Given below are two statements:

Statement I : A second's pendulum has a time period of 1 second.

Statement II : It takes precisely one second to move between the two extreme positions.

In the light of the above statements, choose the correct answer from the options given below:

The angular frequency of a simple pendulum is Ω rad/s. Now the length is made one fourth of the original length, the angular frequency becomes

The motion of a simple pendulum executing S.H.M. is represented by the following equation. $y=Asin\left(\pi t+\varphi \right),$  where time is measured in second.

The length of pendulum is

Time period of a simple pendulum is T inside a lift when the lift is stationary. If the lift moves upwards with an acceleration g / 2, the time period of pendulum will be:

A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 50 cm long. The speed of the bob, when the length makes an angle of 60° to the vertical will be _______ m/s. (g = 10 m/s²)

Time period of a simple pendulum is T. The time taken to complete <!-- [if gte mso 9]><xml> </xml><![endif]-->  $\frac{5}{8}$ oscillations starting from mean position is $\frac{\alpha }{\beta }T.$ <!-- [if gte mso 9]><xml> </xml><![endif]--> The value of a is………..