A particle is executing S.H.M. with an amplitude of 4 cm and time period 12 seconds. The time taken by the particle in going 2 cm from the mean position is T1 seconds. The time taken from this displaced position to reach the extreme position on the same side is T2 seconds. Then T1 /T2 will be
The acceleration of a particle performing S.H.M. is 12 cm/sec2 at a distance of 4 cm from the mean position. Its time period is
2.0 sec
3.14 sec
0.5 sec
1.0 sec
Frequency of a simple pendulum in a freely falling lift is
zero
infinite
finite
cannot say
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
In a simple harmonic motion, if the displacement is half of the amplitude, then which part of total energy will be kinetic energy ?
1/3
3/4
2/3
1/4
A body executes SHM with an amplitude . At what displacement from the mean position is the potential energy of the body is one-fourth of its total energy ?
Some other fraction of a
A system exhibiting S.H.M. must possess
elasticity as well as inertia
elasticity only
inertia only
elasticity, inertia and external force
Two springs of constants k1 and k2 have equal maximum velocities when executing simple harmonic motion. The ratio of their amplitudes (masses are equal) will be
A particle of mass 200 g executes S.H.M. The restoring force is provided by a spring of force constant 80 N/m. The time period of oscillations is
0.15 sec
0.02 sec
0.31 sec
0.05 sec
