A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:
The time period of a simple pendulum is 2 s. If its length is increased by 4 times, then its period becomes:
16 s
12 s
8 s
4 s
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.Select the correct statement :
Maximum potential energy is 160 J
Maximum potential energy is 100 J
Maximum potential energy is zero
Minimum potential energy is 100 J
A system exhibiting S.H.M. must possess:
Elasticity as well as inertia
Elasticity only
Inertia only
Elasticity, inertia and external force
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 particle executes simple harmonic motion with a frequency f. The frequency with which kinetic energy oscillates is:
f
4 f
2 f
f/2
A particle moving along the X-axis executes simple harmonic motion; then the force acting on it is given by:
-AKx
A cos Kx
A exp(-Kx)
AKx
A particle is executing S.H.M of amplitude 4 cm and T = 4 sec. The time taken by it to move from positive extreme position to half the amplitude is:
1 sec
1/3 sec
2/3 sec
When a damped harmonic oscillator completes 100 oscillations, its amplitude is reduced to of its initial value. What will be its amplitude when it completes 200 oscillations ?
The total energy of a body performing S.H.M. depends on:
k,a,m
k,a,x
k,a
k,x
