The time period of a simple pendulum is independent of its:

The length of simple pendulum is increased by 1%. Its time period will

A simple pendulum of length L is taken to a planet where the acceleration due to gravity is four times that of Earth. What will be its new time period?

A particle moves so that its acceleration a is given by $a = - bx$ where $x$ is displacement from equilibrium position and b is a non-negative real constant. The time period of oscillation of the particle is

A child swings on a playground swing, undergoing simple harmonic motion. If the maximum acceleration of the child is $16 \, m/s^2$ when the swing is 4 cm from its equilibrium position, what is the period of the swing's motion?

A disc of radius $R$ and mass $M$ is pivoted at the rim and is set for small oscillations. If simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should be

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

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

A particle of mass m is located in a one dimensional potential field where potential energy is given by $\left( x \right) = A(1 - \cos px)$ where A and p are constants. The period of small oscillations of the particle is

If a simple harmonic is represented by \frac{{{d^2}x}}{{d{t^2}}}$$ + \alpha x = 0,its time period is

A seconds pendulum at the equator $\left[ {g = 9.8\,\,m/{s^2}} \right]$ is taken to the poles $\left[ {g = 9.83\,\,m/{s^2}} \right]$. If it is to act as second’s pendulum, its length should be