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
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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
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Answer:
☆Given:
[tex]\ g_{p} = \dfrac{1}{4}g_{e}[/tex]------(1)
where [tex]\ g_{p}[/tex] is the gravity by planet 'p'.
▪︎ time peiod(T) of oscillation on earth is 2sec.
☆Formula of time period:
[tex]\boxed{ T = 2\pi \sqrt{ \dfrac{l}{g}}}[/tex]
☆Solution:
For earth-
》[tex]\ T_{e} = 2\pi \sqrt{ \dfrac{l}{g_{e}}}[/tex]
》[tex]\ 2 = 2\pi \sqrt{ \dfrac{l}{g_{e}}}[/tex]
》[tex] \sqrt{ \dfrac{l}{g_{e}}} = \dfrac{1}{\pi}[/tex]----(2)
_____
For planet-
》[tex]\ T_{p} = 2\pi \sqrt{ \dfrac{l}{g_{p}}}[/tex]
》[tex]\ T_{p} = 2\pi \sqrt{ \dfrac{l}{\dfrac{1}{4} g_{e}}}[/tex]
》[tex]\ T_{p} = 2\pi \sqrt{ \dfrac{4l}{g_{e}}}[/tex]
》[tex]\ T_{p} = 2×2\pi \sqrt{ \dfrac{l}{g_{e}}}[/tex]
》[tex]\ T_{p} = 4 \pi \dfrac{1}{\pi}[/tex]...(by equation 2)
》[tex]\bf{T_{p} = 4sec}[/tex]
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☆so, the time period of oscillation on planet is 4secs.
The correct answer is option (d) 4s
Given:
[tex]Gp=\frac{1}{4}ge\\[/tex]
where [tex]g_{p}[/tex] is the gravity by planet p
T(time period) on earth is 2sec
Formula of time period:
[tex]T=2\pi \sqrt{\frac{l}{g} }[/tex]