The displacement-time graph of a particle executing SHM is shown in the figure. Its displacement equation will be: (Time period = 2 second)

1. $\mathrm{x}=10$ $\sin \left(\mathrm{\pi t}+\frac{\mathrm{\pi }}{6}\right)$

2. $\mathrm{x}=10$ $\sin \left(\mathrm{\pi t}\right)$

3. $\mathrm{x}=10$ $\cos \left(\mathrm{\pi t}\right)$

4. $x=5$ $\sin \left(\mathrm{\pi t}+\frac{\mathrm{\pi }}{6}\right)$

All the surfaces are smooth and the system, given below, is oscillating with an amplitude \(\mathrm{A}.\) What is the extension of spring having spring constant \(\mathrm{k_1},\) when the block is at the extreme position?
             

The amplitude of a simple harmonic oscillator is \(A\) and speed at the mean position is \(v_0\) .The speed of the oscillator at the position \(x={A \over \sqrt{3}}\) will be: 

A spring having a spring constant of 1200 N/m is mounted on a horizontal table as shown in the figure. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released. The frequency of oscillations will be:
    

Identify the correct definition:

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:

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

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. \( \frac{\pi}{2}\sqrt{\frac{m}{k}}\)

A particle is attached to a vertical spring and pulled down a distance of 0.01 m below its mean position and released. If its initial acceleration is 0.16 $\mathrm{m}/{\mathrm{s}}^2$, then its time period in seconds will be:

1.  $\mathrm{\pi }$

2.  $\frac{\mathrm{\pi }}{2}$

3.  $\frac{\mathrm{\pi }}{4}$

4.  $2\mathrm{\pi }$

The time periods for the figures (a) and (b) are ${\mathrm{T}}_1$ $\mathrm{and}$ ${\mathrm{T}}_2$ respectively. If all surfaces shown below are smooth, then the ratio $\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}$ will be:
   

1.  1: $\sqrt{3}$

2.  1: 1

3.  2: 1

4.  $\sqrt{3}$: 2