The figure given below shows the graphs for amplitudes of forced oscillations in resonance conditions for different damping conditions. 

           

One of the conclusions that can be drawn from the graph above is:

1. As damping increases, amplitude increases

2. As damping increases, the amplitude decreases

3. As damping increases, the amplitude does not change

4. As damping increases, the amplitude may increase or decrease

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 time taken for its mechanical energy to drop to half its initial value is:

1.  3.46 s

2.  4.36 s

3.  6.93 s

4.  0.3 s

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 time taken for its amplitude of vibrations to drop to half of its initial value is:

1. \(5.93\) s
2. \(6.93\) s
3. \(7.93\) s
4. \(0.3\) s

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

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

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. $\frac{1}{9}$

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. \(\frac{1}{\sqrt{3}}\)

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

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.