A simple pendulum of length $L$ has a time period $T$. If the length is increased by $\Delta L$ such that $\Delta L << L$, the change in time period $\Delta T$ is approximately:

The time period of a simple pendulum, when it is made to oscillate on the surface of moon

A uniform rod of length 2.0 $m$ is suspended through an end is set into oscillation with small amplitude under gravity. The time period of oscillation is approximately

How does the time period of pendulum vary with length

A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is $U\left( x \right) = k{\left[ x \right]^3}$, where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simpleharmonic oscillator, The acceleration of the bob of the pendulum is ${\rm{20}}\,\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ at a distance of ${\rm{5}}\,\,{\rm{m}}$ from the meanposition. The time period of oscillation if

If the length of a pendulum is made 9 times and mass of the bob is made 4 times then the value of time period becomes

A second’s pendulum is placed in a space laboratory orbiting around the earth at a height 3$R$, where $R$ is the radius of the earth. The time period of the pendulum is

A simple pendulum is oscillating with an amplitude $A$. At what displacement from the mean position is its kinetic energy equal to its potential energy?

The length of a simple pendulum is increased by 44%. What is the percentage increase in its time period?

To show that a simple pendulum executes simple harmonic motion, it is necessary to assume that