# Question #d58fd

May 14, 2016

$2 \sqrt{2} s$

#### Explanation:

For small angle of deviation $\theta$ from the mean position the time period $T$ of a pendulum is given as

$T = 2 \pi \sqrt{\frac{L}{g}}$ .......(1)
where $L$ is the length of the pendulum and $g$ is the local gravity.

1. We know that seconds pendulum is a pendulum which has its period of two seconds. Its each swing takes one second, $T = 2 s$
2. Also local acceleration due to gravity $g = G \frac{{M}_{e}}{R} _ {e}^{2}$.
$G$ is constant $= 6.67408 \times {10}^{-} 11 {m}^{3} k {g}^{-} 1 {s}^{-} 2$. ${M}_{e}$ is mass and ${R}_{e}$ is radius of earth respectively.

Now the acceleration due gravity of the planet is

${g}_{p} = G \frac{{M}_{p}}{R} _ {p}^{2}$

Inserting given quantities we get

${g}_{p} = G \frac{2 {M}_{e}}{2 {R}_{e}} ^ 2$
$\implies {g}_{p} = G \frac{{M}_{e}}{2 {\left({R}_{e}\right)}^{2}}$,
in terms of $g$
${g}_{p} = \frac{g}{2}$

Inserting in (1)

${T}_{p} = 2 \pi \sqrt{\frac{L}{g} _ p}$ ........(2)

in terms of $g$

${T}_{p} = 2 \pi \sqrt{2 \frac{L}{g}}$

Using (1)

${T}_{p} = \sqrt{2} T$
$\implies {T}_{p} = 2 \sqrt{2} s$