Question

A man measures the period of a simple pendulum inside a stationary lift and finds it to be $T$ second. If the lift accelerates downwards with acceleration of $\dfrac{g}{4}$, the period of oscillation will be:
(A) $T \times \dfrac{{\sqrt 3 }}{2}{\text{s}}$
(B) $T \times \dfrac{2}{{\sqrt 3 }}{\text{s}}$
(C) $\dfrac{T}{2}{\text{s}}$
(D) $\sqrt T {\text{s}}$

Answer 1

First of all, we find the resultant acceleration of the pendulum under the said motion and find the expression of time period of a simple pendulum. The new acceleration gives a whole new time period. We compare the two different time periods obtained from the two expressions and manipulate accordingly.

Complete step by step answer:
Given, the time period which is measured in a stationary lift is $T$ second.
When the lift starts moving down with acceleration of $\dfrac{g}{4}$, then the effective acceleration due to gravity is reduced, and there will be an impact on the time period of the pendulum.
We know,
When the lift is accelerating downward, the resultant acceleration due to gravity can be written as:
$g' = g - a$ …… (1)
Where,
$g'$ indicates resultant acceleration due to gravity.
$g$ indicates acceleration due to gravity.
$a$ indicates the acceleration with which the lift is going downwards.
Equation (1) can be re-written as:

$$g' = g - a \\\ \Rightarrow g' = g - \dfrac{g}{4} \\\ \Rightarrow g' = \dfrac{{3g}}{4} \\\$$

We know, the time period of a simple pendulum is given by:
$T = 2\pi \sqrt {\dfrac{l}{g}}$ …… (2)
Where,
$T$ indicates the time period of the simple pendulum.
$l$ indicates the length of the pendulum.
$g$ indicates the acceleration due to gravity.
But when the lift is accelerating downwards, the time period will change which can be written as:
$T' = 2\pi \sqrt {\dfrac{l}{{g'}}}$ …… (3)
$T'$ indicates the new time period of the simple pendulum.
$l$ indicates the length of the pendulum.
$g'$ indicates the acceleration with which the person is going down.
Substituting, $g' = \dfrac{{3g}}{4}$ in the equation (3), we get:

T' = 2\pi \sqrt {\dfrac{l}{{g'}}} \\\ \Rightarrow T' = 2\pi \sqrt {\dfrac{l}{{\left( {\dfrac{{3g}}{4}} \right)}}} \\\ \Rightarrow T' = 2\pi \sqrt {\dfrac{{4l}}{{3g}}} \\\ \Rightarrow T' = 2\pi \times \dfrac{2}{{\sqrt 3 }} \times \sqrt {\dfrac{l}{g}} \\\ $$ ……(4) Now, dividing the equation (4) by equation (3), we get:

\dfrac{{T'}}{T} = \dfrac{{2\pi \times \dfrac{2}{{\sqrt 3 }} \times \sqrt {\dfrac{l}{g}} }}{{2\pi\sqrt{\dfrac{l}{g}} }} \\
\Rightarrow\dfrac{{T'}}{T} = \dfrac{{\dfrac{2}{{\sqrt 3 }}}}{1} \\
\Rightarrow\dfrac{{T'}}{T} = \dfrac{2}{{\sqrt 3 }} \\
\therefore T' = T \times \dfrac{2}{{\sqrt 3 }},{\text{s}} \\

$$**Hence, the correct option is (B).** **Note:** To solve this problem, it would require the concept of apparent acceleration due to gravity. Resultant acceleration will decrease if a body is going down while it will increase if a body is going up with certain acceleration. Remember, when the acceleration increases, time period decreases, or vice-versa.$$