Question

a seconds pendulum is taken to a planet where acceleration due to gravity is 9 times that on the earth what is the time period of the pendulum and that planet​

Answer 1

Answer:

If the acceleration due to gravity become 1/9th then the time period will be Triple ,i.e., 6 seconds. T' = 6 seconds. Thus, time periods of the second pendulum on the planet where acceleration due to gravity is 1/9th of the earth is 6 seconds.

Answer 2

Given :

➳ Acceleration due to gravity is 9 times that on the earth.

To Find :

⟶ Length of second pendulum on that planet.

SoluTion :

➠ Second pendulum is the simple pendulum, having a time period of 2 second. Its effective length is 99.992 cm or approximate one metre on earth.

➠ Formula of time period in terms of length of simple pendulum and acceleration due to gravity is given by

\bigstar\:\boxed{\bf{\gray{T=2\pi\sqrt{\dfrac{L}{g}}}}}★

T=2π

g

L

➠ ATQ, length of simple pendulum remains same on both planets, so we can say that time period of simple pendulum is inversely proportional to the square root of acceleration due to gravity.

\bigstar\:\boxed{\bf{\gray{T\propto\:\dfrac{1}{\sqrt{g}}}}}★

T∝

g

1

Let,

▪ Time period on earth = T

▪ Acc. due to gravity of earth = g

▪ Time period on planet = T'

▪ Acc. due to gravity of planet = g'

⇒ T/T' = √g'/√g

⇒ 2/T' = √9g/√g

⇒ T' = 2/√9

⇒ T' = 2/3

⇒ T' = 0.66 s