A partricle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β. Then, its time period of vibration will be
(1)β2/α2
(2)α/β
(3)β2/α
(4)2πβ/α
1. | simple harmonic motion of frequency \(\frac{\omega}{\pi}\). |
2. | simple harmonic motion of frequency \(\frac{3\omega}{2\pi}\). |
3. | non-simple harmonic motion. |
4. | simple harmonic motion of frequency \(\frac{\omega}{2\pi}\). |
A point performs simple harmonic oscillation of period T and the equation of motion is given by x= a sin .After the elapse of what fraction of the time period the velocity of the point will be equal to half to its maximum velocity?
1.
2.
3.
4
The angular velocities of three bodies in simple harmonic motion are with their respective amplitudes as . If all the three bodies have same mass and maximum velocity, then
1.
2.
3.
4.
The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is 16 cm/sec. The distance of the particle from the mean position at which the speed of the particle becomes will be
1.
2.
3. 1 cm
4. 2 cm
The maximum velocity of a simple harmonic motion represented by is given by
1. 300
2.
3. 100
4.
The displacement equation of a particle is The amplitude and maximum velocity will be respectively
1. 5, 10
2. 3, 2
3. 4, 2
4. 3, 4
The instantaneous displacement of a simple pendulum oscillator is given by . Its speed will be maximum at time
1.
2.
3.
4.