Two simple harmonic motions, ${\mathrm{y}}_1$ $=$ $\mathrm{a}$ $\sin$ $\mathrm{\omega t}$ and ${\mathrm{y}}_2$ $=$ $2\mathrm{a}$ $\sin \left(\mathrm{\omega t}\right)$ $+$ $\frac{2\mathrm{\pi }}{3}$  are superimposed on a particle of mass m. The maximum kinetic energy of the particle will be:

1.  $\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{a}}^2$

2.  $\frac{5{\mathrm{m\omega }}^2{\mathrm{a}}^2}{4}$

3.  $\frac{3}{2}{\mathrm{m\omega }}^2{\mathrm{a}}^2$

4.  Zero

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

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:

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.

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:

Two simple pendulums of length 1 m and 16 m are in the same phase at the mean position at any instant. If T is the time period of the smaller pendulum, then the minimum time after which they will again be in the same phase will be:

1.  $\frac{3\mathrm{T}}{2}$

2.  $\frac{3\mathrm{T}}{4}$

3.  $\frac{2\mathrm{T}}{3}$

4.  $\frac{4\mathrm{T}}{3}$

A particle executes SHM with a time period of 4 s. The time taken by the particle to go directly from its mean position to half of its amplitude will be:

1.  $\frac{1}{3}$ s

2.  1 s

3.  $\frac{1}{2}$ s

4.   2 s

The graph between the velocity (v) of a particle executing S.H.M. and its displacement (x) is shown in the figure. The time period of oscillation for this SHM will be

1.  $\sqrt{\frac{\mathrm{\alpha }}{\mathrm{\beta }}}$

2.  $2\mathrm{\pi }\sqrt{\frac{\mathrm{\alpha }}{\mathrm{\beta }}}$

3.  $2\mathrm{\pi }\left(\frac{\mathrm{\beta }}{\mathrm{\alpha }}\right)$

4.  $2\mathrm{\pi }\left(\frac{\mathrm{\alpha }}{\mathrm{\beta }}\right)$