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Define second’s pendulum. Hence calculate the length of second’s pendulum.

Last updated date: 23rd Apr 2024
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Hint: Second’s pendulum takes one second to move from its mean position to one of its extreme position and use the expression of the time period of a simple pendulum.

Complete answer:
A second’s pendulum is a type of a simple pendulum whose time period of vibration is two seconds, that is it takes one second to move from a mean position to its extreme position and one second to move to another extreme point.
We can also define as the bob of the second’s pendulum takes exactly one second while oscillating through the mean position.
Let us consider a bob of mass $m$ is suspended by a weightless, inflexible and inelastic string of length $l$ from a rigid support, and then the expression for the time period of the simple pendulum is,
$T = 2\pi \sqrt {\dfrac{l}{g}} $ ... (1)
Here, $g$ is the acceleration due to gravity and $l$ is the length of the pendulum.
We know that the time period of the vibration of the second's pendulum is $T = 2\;{\rm{s}}$.
Let us rewrite the equation (1),
$l = g{\left( {\dfrac{T}{{2\pi }}} \right)^2}$

Now we substitute the values $T$ as $2\;{\rm{s}}$ and $g$ as $9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ in the above expression, we get,
l = 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}{\left( {\dfrac{{2\;{\rm{s}}}}{{2\pi }}} \right)^2}\\
 = 0.993\;{\rm{m}}
$l = 99.3\;{\rm{cm}}$

Hence, the length of the second’s pendulum is 99.3 cm.

Additional information: The frequency of the second’s pendulum is equal to $\dfrac{1}{2}\;{\rm{Hz}}$.

Note: The assumptions we take while obtaining the expression of time period are:
1. Air resistance is negligible.
2. The bob of the pendulum swings in a perfect plane.