Simple Harmonic Motion

The maximum displacement from equilibrium of a particle is 1 m and its maximum acceleration is

1.57 m s^-2. The period of the particle will be

• 1.25 s

• 0.63 s

• 0.20 s

• 5.03 s

If a simple harmonic oscillator has a displacement of 0.02 m and acceleration equal to 2.00 m s^-2, the angular frequency is equal to:

• 10.0 rad s^-1

• 0.10 rad s^-1

• 100 rad s^-1

• 1.00 rad s^-1

Given equation of SHM, x=0.34 cos 3000t where x in mm and t in second. The frequency is:

• 3000 Hz

• 3000/2π Hz

• 0.74/2π Hz

• 3000/π Hz

The acceleration of a particle in SHM is:

• always zero

• always constant

• maximum at amplitude

• maximum at the equilibrium position

The amplitude and period in a SHM is 0.5 m and 0.4 s respectively. The equation of SHM will be:

• x=0.5 sin 5πt

• x=0.5 sin 4πt

• x=0.5 sin 2.5πt

• x=0.5 sin 0.8πt

Which of the following equation represent the displacement of SHM:

• x= A tan ωt

• x= A sin ωt

• x= A sin ωt cos kx

• x= A sin (ωt ± kx)

A particle undergoes simple harmonic motion with angular velocity of 5 rad/s and amplitude of 50 cm. Find the displacement, x at time t = 10 s.

• 0.500 m

• 0.383 m

• -0.131 m

• 0.131 m

The variation of the acceleration, a of a particle executing SHM with displacement, x is as shown in figure:

• 
• 
• 
• 

The graph in FIGURE 1 shows the relationship between the acceleration, a and its displacement, x for an object of mass 2 kg which is in simple harmonic motion. Determine the frequency of oscillation.

• 1.6 Hz

• 1.26 Hz

• 0.20 Hz

• 5.00 Hz

The displacement x, for a particle at time t in simple harmonic motion is given by the equation, x = 10 sin 20t where x is in meters and t is in seconds. Determine the displacement at t=10 s

• 91.3 m

• 8.73 m

• -91.3 m

• -8.73 m