Question 104: Oscillations

Question

The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be :

Accepted Answer

$$ \because $$    g = $${{GM} \over {{R^2}}}$$
$${{{g_p}} \over {{g_e}}}$$ = $${{{M_e}} \over {{M_e}}}{\left( {{{{{\mathop{m R}
olimits} _e}} \over {{R_p}}}} ight)^2}$$ = 3$${\left( {{1 \over 3}} ight)^2}$$ = $${{1 \over 3}}$$
Also  T $$ \propto $$ $${1 \over {\sqrt g }}$$
$$ \Rightarrow $$  $${{{T_p}} \over {{T_e}}}$$ = $$\sqrt {{{{g_e}} \over {{g_p}}}} $$ = $$\sqrt 3 $$
$$ \Rightarrow $$  T
p
= 2$$\sqrt 3 $$ s

Answer

$${{\sqrt 3 } \over 2}$$ s

Answer

$${3 \over 2}$$ s

Answer

$${2 \over {\sqrt 3 }}$$ s

Answer

$$2\sqrt 3 $$ s

Oscillations

The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be :

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