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))`

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 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)) ,

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.

The bob of mass 50 gm of a pendulum is released from a horizontal position A as shown in the figure . .If the length of the pendulum is 1.5 m , what is its kinetic energy at the lower most point B ? (g = 10 m/(s^(2)))

A simple pendulum is released from A as shown. If m and 1 represent the mass of the bob and length of the pendulum, the gain kinetic energy at B is

A simple pendulum is released from A as shown. If m and 1 represent the mass of the bob and length of the pendulum, the gain kinetic energy at B is

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) .

the bob of simple pendulum is relasesd when string makes an angle theta with the vertical . If m be the mass of the bob then find the tension when the bob is at the bottiommost position .