What happens to the time period of a simple pendulum when it is taken to moon's surface from earth's surface ?
What happens to the time period of a simple pendulum when it is taken to moon's surface from earth's surface ?
The value of $g$ on Moon is less than that on Earth and according to $T \propto \frac{1}{\sqrt{g}}$ hence, time period is more on Moon.
Similar Questions
The length of a second's pendulum on the surface of earth is $1\, m$. What will be the length of a second's pendulum on the moon ?
The length of a second's pendulum on the surface of earth is $1\, m$. What will be the length of a second's pendulum on the moon ?
A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle $(\theta)$ of thread deflection in the extreme position will be :
A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle $(\theta)$ of thread deflection in the extreme position will be :
- [JEE MAIN 2024]
A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration $a$, then the time period is given by $T = 2\pi \sqrt {\frac{l}{{g'}}} $, where $g'$ is equal to
A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration $a$, then the time period is given by $T = 2\pi \sqrt {\frac{l}{{g'}}} $, where $g'$ is equal to
- [AIPMT 1991]
In case of a simple pendulum, time period versus length is depicted by
In case of a simple pendulum, time period versus length is depicted by
What is the length of a simple pendulum, which ticks seconds?
What is the length of a simple pendulum, which ticks seconds?