Simple Pendulum

The length of a seconds-pendulum on the surface of earth is $1 \mathrm{m}.$ Its length on the surface of the moon would be $\underline{ }$

A simple pendulum, suspended from the ceiling of a lift, has a period of oscillation $T,$ when the lift is at rest. If the lift starts moring upwards with an acceleration $a=3\mathrm{g},$ then the new period will be

What is the number of degrees of freedom for an oscillating simple pendulum?

A simple pendulum is executing simple harmonic motion with a time period T. If the length of the pendulum is increased by 21%, the increase in the time period of the pendulum of increased length is 

In an experiment to determine the gravitational acceleration $g$ of a place with the help of a simple pendulum, the measured time period squared is plotted against the string length of the pendulum in the figure. What is the value of $g$ at the place?

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

As shown in the above figure, a cart is accelerating in the horizontal direction with a constant acceleration of $\sqrt{3}g$ and a $50 \mathrm{cm}$ long simple pendulum is suspended from its roof. Now, find the time period of small oscillations of this pendulum in second about its equilibrium position taking $g={\mathrm{\pi }}^2 \mathrm{m} {\mathrm{s}}^{-2}$:

Time period of a simple pendulum is $T.$ The time taken to complete $\frac{5}{8}$ oscillations starting from mean position is $\frac{\alpha }{B}T.$ The value of $\alpha$ is _______ .

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 

If the inertial mass and gravitational mass of the simple pendulum of length $l$ are not equal, then the time period of the simple pendulum is

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

A simple pendulum with a bob of mass $40 \mathrm{g}$ and charge $+2 \mathrm{\mu C}$ makes $20$ oscillations in $44$ seconds. A vertical electric field of magnitude $4.2\times {10}^4 {\mathrm{NC}}^{-1}$ pointing downward is applied. The time taken by the pendulum to make $15$ oscillations in the electric field is (Acceleration due to gravity $=10 {\mathrm{ms}}^{-2}$)