Figure shown the kinetic energy `K` of a pendulum versus. its angle `theta` from the vertical. The pendulum bob has mass `0.2kg` The length of the pendulum is equal to `(g = 10m//s^(2))`
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Figure shown the kinetic energy K of a pendulum versus. its angle theta from the verticle. The pendulum bob has mass 0.2kg The length of the pendulum is equal to (g = 10m//s^(2)) ,

Figure shows the kinetic energy K of a simple pendulum versus its angle theta from the verticle. The pendulum bob has mass 0.2 kg . If the length of the pendulum is equal to (n)/(g) meter. Then find n (g = 10 m//s^(2))

The given figure shows the variation of the kinetic energy of a simple pendulum with its angular displacement (theta) from the vertical. Mass of the pendulum bob is m = 0.2 kg. Find the time period of the pendulum. Take g = 10 ms^(-2) .

Assertion : If the length of the simple pendulum is equal to the radius of the earth, the period of the pendulum is T=2pisqrt(R//g) Reason : Length of this pendulum is equal to radius of earth.

A 1N pendulum bob is held at an angle theta from the vertical by a 2N horizontal force F as shown in figure. The The tension in the string supporting the pendulum bob (in newton) is .

Identify wrong statements among the following The greater the mass of a pendulum bob, the shorter is its frequency of oscillation A simple pendulum with a bob of mass M swings with an angular amplitude of 40^(@) . When its angular amplitude is 20^(@) , the tension in the string is less than Mgcos 20^(@) . (3) The fractional change in the time period of a pendulum on changing the temperature is independent of the length of the pendulum. As the length of a simple pendulum is increased, the maximum velocity of its bob during its oscillation will also decreases.

Instantaneous angle (in radian) between string of a simple pendulum and verical is given by theta = (pi)/(180) sin2pit . Find the length of the pendulum if g = pi^(2) m//s^(2) .