Question

Two simple pendulums of length 5cm and 20cm respectively are given small linear displacement in one direction at the same time. They will again be in phase when the pendulum of shorter length has compared ‘n’ oscillations the value of n is
A. $5$
B. $1$
C. $2$
D. $3$

Answer 1

First find the time periods of oscillations of the two pendulums. Then when they again come in phase, write the expressions for the time taken by each. Equate the two expressions and find the relation between the number of oscillations that each completes.

Formula used:
The time period of a simple pendulum for small linear oscillations is given as,
$T=\sqrt{\dfrac{l}{g}}$
where l is the length of the pendulum and g is the acceleration due to gravity.

Complete step by step answer:
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 pendulum of shorter length (${{l}_{1}}$) be ${{T}_{1}}$ and the time period of the pendulum of longer length (${{l}_{2}}$) be ${{T}_{2}}$. This means that ${{T}_{1}}=\sqrt{\dfrac{{{l}_{1}}}{g}}$ and ${{T}_{2}}=\sqrt{\dfrac{{{l}_{2}}}{g}}$.

It is given that the two pendulums are given the same linear displacement at the same time.It is said that both of them come into phase again when the pendulum of shorter length completes ‘n’ oscillations. Let they come into phase after time t and let the longer pendulum complete (n’) oscillations within this time.Therefore, we can say that the time taken to complete n oscillations by the shorter pendulum is $t=n{{T}_{1}}$.And the time taken by the longer pendulum to complete n’ oscillations is $t=n'{{T}_{2}}$.
$t=n{{T}_{1}}\\\ \Rightarrow t =n'{{T}_{2}}$
$\Rightarrow \dfrac{n'}{n}=\dfrac{{{T}_{1}}}{{{T}_{2}}}$.
Substitute the values of ${{T}_{1}}$ and ${{T}_{2}}$.
$\dfrac{n'}{n}=\dfrac{\sqrt{\dfrac{{{l}_{1}}}{g}}}{\sqrt{\dfrac{{{l}_{2}}}{g}}}\\\ \Rightarrow\dfrac{n'}{n}=\sqrt{\dfrac{{{l}_{1}}}{{{l}_{2}}}}$.
It is given that ${{l}_{1}}=5cm$ and ${{l}_{2}}=20cm$.
Substitute these values in the above equation.
$\dfrac{n'}{n}=\sqrt{\dfrac{5}{20}}\\\ \Rightarrow \dfrac{n'}{n} =\sqrt{\dfrac{1}{4}}\\\ \therefore \dfrac{n'}{n} =\dfrac{1}{2}$.
This means that when the longer pendulum completes 1 oscillation, the shorter pendulum will complete 2 oscillations and this will be the first time that they will be in phase after the start they left. Hence, the value of n is 2.

Therefore, the correct option is C.

Note: It is not the two pendulums will come in phase only when $n=2$ and $n'=1$.They will come in phase again and again after equal time intervals. For example, the next time they will be in phase is when $n=4$ and $n'=2$.Note that n must be a natural number because we are concerned with complete oscillations.