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The relation between \[T\] and $g$ by
A. $T \propto g$
B. $T \propto {g^2}$
C. ${T^2} \propto {g^{ - 1}}$
D. $T \propto \dfrac{1}{g}$

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Last updated date: 22nd Feb 2024
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IVSAT 2024
Answer
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Hint: We have to find the relation between \[T\]and $g$. This is a question from the topic of the simple pendulum. A simple pendulum consists of a mass suspended by an inextensible massless string of a length $l$. Here we have to find the relation between the period of oscillation of the simple pendulum and the acceleration due to gravity.

Complete step by step answer:
The time taken by an oscillating body to repeat its periodic motion is called the period of oscillation. Here we consider \[T\] as the period of oscillation of the simple pendulum. $g$ stands for the acceleration due to gravity.
To find the relation between the period of oscillation and the acceleration due to gravity, let us consider the expression for the period of oscillation of a simple pendulum,
The Time period of a simple pendulum is given by,
$T = 2\pi \sqrt {\dfrac{l}{g}} $
To eliminate the square root on the RHS, we have to square both sides of the equation
We get,
${T^2} = 4{\pi ^2}\dfrac{l}{g}$
From this we get that, for a constant length of the pendulum the square of the time period will be inversely proportional to the acceleration due to gravity, i.e.
${T^2} \propto \dfrac{1}{g}
\Rightarrow {T^2} \propto {g^{ - 1}}$

So, the correct answer is “Option C”.

Note:
The time period of the oscillation of a pendulum is independent of the mass of the bob. When a body is allowed to oscillate freely it will oscillate with a particular frequency. Such oscillations are called free oscillations. The frequency of free oscillations is called natural frequency. A pendulum that is adjusted in a way that it has a period of two seconds is called a second’s pendulum.
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