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# If the length of a simple pendulum is increased by $44\%$ then what is the change in the time period of the pendulum?

Last updated date: 09th Aug 2024
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Hint: A simple pendulum of length l is increased by $44\%$. We have to find the change in the time period of the pendulum. The change in the time period of the pendulum in percentage is given by $\Rightarrow \left( {\dfrac{{T'}}{T} - 1} \right) \times 100$
Where
$T$ is the actual time period of the pendulum.
$T'$ is the time period of the pendulum after increasing the length.

Complete step by step solution:
A simple pendulum has a string with a very small mass (sphere ball). It is small but strong enough not to stretch. The mass suspended from the light string that can oscillate when displaced from its rest position.
Pendulums are in common usage such as in clocks, child’s swing and some are for fun. For small displacements, a pendulum is a simple harmonic oscillator.
The pendulum’s time period is proportional to the square root of the length of string and inversely proportional to the square root of acceleration due to gravity.
The time period of a simple pendulum is given by
$T \propto \sqrt {\dfrac{l}{g}}$
$T = 2\pi \sqrt {\dfrac{l}{g}}$
Where,
T is the time period
l is the length of the string
g is the acceleration due to gravity
Given,
Length of simple pendulum is increased by $44\%$

Which means the length of the string, $l = l + \dfrac{{44l}}{{100}}$
The time period of a simple pendulum is given by
$T = 2\pi \sqrt {\dfrac{l}{g}} {\text{ }} \to {\text{1}}$
Let the increase in time period be $T'$
Substitute the known value
$T' = 2\pi \sqrt {\dfrac{{l + \dfrac{{44l}}{{100}}}}{g}}$
$T' = 2\pi \sqrt {\dfrac{{l\left( {1 + \dfrac{{44}}{{100}}} \right)}}{g}}$
$T' = 2\pi \sqrt {\dfrac{{l\left( {\dfrac{{144}}{{100}}} \right)}}{g}}$
$T' = 2\pi \left( {\dfrac{{12}}{{10}}} \right)\sqrt {\dfrac{l}{g}}$
$T' = (1.2) \times 2\pi \sqrt {\dfrac{l}{g}}$
From equation 1 we get
$T' = (1.2) \times T$
$\dfrac{{T'}}{T} = (1.2){\text{ }} \to {\text{2}}$
The change in the time period of the pendulum is given by
$\Rightarrow \left( {\dfrac{{T'}}{T} - 1} \right) \times 100$
From equation 2 we get
$\Rightarrow \left( {\dfrac{{T'}}{T} - 1} \right) \times 100 = \left( {1.2 - 1} \right) \times 100$
$\Rightarrow \left( {\dfrac{{T'}}{T} - 1} \right) \times 100 = 20\%$
The change in the time period of the pendulum is $20\%$

Note: In percentage we write the change in the time period of the pendulum as $20\%$. If we write it numerically, the change in the time period of the pendulum is 0.2s.
The percentage change in time period of the pendulum is $20\%$
Numerically, it is$\Rightarrow \dfrac{{20}}{{100}} = 0.2s$