Which of the following figure represents damped harmonic motion?

(i)
(ii)
(iii)
(iv)

1. (i) and (ii)

2. (iii) and (iv)

3. (i), (ii), (iii), and (iv)

4. (i) and  (iv)

The damping force of an oscillator is directly proportional to the velocity. The units of the constant of proportionality are:
1. kg-msec^-1
2. kg-msec^-2
3. kg-sec^-1
4. kg-sec

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

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.

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

1.		2.
3.		4.

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

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

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 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

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. $\frac{\sqrt{3}{A}_0}{2}$