In damped oscillations, the damping force is directly proportional to the speed of the oscillator. If amplitude becomes half of its maximum value in 1 sec, then after 2 sec, the amplitude of the damped oscillation for which data is given, will be: (Initial amplitude = $A_0$)

1. $\frac{1}{4}{A}_0$

2. $\frac{1}{2}{A}_0$

3. $A_0$

4. $\frac{\sqrt{3}{A}_0}{2}$

For forced oscillations, a particle oscillates in a simple harmonic fashion with a frequency equal to:

1. the frequency of driving force.

2. the mean of frequency of driving force and natural frequency of the body.

3. the difference of frequency of driving force and natural frequency of the body.

4. the natural frequency of the body.

In damped oscillation, mass is 2 kg and spring constant is 500 N/m and damping coefficient is 1 kg s^–1. If the mass is displaced by 20 cm from its mean position and released, then what will be the value of its mechanical energy after 4 seconds?

1. 2.37 J

2. 1.37 J

3. 10 J

4. 5 J

A particle, with restoring force proportional to the displacement and resisting force proportional to velocity is subjected to a force F sin ωt. If the amplitude of the particle is maximum for $\omega$ $={\omega }_1$ and the energy of the particle is maximum for $\omega$ $={\omega }_2$, then:
1. ${\omega }_1$ $\ne {\omega }_0$ $and$ ${\omega }_2$ $=$ ${\omega }_0$

2. ${\omega }_1$ $={\omega }_0$ $and$ ${\omega }_2$ $=$ ${\omega }_0$

3. ${\omega }_1$ $={\omega }_0$ $and$ ${\omega }_2$ $\ne$ ${\omega }_0$

4. ${\omega }_1$ $\ne {\omega }_0$ $and$ ${\omega }_2$ $\ne$ ${\omega }_0$

The amplitude of an S.H.M. reduces to \(1/3\) in first \(20\) s, then in first \(40\) s its amplitude becomes:
1. \(1\over 3\)
2. \(1\over 9\)
3. \(1\over 27\)
4. \(\frac{1}{\sqrt{3}}\)

When an oscillator completes 100 oscillations, its amplitude is reduced to $\frac{1}{3}$ of the initial value. What will be its amplitude, when it completes 200 oscillations?

1. $\frac{1}{8}$

2. $\frac{2}{3}$

3. $\frac{1}{6}$

4. $\frac{1}{9}$

In case of a forced vibration, the resonance wave becomes very sharp when the:
1. Damping force is small
2. Restoring force is small
3. Applied periodic force is small
4. Quality factor is small

For the damped oscillator shown in the figure, the mass \(m\) of the block is \(200\) g, \(k=90\) N/m  and the damping constant \(b\) is \(40\) g/s. The period of oscillation is:

                

1. \(0.03\) s
2. \(0.3\) s
3. \(3\) s
4. \(3.4\) s

A body executes oscillations under the effect of a small damping force. If the amplitude of the body reduces by 50% in 6 minutes, then amplitude after the next 12 minutes will be [initial amplitude is ${\mathrm{A}}_0$] -

1.  $\frac{{\mathrm{A}}_0}{4}$

2.  $\frac{{\mathrm{A}}_0}{8}$

3.  $\frac{{\mathrm{A}}_0}{16}$

4.  $\frac{{\mathrm{A}}_0}{6}$

In a forced oscillation, when the system oscillates under the action of the driving force $\mathrm{F}$ $=$ ${\mathrm{F}}_0$ $\sin$ $\mathrm{\omega t}$ in addition to its internal restoring force, the particle oscillates with a frequency equal to

1.  The natural frequency of the body

2.  Frequency of driving force

3.  The difference in frequency of driving force and natural frequency

4.  Mean of the driving frequency and natural frequency