All the surfaces are smooth and springs are ideal. If a block of mass \(m\) is given the velocity \(v_0\) in the right direction, then the time period of the block shown in the figure will be:

                       
1. \(\frac{12l}{v_0}\)
2. \(\frac{2l}{v_0}+ \frac{3\pi}{2}\sqrt{\frac{m}{k}}\)
3. \(\frac{4l}{v_0}+ \frac{3\pi}{2}\sqrt{\frac{m}{k}}\)
4. \( \frac{\pi}{2}\sqrt{\frac{m}{k}}\)

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.

In a spring pendulum, in place of mass, a liquid is used. If liquid leaks out continuously, then the time period of the spring pendulum:

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

1.		2.
3.		4.

Equation of a simple harmonic motion is  given by x = asin$\mathrm{\omega }$t. For which value of x, kinetic energy is equal to the potential energy?

1.  $\mathrm{x}$ $=$ $\pm$ $\mathrm{a}$

2.  $\mathrm{x}$ $=$ $\pm$ $\frac{\mathrm{a}}{2}$

3.  $\mathrm{x}$ $=$ $\pm$ $\frac{\mathrm{a}}{\sqrt{2}}$

4.  $\mathrm{x}$ $=$ $\pm$ $\frac{\sqrt{3}\mathrm{a}}{2}$

The displacement \( x\) of a particle varies with time \(t\) as \(x = A sin\left (\frac{2\pi t}{T} +\frac{\pi}{3} \right)\)$\mathrm{}$. The time taken by the particle to reach from \(x = \frac{A}{2} \) to \(x = -\frac{A}{2} \) will be:

Force on a particle F varies with time t as shown in the given graph. The displacement x vs time t graph corresponding to the force-time graph 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 particle executes linear SHM between \(x=A.\) The time taken to go from \(0\) to \(A/2\) is ${\mathrm{T}}_1$ and to go from \(A/2\) to \(A\) is ${\mathrm{T}}_2$, then: