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 time period of a simple pendulum in a stationary trolley is \(T_1.\) If the trolley is moving with a constant speed, then the time period is \(T_2.\) Then: 
1. \(T_1>T _2\)
2. \(T_1<T _2\)
3. \(T_1=T _2\)
4. \(T_2= \infty \)

 A particle is moving along the x-axis. The speed of particle v varies with position x as $\frac{{\mathrm{v}}^2}{144}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\frac{{\mathrm{x}}^2}{9}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>1$. The time period of S.H.M is

1.  $\mathrm{\pi }<mspace\; width="1em"></mspace>\mathrm{unit}$

2.  $\frac{3\mathrm{\pi }}{2}<mspace\; width="1em"></mspace>\mathrm{unit}$

3.  $\frac{\mathrm{\pi }}{2}<mspace\; width="1em"></mspace>\mathrm{unit}$

4.  $\frac{\mathrm{\pi }}{4}<mspace\; width="1em"></mspace>\mathrm{unit}$

A block of mass \(m\) is attached to a massless spring of spring constant \(k,\) with the other end of the spring fixed to the wall of a trolley. Initially, the spring is unstretched, and the trolley begins to move rightward with a velocity that increases linearly with time, as shown in the velocity-time \((v\text-t)\) graph.
            
The total energy of oscillation of the block as observed from the trolley frame is:
1. \(\dfrac{32m^2}{6k}\)
2. \(\dfrac{9m^2}{8k}\)
3. \(\dfrac{9m^2}{32k}\)
4. \(\dfrac{8m^2}{9k}\)
 

A body of mass \(20\) g is executing SHM with amplitude \(5\) cm. When it passes through the equilibrium position its speed is \(20\) cm/s. What would be the distance from equilibrium when its speed becomes \(10\) cm/s?
1. \(\frac{5\sqrt{3}}{4}\) cm

2. \(\frac{5\sqrt{3}}{2}\) cm

3. \(\frac{25\sqrt{7}}{2}\) cm

4. \(5\sqrt{3}\) cm

For a simple harmonic oscillator, a velocity-time diagram is shown. The angular frequency of oscillation is:

1.  $\frac{24}{2}$ rad/s

2.  $\frac{25\mathrm{\pi }}{4}$ rad/s

3.  25 rad/s

4.  25$\mathrm{\pi }$ rad/s

In an S.H.M, when the displacement is one-fourth of the amplitude, the ratio of total energy to the potential energy is 

1.  16: 1

2.  1: 16

3.  1: 8

4.  8: 1

A particle executing S.H.M. Its displacement x varies with time t as $x=8\sin 4t+6\cos 4t.$  The maximum  velocity of the particle will be:

1.  20 unit

2.  10 unit

3.  40 unit

4.  5 unit

The amplitude of oscillation for a particle executing S.H.M with angular frequency $\mathrm{\pi }$rad/s and maximum acceleration $5{\mathrm{\pi }}^2<mspace\; width="1em"></mspace>\mathrm{m}/{\mathrm{s}}^2$ is

1.  25 cm

2.  25 m

3.  5 cm

4.  5 m