
The time period of a simple pendulum at the centre of earth is:
A 0
B ∞
C Less than 0
D none
Answer
578.1k+ views
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.
Complete step by step answer:
To simply dene simple pendulum, it is a bob that comprises of a molecule of mass ‘m’ suspended by a massless non-expandable string of length ‘L’ in an area of space in which there is a consistent, uniform gravitational field, e.g. near the surface of the earth. The suspended molecule is called the pendulum bob.
We know that the time period of a simple pendulum is given by:
$T = 2\pi \sqrt {\dfrac{L}{g}} $
Here, T is the time period, L is the length of the pendulum and g is the acceleration due to gravity.
Here, at the centre of the earth the value g is equal to zero. Therefore, the time period becomes:
$\begin{array}{c}
T = 2\pi \sqrt {\dfrac{L}{0}} \\
= \infty
\end{array}$
Therefore, the correct option is B.
Additional Information:If we pull the pendulum bob to one side and let it go, we find that it swings to and fro, meaning that it oscillates. At this point, we are not aware of whether or not the bob goes through simple harmonic motion, but we know that it oscillates.
Note:When the spherical bob of the pendulum supplants at the primary angle and then liberated, the pendulum is observed to move in a repeated back-and-forth motion periodically. This motion is called the oscillatory motion. The central point of oscillation is termed as an equilibrium position.
Complete step by step answer:
To simply dene simple pendulum, it is a bob that comprises of a molecule of mass ‘m’ suspended by a massless non-expandable string of length ‘L’ in an area of space in which there is a consistent, uniform gravitational field, e.g. near the surface of the earth. The suspended molecule is called the pendulum bob.
We know that the time period of a simple pendulum is given by:
$T = 2\pi \sqrt {\dfrac{L}{g}} $
Here, T is the time period, L is the length of the pendulum and g is the acceleration due to gravity.
Here, at the centre of the earth the value g is equal to zero. Therefore, the time period becomes:
$\begin{array}{c}
T = 2\pi \sqrt {\dfrac{L}{0}} \\
= \infty
\end{array}$
Therefore, the correct option is B.
Additional Information:If we pull the pendulum bob to one side and let it go, we find that it swings to and fro, meaning that it oscillates. At this point, we are not aware of whether or not the bob goes through simple harmonic motion, but we know that it oscillates.
Note:When the spherical bob of the pendulum supplants at the primary angle and then liberated, the pendulum is observed to move in a repeated back-and-forth motion periodically. This motion is called the oscillatory motion. The central point of oscillation is termed as an equilibrium position.
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