Question

time periods of a pendulum on two planets are in the ratio 3:4.the accleration due to gravity on them are in the ratio is​

Answer 1

Given :

➳ Time period of a pendulum on two planets are in the ratio 3:4.

To Find :

⟶ The ratio of acceleration due to gravity on that planets.

SoluTion :

⇒ Formula of time period of a simple pendulum in terms of acceleration due to gravity is given by

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

• T denotes time period
• L denotes length of pendulum
• g denotes acc. due to gravity

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

$\implies\sf\:\dfrac{T}{T'}=\sqrt{\dfrac{g'}{g}}$

$\implies\sf\:\dfrac{3}{4}=\sqrt{\dfrac{g'}{g}}$

$\implies\sf\:\dfrac{g}{g'}=\dfrac{4^2}{3^2}$

$\implies\boxed{\bf{g:g'=16:9}}$

Answer 2

$\huge\sf\pink{Answer}$

☞ Their Ratio is 16:9

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$\huge\sf\blue{Given}$

✭ Time period of a pendulum on two planets are of the ratio 3:4

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$\huge\sf\gray{To \:Find}$

◈ Ratio of their Gravitational Acceleration?

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$\huge\sf\purple{Steps}$

The formula that is to be used here is,

$\underline{\boxed{\red{\sf T = 2 \pi \sqrt{\dfrac{L}{g}}}}}$

Where,

◕ T = Time

◕ L = Length of pendulum

◕ g = Gravitational Acceleration

So as we are given that the length of the pendulum is the same them the time period is inversely

proportional to the square root of Gravitational Acceleration, which means,

$\underline{\boxed{\green{\sf \dfrac{T}{T'} = \sqrt{\dfrac{g'}{g}}}}}$

Substituting the given values,

➳ $\sf\dfrac{3}{4} = \sqrt{\dfrac{g'}{g}}$

➳ $\sf\dfrac{3^2}{4^2} = \dfrac{g'}{g}$ « Squaring both sides »

➳ $\sf \dfrac{16}{9} = \dfrac{g}{g'}$

➳ $\sf\orange{ g:g' = 16:9}$

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