Question

Find the diamensional formula for the period of oscillation of a simple pendulum

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Time}\mathrm{period}\mathrm{of}\mathrm{simple}\mathrm{pendulum}\mathrm{is}, \\ \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}}} \\ \mathrm{where},\mathrm{l}=\mathrm{length} \\ \mathrm{g}=\mathrm{acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity} \\ \mathrm{Dimension}\mathrm{of}\mathrm{length}=\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right] \\ \mathrm{Dimension}\mathrm{of}\mathrm{g}=\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^{-2}\right] \\ \mathrm{Dimension}\mathrm{of}\mathrm{time},\mathrm{T}={\left(\frac{\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right]}{\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^{-2}\right]}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \mathrm{here},\mathrm{dimension}\mathrm{of}2\mathrm{\pi }\mathrm{is}\mathrm{neglected}\mathrm{because}\mathrm{it}\mathrm{has}\mathrm{no}\mathrm{dimension}. \\ \mathrm{Dimension}\mathrm{of}\mathrm{time},\mathrm{T}={\left[{\mathrm{M}}^{0-0}{\mathrm{L}}^{1-1}{\mathrm{T}}^{0+2}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \mathrm{Dimension}\mathrm{of}\mathrm{time},\mathrm{T}={\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \mathrm{Dimension}\mathrm{of}\mathrm{time},\mathrm{T}=\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^1\right] \end{gathered}$$