Simple Pendulum

What is Foucault's pendulum?

At what height from the surface of earth, the value of 'g' is the same as that at a depth of $160 \mathrm{km}$ below the surface of the earth? The radius of the earth is $6400 \mathrm{km}$.

Explain how the value of 'g' changes with the depth .

A simple pendulum is set up in a trolly which moves to the right with an acceleration $\mathrm{a}$ on a horizontal plane. Calculate the angle, which the thread of the pendulum makes with the vertical 

What is horizontal and vertical acceleration ?

A simple pendulum has a time period $T_1$, when on earth's surface, and $T_2$ when taken at a height $\mathrm{R}$ above the earth's surface where $R$ is the radius of earth. Find the ratio of two time periods.

At what depth below the surface of the earth, The acceleration due to gravity becomes (i) half and (1ii) $\frac{1}{4}\mathrm{th}$ times that of the earth. Given the radius of the earth =$6400 \mathrm{km}$

Calculate the height from the surface of the earth, at which the value of 'g' decreases by $64\%$ as compared to that at the earth?

( radius of the earth =$64000 \mathrm{km}$)

The value of acceleration due to gravity at a planet is $\frac{1}{4}\mathrm{th}$ that of earth. If the time period of simple pendulum is $2 \mathrm{s}$ on the earth, then what will be the time period on the planet?

A uniform metal rod is used as a bar pendulum. If the room temperature rises by $10 °\mathrm{C}$ and the coefficient of linear expansion of the metal of the rod is $2\times {10}^{-6}$ per $°\mathrm{C}$, the period of the pendulum will have percentage increase of

Two pendulums begin to swing simultaneously. The first pendulum makes nine full oscillations when the other makes seven. The ratio of the lengths of the two pendulums is

A person measures a time period of a simple pendulum inside a stationary lift and finds it to be $T$. If the lift starts accelerating upwards with an acceleration $\left(\frac{g}{3}\right)$, the time period of the pendulum will be

A vertical spring mass system has the same time period as simple pendulum undergoing small oscillations. Now, both of them are put in an elevator going downwards with an acceleration $5 \mathrm{m} {\mathrm{s}}^{-2}$. The ratio of time period of the spring mass system to the time period of the pendulum is (assume that acceleration due to gravity, $g=10 \mathrm{m} {\mathrm{s}}^{-2}$),

A simple pendulum is placed inside a lift which is moving with a uniform acceleration. If the time periods of the pendulum while the lift is moving upwards and downwards are in the ratio $1:2$, then the acceleration of the lift is (acceleration due to gravity, $g=10 \mathrm{m} {\mathrm{s}}^{-2}$),

A cabin is falling freely under gravity, what is the time period of a pendulum attached to its ceiling?

The time period of a simple pendulum inside a stationary lift is $\sqrt{5} \mathrm{s}$. What will be the time period when the lift moves upwards with an acceleration $\frac{g}{4}?$

The length of a pendulum is decreased by $10⁒$. What will be the change in its frequency?

If the length of a simple pendulum is increased by $25\%$, then what is the change in its time period?

The time period of a simple pendulum depends on the 

A simple pendulum of length $1 \mathrm{m}$ has mass $10 \mathrm{g}$ and oscillates freely with amplitude of $5 \mathrm{cm}$. Its potential energy at extreme position is