A simple pendulum of length `l` is made to oscillate with an amplitude of 45 degrees. The acceleration due to gravity is g. Let `T_(0) = 2pi sqrt(l//g)` . The time period of oscillation of this pendulum will be -

A simple pendulum of length 'L' has mass 'M' and it oscillates freely with amplitude A, Energy is (g = acceleration due to gravity)

A simple pendulum of length 'L' has mass 'M' and it oscillates freely with amplitude energy is (g = acceleration due to gravity)

A simple pendulum of length 'L' has mass 'M' and it oscillates freely with amplitude energy is (g = acceleration due to gravity)

A simple pendulum of length 1m is allowed to oscillate with amplitude 2^(@) . it collides at T to the vertical its time period will be (use g = pi^(2) )

A simple pendulum of length 1m is allowed to oscillate with amplitude 2^(@) . it collides at T to the vertical its time period will be (use g = pi^(2) )

A simple pendulum of length 1m is allowed to oscillate with amplitude 2^(@) . it colliates at T to the verticle its time period will be (use g = pi^(2) )

A simple pendulum oscillates in air with time period T and amplitude A . As the time passes

The period of oscillation of a simple pendulum at a given place with acceleration due to gravity 'g' depends on

A simple pendulum designed on the moon as a seconds pendulum is taken to a planet where the acceleration due to gravity on the surface is twice that on the Earth . If g_("earth") : g_("moon") =6 :1 find the period of oscillation of the pendulum on the planet mentionedabove.