If the length of seconds pendulum on the surface of the earth is 1 m. The length of the seconds pendulum on the surface of the planet is { g pl an e t = 2 7 g e a r t h } {periodic time of the s

Question

If the length of seconds pendulum on the surface of the earth is 1 m. The length of the seconds pendulum on the surface of the planet is { g pl an e t ​ = 2 7 ​ g e a r t h ​ } {periodic time of the seconds pendulum is same on both planets}

TestHub · Accepted Answer

The time period of a simple pendulum is given by $T = 2 π \frac{l}{g}$ .

Since the time period of a seconds pendulum is constant ( $T = 2 s$ ), we have $\frac{l}{g} = constant$ .

$\frac{l _{e a r t h}}{g _{e a r t h}} = \frac{l _{pl an e t}}{g _{pl an e t}}$

Substituting the given values $l_{e a r t h} = 1 m$ and $g_{pl an e t} = \frac{7}{2} g_{e a r t h}$ :

$\frac{1}{g _{e a r t h}} = \frac{l _{pl an e t}}{\frac{7}{2} g _{e a r t h}}$

$l_{pl an e t} = \frac{7}{2} m$

Answer: $\frac{7}{2} m$

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