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

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.  $\frac{{\mathrm{A}}_0}{6}$

A particle executes simple harmonic oscillations under the effect of small damping. If the amplitude of oscillation becomes half of the initial value of 16 mm in five minutes, then what will be the amplitude after fifteen minutes?

1.  8 mm

2.  4 mm

3.  2 mm

4.  1 mm

The amplitude of a damped oscillator becomes one-third in 10 minutes and $\frac{1}{\mathrm{n}}$ times of the original value in 30 minutes. The value of n is:

1.  81

2.  3

3.  9

4.  27

For damped oscillations, the graph between energy and time will be:

1.		2.
3.		4.

Which of the following is not true for damped oscillations with time period T and an initial amplitude a?
1.  Angular frequency is slightly less than the natural frequency.
2.  Force remains constant in time interval t = 0 to $\mathrm{t}$ $=$ $\frac{\mathrm{T}}{8}$.
3.  If amplitude after time t is $\frac{\mathrm{a}}{\mathrm{N}}$, then the amplitude after time 2t will be $\frac{\mathrm{a}}{{\mathrm{N}}^2}$.
4.  Total mechanical energy decreases exponentially.

The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5 s. In another 10 s, it will decrease to $\mathrm{\alpha }$ times its original magnitude, where $\mathrm{\alpha }$ equals

1. 0.7

2. 0.81

3. 0.729

4. 0.6

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, $\mathrm{F}={\mathrm{F}}_{0<mspace\; width="1em"></mspace>}\sin <mspace\; width="1em"></mspace>\mathrm{\omega t}$

If, the amplitude of the particle is maximum for $\mathrm{\omega }={\mathrm{\omega }}_1$ and the energy of the particle is maximum for $\mathrm{\omega }={\mathrm{\omega }}_2$, then

1. ${\mathrm{\omega }}_1={\mathrm{\omega }}_0<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{\omega }}_2\ne {\mathrm{\omega }}_0$

2. ${\mathrm{\omega }}_1={\mathrm{\omega }}_0<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{\omega }}_2={\mathrm{\omega }}_0$

3. ${\mathrm{\omega }}_1\ne {\mathrm{\omega }}_0<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{\omega }}_2={\mathrm{\omega }}_0$

4. ${\mathrm{\omega }}_1\ne {\mathrm{\omega }}_0<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{\omega }}_2\ne {\mathrm{\omega }}_0$

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. $\frac{\sqrt{3}{A}_0}{2}$