The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle `theta` and then released. The bob will then pass through its equilibrium position with a speed v, where v equals

The bob of pendulum of length l is pulled aside from its equilibrium position through an angle theta and then released. Find the speed v with which the bob passes through the equilibrium position.

The bob of a simple pendulum of length 1 m is' pulled aside from its equilibrium’ position through an angle 60^@ and released. Calculate the speed at which the bob passes the equilibrium position.

Show that a simple pendulum bob which has been pulled aside from its equilibrium position through an angle theta and the released will pass through an equilibrium position with speed v=sqrt(2gl(1-cos theta)) , where/is the length of the pendulum.

A pendulum bob of mass m is displaced through an angle theta from its equilibrium position and then released. What will be the tension in the string when the bob crosses the equilibrium position?

The mass of a simple pendulum bob is 100 gm. The length of the pendulum is 1 m. The bob is drawn aside from the equilibrium position so that the string makes an angle of 60^(@) with the vertical and let go. The kinetic energy of the bob while crossing its equilibrium position will be

A pendulum is displaced to an angle theta from its equilibrium position, then it will pass through its mean position with a velocity v equal to

A pendulum is displaced to an angle theta from its equilibrium position, then it will pass through its mean position with a velocity v equal to