
The time period of a simple pendulum at the centre of earth is:
A 0
B ∞
C Less than 0
D none
Answer
516.9k+ 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.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE
