The angular velocity and the amplitude of a simple pendulum is and respectively. At a displacement x from the mean position, if its kinetic energy is T and potential energy is U, then the ration of T to U is
A simple pendulum with a bob of mass m oscillates from A to C and back to A [See Fig.] such that PB is H. If the acceleration due to gravity is g then velocity of the bob as it passes through B is
zero
2gH
mgH
The kinetic energy of a particle executing S.H.M. is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particle is 5.12 kg, the time period of oscillations is
A body is executing SHM. When the displacements from the mean position is 4 cm and 5 cm, the corresponding velocities of the body is 10 cm/s and 8cm/s. Then the time period of the body is
A body is executing S.H.M When its displacement from the mean position is 4 cm and 5 cm, the corresponding velocities are 10 cm/sec and 8cm/sec. Then time period of the body is
A particle of mass 10 g is executing S.H.M, with an amplitude of 0.5 m and periodic time of second. The maximum value of force acting on the particle is
25 N
5 N
2.5 N
0.5 N
The displacement x (in metres) of a particle performing S.H.M is related to time t (in second) as :
A linear harmonic oscillator of force constant 2 x 106 N/m and amplitude 0.01 m has a total mechanical energy of 160 J. Its
maximum potential energy is 160 J
maximum potential energy is 100 J
maximum potential energy is zero
minimum potential energy is 100 J
A particle executes simple harmonic motion of amplitude A. At what distance from the mean position its kinetic energy is equal to its potential energy ?
0.51 A
0.61 A
0.71 A
0.81 A
A particle is performing simple harmonic motion along x axis with amplitude 4 cm and time period 1.2 sec. The minimum time taken by the particle to move from x = + 2 cm to x = + 4 cm and back again is
0.4 s
0.3 s
0.2 s
0.6 s