L^0T^{1}]\) R.H.S = L.H.S So it is dimensionally correct.

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

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

Then

\([T] = \sqrt{\frac lg}\)

\([M^0L^0T] = \frac{[M^0LT^0]^{^1/_2}}{[M^0LT^{-2}]^{^1/_2}}\)

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

R.H.S = L.H.S

So it is dimensionally correct.