Question

Acceleration due to gravity of a planet is one fourth that of earth. What is
period of oscillation of a simple pendulum on this planet if it is 2 s on
earth?
a) 1 s
b) 2 s
c) 3 s
d) 4 s​

Answer 1

Answer:

☆Given:

$\ g_{p} = \dfrac{1}{4}g_{e}$------(1)

where $\ g_{p}$ is the gravity by planet 'p'.

▪︎ time peiod(T) of oscillation on earth is 2sec.

☆Formula of time period:

$\boxed{ T = 2\pi \sqrt{ \dfrac{l}{g}}}$

☆Solution:

For earth-

》$\ T_{e} = 2\pi \sqrt{ \dfrac{l}{g_{e}}}$

》$\ 2 = 2\pi \sqrt{ \dfrac{l}{g_{e}}}$

》$\sqrt{ \dfrac{l}{g_{e}}} = \dfrac{1}{\pi}$----(2)

_____

For planet-

》$\ T_{p} = 2\pi \sqrt{ \dfrac{l}{g_{p}}}$

》$\ T_{p} = 2\pi \sqrt{ \dfrac{l}{\dfrac{1}{4} g_{e}}}$

》$\ T_{p} = 2\pi \sqrt{ \dfrac{4l}{g_{e}}}$

》$\ T_{p} = 2×2\pi \sqrt{ \dfrac{l}{g_{e}}}$

》$\ T_{p} = 4 \pi \dfrac{1}{\pi}$...(by equation 2)

》$\bf{T_{p} = 4sec}$

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☆so, the time period of oscillation on planet is 4secs.

Answer 2

The correct answer is option (d) 4s

Given:

$Gp=\frac{1}{4}ge$  

where $g_{p}$ is the gravity by planet p

T(time period) on earth is 2sec

Formula of time period:

$T=2\pi \sqrt{\frac{l}{g} }$