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PhysicsSHMCalculation of time periodEasy2 minQB

PhysicsEasysingle choice

What is the time period of the seconds pendulum, on a planet similar to earth, but mass and radius three times as that of the earth?

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Answer:

Solution:

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

$\frac{g_{p}}{g_{e}} = \frac{M_{p}}{M_{e}} \times {(\frac{R_{e}}{R_{p}})}^{2} = 3 \times \frac{1}{9} = \frac{1}{3}$

$T = 2 π \sqrt{\frac{l}{g}}$

$\frac{T_{p}}{T_{e}} = \sqrt{\frac{g_{e}}{g_{p}}} = \sqrt{3}$

$T_{p} = \sqrt{3} T_{e}$

Time period at earth for seconds pendulum $= 2 s$

$∴ T_{p} = 2 \sqrt{3} s$

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Calculation of time period

⏱ 2mℹ️ Source: QB

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