Question

Two pendulums differ in lengths by $\;22cm$ . They oscillate at the same place such that one of them makes $\;15$ oscillations and the other makes $\;18$ oscillations during the same time. The lengths (in $\;cm$ ) of the pendulums are:-
(A) $\;72$ and $\;50$
(B) $\;60$ and $\;38$
(C) $\;50$ and $\;28$
(D) $\;80$ and $\;58$

Answer 1

Use the formula for the time period of a simple pendulum. This formula gives the relation between the time period of a simple pendulum, length of the simple pendulum, and acceleration due to gravity. Recall the concept of the second’s pendulum and the time period of this second’s pendulum. Substitute this value in the formula and determine the length of the second’s pendulum.

Formula used:
The time period $T$ of a simple pendulum is known by
$T = 2\pi \sqrt {\dfrac{L}{g}}$ ........... $\left( 1 \right)$
Here, $L$ is the length of the simple pendulum and $g$ is the acceleration due to gravity.

Complete step-by-step solution:
Let us assume that the two given pendulums are at the same altitude due to which the value of acceleration due to the gravitational force will be the same. Let the time period of the first pendulum having length say ${L_1}$ be ${T_1}$ and the time period of the other pendulum having length say ${L_2}$ be ${T_2}$ . This means that
${T_1} = 2\pi \sqrt {\dfrac{{{L_1}}}{g}}$ and ${T_2} = \sqrt {\dfrac{{{L_2}}}{g}}$ .
It is given that the two pendulums are given the same displacement at the same time. It is said that both of them come into phase again when the pendulum of length ${L_1}$ completes $\;15$ oscillations. Let them come into phase after time $t$ and the other pendulum completes $\;18$ oscillations within this time. Therefore, we can say that the time taken to complete $\;15$ oscillations by the first pendulum is $t = 15{T_1}$ and the time taken by the other pendulum to complete $\;18$ oscillations is $t = 18{T_2}$ .
$t = 15{T_1}$ ……… $\left( 2 \right)$
$\Rightarrow t = 18{T_2}$ ………. $\left( 3 \right)$
Divide both the equations to get the ratio,
$\Rightarrow \dfrac{{18}}{{15}} = \dfrac{{{T_1}}}{{{T_2}}}$
Substitute the values of ${T_1}$ and ${T_2}$ .
$\dfrac{{18}}{{15}} = \dfrac{{\sqrt {\dfrac{{{L_1}}}{g}} }}{{\sqrt {\dfrac{{{L_2}}}{g}} }}$
$\Rightarrow \dfrac{{18}}{{15}} = \sqrt {\dfrac{{{L_1}}}{{{L_2}}}}$
On further simplifying the above equation we get,
$\Rightarrow \dfrac{6}{5} = \sqrt {\dfrac{{{L_1}}}{{{L_2}}}}$
On squaring both the sides we get,
$\Rightarrow \dfrac{{{L_1}}}{{{L_2}}} = \dfrac{{36}}{{25}}$ …………. $\left( 4 \right)$
As the question says that two pendulums differ by lengths by $\;22cm$ then let us assume that the length of the first pendulum is larger than the other one. Therefore,
${L_1} - {L_2} = 22cm$ ………… $\left( 5 \right)$
Using the dividendo rule in the equation $\left( 4 \right)$ ,
$\Rightarrow \dfrac{{11}}{{25}} = \dfrac{{{L_1} - {L_2}}}{{{L_2}}}$
Using the equation $\left( 5 \right)$ we can write,
$\Rightarrow \dfrac{{11}}{{25}} = \dfrac{{22}}{{{L_2}}}$
$\Rightarrow {L_2} = 50cm$
Put the value of ${L_2}$ in equation $(4)$ we get,
$\Rightarrow \dfrac{{{L_1}}}{{50}} = \dfrac{{36}}{{25}}$
$\Rightarrow {L_1} = \dfrac{{50 \times 36}}{{25}}$
On further solving the equation we get,
$\Rightarrow {L_1} = 72cm$

Hence, the correct option is (A).

Note: To complete one whole swing pendulum requires a certain period of time that is called a time period of the pendulum. The time period is usually concerned and gets affected by the length and the strength of gravity. Remember there is no effect in the time period if we change the mass of the hanging object.