Physics, asked by abhishek433049, 9 months ago

A simple pendulum makes 20 oscillations in one sec-
ond on the surface of the earth. Determine the time
period of the simple pendulum on the surface of a
planet where the acceleration due to gravity is one
fourth of the acceleration due to gravity on the sur-
face of the earth.​

Answers

Answered by rounaquejahan007
0

Answer:

On surface of moon, acceleration due to gravity is g/6. Time period of seconds pendulum on earth is 2 seconds.

On Earth, 2=2π

g

l

On moon,T

moon

=2π

6

g

l

Dividing the above two equations, we get, T

moon

=2

6

s

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