Question

If the effective length of a simple pendulum is equal to the radius of the Earth $\left(R\right)$, the time period will be,
A. $T=\pi \sqrt{{\displaystyle \frac{R}{g}}}$
B. $T=2\pi \sqrt{{\displaystyle \frac{2R}{g}}}$
C. $T=2\pi \sqrt{{\displaystyle \frac{R}{g}}}$
D. $T=2\pi \sqrt{{\displaystyle \frac{R}{2g}}}$

Answer 1

Hint: This problem can be solved by using the formula for the time period of the pendulum, when the effective length of the pendulum is comparable to the radius of the Earth. By using this formula and plugging in the effective length of the pendulum as given in the question, we can get the required time period.

Formula used:
For a pendulum with effective length $l$, comparable to the radius of the earth $R$, the time period $T$ is given by
$T=2\pi \sqrt{{\displaystyle \frac{1}{g({\displaystyle \frac{1}{l}}+{\displaystyle \frac{1}{R}})}}}$
where $g$ is the acceleration due to gravity.

Complete step by step answer:
When the effective length of a pendulum is comparable to the radius of the earth, then its tangential acceleration cannot be considered the same throughout its motion, since the value of acceleration due to gravity $g$ varies with height above the surface of the earth.
Hence, a different general formula exists for the time period in this case.
For a pendulum with effective length $l$, comparable to the radius of the earth $R$, the time period $T$ is given by
$T=2\pi \sqrt{{\displaystyle \frac{1}{g({\displaystyle \frac{1}{l}}+{\displaystyle \frac{1}{R}})}}}$ --(1)
where $g$ is the acceleration due to gravity.
Now, let us analyze the question.
The radius of the earth is $R$
Let the required time period of the pendulum be $T$.
We are given the effective length of the pendulum $\left(l\right)$ is equal to the radius of the earth. Hence,
$l=R$
Therefore, putting this value and using equation (1), we get,
$T=2\pi \sqrt{{\displaystyle \frac{1}{g({\displaystyle \frac{1}{R}}+{\displaystyle \frac{1}{R}})}}}=2\pi \sqrt{{\displaystyle \frac{1}{g\left({\displaystyle \frac{2}{R}}\right)}}}=2\pi \sqrt{{\displaystyle \frac{R}{2g}}}$
Hence, the time period of the pendulum is $2\pi \sqrt{{\displaystyle \frac{R}{2g}}}$.
Therefore, the correct option is D) $T=2\pi \sqrt{{\displaystyle \frac{R}{2g}}}$.

Note: The general formula (1) for the time period of a pendulum actually becomes the well known formula for the time period of a pendulum, that is, $T=2\pi \sqrt{{\displaystyle \frac{l}{g}}}$. This happens when $l$ is so small in comparison to $R$ that ${\displaystyle \frac{1}{l}}\gg {\displaystyle \frac{1}{R}}$. Hence, it follows that ${\displaystyle \frac{1}{l}}+{\displaystyle \frac{1}{R}}\approx {\displaystyle \frac{1}{l}}$. This approximation leads to the well known formula for the time period of a simple pendulum.
Therefore, students must be careful and judge which formula must be used after analyzing the question. If the effective length of the pendulum is comparable to the radius of the earth, then the student should employ formula (1) and not the well-known approximation. Doing otherwise, will lead to an error in the answer.