Question

The period of simple pendulum is found to increase by $50$% when the length of the pendulum is increased by $0.6m$. Calculate the initial length and initial period of oscillation at a place where $g=9.8m/s^2$

Answer 1

Original time period $T=T$
Original length $L=L$
$T=2\pi \sqrt{\frac{L}{g}}...\left(1\right)$

$T^{\prime }=T+\frac{T}{2}$

$T^{\prime }=\frac{3T}{2}$

$T^{\prime }=2\pi \sqrt{\frac{L^{\prime }}{g}}....\left(2\right)$

$L^{\prime }=L+0.6$

Eq. $\left(1\right)$ by eq. $\left(2\right)$

$\frac{T}{T^{\prime }}=\sqrt{\frac{L}{L^{\prime }}}$

$\frac{T\times 2}{3T}=\sqrt{\frac{L}{L+0.6}}$

${\left(\frac{2}{3}\right)}^2=\frac{L}{L+0.6}$

$\frac{4}{9}=\frac{L}{L+0.6}$

$4L+2.4=9L$

$2.4=5L$

$L=0.48m$

$T=2\pi \sqrt{\frac{L}{g}}=2\pi \sqrt{\frac{0.48}{9.8}}$

$T=1.38seconds$.