The length of a second pendulum is $1 m$ [ where $g = 9.8 m / s^{2}$ ]. What will be the length of the second pendulum on any planet where $g = 4.9 m / s^{2}$ ?

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Solution

Given
$T = 2 s; g_{1} = 9.8 m s^{- 2}, l_{1} = 1 m$
$g_{2} = 4.9 m s^{- 2}, l_{2} = ?$
$∵$ Time period of a simple pendulum
$T = 2 π \sqrt{\frac{l}{g}}$
$∴$ For second pendulum
$2 = 2 π \sqrt{\frac{l}{g_{1}}} = 2 π \sqrt{\frac{l_{2}}{g_{2}}}$
or $\sqrt{\frac{l_{1}}{g_{1}}} = \sqrt{\frac{l_{2}}{g_{2}}} \Rightarrow \frac{l_{1}}{g_{1}} = \frac{l_{2}}{g_{2}}$
$\Rightarrow l_{2} = l_{1} \times \frac{g_{2}}{g_{1}} = 1 \times \frac{4.9}{9.8} = \frac{1}{2}$
or $l_{2} = 0.5 m$

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The length of a second pendulum is 1m [ where g=9.8m/s2]. What will be the length of the second pendulum on any planet where g=4.9m/s2?

Given T=2s;g1=9.8ms−2,l1=1mg2=4.9ms−2,l2=?∵ Time period of a simple pendulum T=2π√lg∴ For second pendulum2=2π√lg1=2π√l2g2or √l1g1=√l2g2⇒l1g1=l2g2⇒l2=l1×g2g1=1×4.99.8=12or l2=0.5m

The length of a second pendulum is $1 m$ [ where $g = 9.8 m / s^{2}$ ]. What will be the length of the second pendulum on any planet where $g = 4.9 m / s^{2}$ ?