Simple Pendulum

What is the second's pendulum? Find its length and frequency.

Derive the expression for the time period of a pendulum of infinite length. Show that this time is finite.

What is the time period of a second pendulum?
(Write the numerical value in terms of SI unit of time).

In case of a simple pendulum, time period versus length is depicted by

When a simple pendulum is taken from equator to poles, its period

A simple pendulum is suspended from the ceiling of a stationary lift. Its time period is measured as $\mathrm{T}$. If the lift accelerates downward, its time period will be

A simple pendulum is suspended from the ceiling of a stationary lift. Its time period is measured as $T$. If the lift accelerates upward its time period will be 

Obtain an expression for period of simple pendulum. On which factors it depends?

Show that under certain conditions, a simple pendulum performs linear $\mathrm{S}.\mathrm{H}.\mathrm{M}.$

A simple pendulum is used in physics laboratory for experiment to obtain experimental value of  gravitational acceleration $\mathrm{g}.$ A student measures the length of pendulum $0.51\mathrm{m},$ displaces it through $10°$ from equilibrium position and released it. Using a stopwatch, the student measures period of pendulum as $1.44\mathrm{s}.$ Determine the experimental value of the gravitational acceleration.

The period of oscillation of simple pendulum increases by $20\%$ when its length is increased by $44 \mathrm{cm}$. Find its (i) initial length (ii) initial period of oscillation.

Define conical pendulum. Obtain an expression for the angle made by string of conical pendulum with vertical. Hence deduce the expression for linear speed of bob of the conical pendulum.

Find the time period of a simple pendulum when it is taken $64 \mathrm{km}$ above the earth's surface.

The time period of vibrations of a simple pendulum varies directly as the _____ of the length of the pendulum. (square / cube/ square root/ cube root).

A pendulum beats seconds. Its time period is _____ (2/6/5/9)seconds.

What is the time period of a second pendulum?
(Write the numerical value in terms of SI unit of time).

The time period of a simple pendulum on the surface of the earth is $4\mathrm{ }\mathrm{s}$. Its time period on the surface of the moon is (consider the acceleration due to gravity on moon to be $\frac{1}{6}$ times the acceleration due to gravity on Earth).

The acceleration due to gravity on the surface of the moon is $1.7\mathrm{ }\mathrm{m}\mathrm{ }{\mathrm{s}}^{-2}$. The time period of a simple pendulum on the moon if its time period on the earth is $3.5\mathrm{ }\mathrm{s}$ is $\left(\mathrm{G}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n},\mathrm{ }g=9.8\mathrm{ }\mathrm{m}\mathrm{ }{\mathrm{s}}^{-2}\right)$

A simple pendulum is set up in a trolly that moves to the right with an acceleration $a$ on a horizontal plane. Then, the thread of the pendulum in the mean position makes an angle with the vertical

If the displacement of the simple pendulum at any time is $0.02 \mathrm{m}$ and acceleration is $2\mathrm{ }\mathrm{m} {\mathrm{s}}^{-2}$, then in this time angular velocity will be