Question

Calculate the time period of a simple pendulum of length $1.12m,$ when acceleration due to gravity is $9.8m{s^{ - 2}}.$
A. $0$
B. $1.2$
C. $2.12$
D. $4.24$

Answer 1

The time periodic of a simple pendulum and its dependence on length is to be used to solve this also the motion of a simple pendulum is SHM.
Time period of a simple pendulum is T$= 2\pi \sqrt {\dfrac{\ell }{g}}$
Where $\ell$the length of the pendulum and g is acceleration due to gravity.

Complete step by step answer:
An ideal simple pendulum consists of a point mass suspended by a flexible, inelastic and weightless string of rigid support of infinite mass.
We know that in SHM, the equation is given
a$= - {\omega ^2}x\,\,..........\left( 1 \right)$
And Time period, $T = \dfrac{{2\pi }}{\omega }.........\left( 2 \right)$
Where a is the acceleration
$\omega$ is the angular frequency
x is displacement from mean position.
Now, as Simple pendulum also performs SHM and its equation is given by
a$= \dfrac{{ - gx}}{\ell }\,..........\left( 3 \right)$
Where $\ell$is length of pendulum
On comparing (1) and (3), we have
${\omega ^2} = \dfrac{g}{\ell }$
$\implies \omega = \sqrt {\dfrac{g}{\ell }}$
So, Time period, $T = \dfrac{{2\pi }}{\omega } = 2\pi \sqrt {\dfrac{\ell }{g}}$
Here, length of pendulum is given to be
$\ell = 1.12m$
So, Time period becomes
$T = 2\pi \sqrt {\dfrac{{1.12}}{{9.8}}}$
$\implies T = 2 \times \dfrac{{22}}{7}\sqrt {\dfrac{{1.12}}{{9.8}}} \,\,\,\,\,\,\,\,\,\left( {as\,\,\pi = \dfrac{{22}}{7}} \right)$
$\implies T = \dfrac{{44}}{7} \times \sqrt {0.114}$
$\implies T = \dfrac{{44}}{7} \times 0.338$
So, $T = 2.12\sec$
Therefore, Time period of simple pendulum is $2.12\sec$

So, the correct answer is “Option C”.

Note:
Remember that simple pendulum performs SHM can be made clear from its equation $a = \dfrac{{ - g}}{\ell }x.$ From this it is clear that acceleration is proportional to its displacement and is directed opposite to that of displacement which is nothing but SHM.