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
The total energy of a body performing S.H.M. depends on:
k,a,m
k,a,x
k,a
k,x
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:
Two springs A and B (kA =3kB )are stretched by the same suspended weight. Then the ratio of work done in stretching is:
1 : 2
1 : 3
1 : 4
2 : 1
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:
If a simple harmonic oscillator has got a displacement of 0.02 m and acceleration equal to 2.0m/s2 at any time, the angular frequency of the oscillator is equal to:
10 rad/s
0.1 rad/s
100 rad/s
1 rad/s
Two simple harmonic motions of angular frequency 100 and 1000 rad s-1 have the same displacement amplitude. The ratio of their maximum accelerations is:
1:10
1:102
1:103
1:104
A hollow sphere filled with water forms the bob of a simple pendulum. A small hole at the bottom of the bob allows the water to slowly flow out as it is set into small oscillations and its period of oscillations is measured. The time period will:
Increase
Decrease
Remain constant
First increases , then decreases
A particle is executing S.H.M. with an amplitude of 4 cm and the 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:
2
1
1/2
A particle executing S.H.M has amplitude 0.01 m and frequency 60 Hz. The maximum acceleration of the particle is:
