Question

The period of oscillation of a simple pendulum is expected to depend upon the length of the pendulum $\left(l\right)$, and acceleration due to gravity $\left(g\right)$. The constant of proportionality is $2\pi$. Then $T$ $=$

• A. $\frac{2\pi l}{g}$
• B. $2\pi \sqrt{\frac{g}{l}}$
• C. $2\pi \sqrt{\frac{l}{g}}$
• D. $\frac{2\pi g}{l}$

Answer 1

Answer: (C)

$2\pi \sqrt{\frac{l}{g}}$

$T\propto L^{\alpha }{g}^{\beta }$

$M^0{L}^0{T}^1=L^{\alpha }(L{T}^{-2}{)}^{\beta }$
$M^0{L}^0{T}^1=L^{\alpha +\beta }{T}^{-2\beta }$
$\Rightarrow \alpha +\beta =0,-2\beta =+1$
$\Rightarrow \alpha =\frac{1}{2},\beta =-\frac{1}{2}$
$\Rightarrow T=2\pi \sqrt{\frac{l}{g}}$