Question

The periodic time of a simple pendulum of length 1 m and amplitude 2 cm is 5 seconds. If the amplitude is made 4 cm, its periodic time in seconds will be
\[\begin{align}
  & \text{A}.\text{ }2.5 \\
 & \text{B}.\text{ }5 \\
 & \text{C}.10 \\
 & \text{D}.\text{ }5\sqrt{2} \\
\end{align}\]

Answer 1

Hint: The time period of a simple pendulum is the time taken by a pendulum to complete one full oscillation. The maximum displacement of the bob in the pendulum is the amplitude of that pendulum.

Formula used:
Time period of a simple pendulum,
$T=2\pi \sqrt{\dfrac{l}{g}}$ .

Complete step by step answer:
In the question we are given the length of the pendulum,$l$ = 1 m
Amplitude of the pendulum, A = 2cm
And the time period of this pendulum, T = 5 seconds.
We have to find the time period of this pendulum, when its amplitude becomes 4 cm.
We know, time period of a simple pendulum is given by the equation,
$T=2\pi \sqrt{\dfrac{l}{g}}$ , where ‘$l$’ is the length of the pendulum and ‘g’ is acceleration due to gravity.
From this equation, it is clear that the time period of a simple pendulum does not depend on its amplitude.
 In the question, we change the amplitude of the pendulum from 2 cm to 4 cm. Length of the pendulum remains the same and acceleration due to gravity; ‘g’ is a constant.
Therefore, the time period of the pendulum when its amplitude = 4cm, length ‘$l$’=1 m will be 5 seconds.
So, the correct answer is “Option B”.

Note:
Time period of simple pendulum
For a simple pendulum, we know its angular frequency $\omega $ is given by
$\omega =\sqrt{\dfrac{g}{l}}$ , where ‘$l$’ is the length of the pendulum and ‘g’ is acceleration due to gravity.
Time period of an oscillation is generally expressed as,
$T=\dfrac{2\pi }{\omega }$ , where ‘T’ is the time period and ‘$\omega $’ is the angular frequency of the pendulum.
By substituting the value of angular frequency (ω) in the above equation, we get
$T=2\pi \sqrt{\dfrac{l}{g}}$
Therefore the time period of a pendulum is, $T=2\pi \sqrt{\dfrac{l}{g}}$