If the vertical spring-mass system is dipped in a non-viscous liquid, then:

1.	only mean position is changed.
2.	only the time period is changed.
3.	time period and mean position both are changed.
4.	time period and mean position both remain the same.

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 time period of a simple pendulum in a stationary trolley is \(T_1.\) If the trolley is moving with a constant speed, then 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 executes SHM with a frequency of \(20\) Hz. The frequency with which its potential energy oscillates is:
1. \(5\) Hz
2. \(20\) Hz
3. \(10\) Hz
4. \(40\) Hz

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

1.  $\mathrm{\pi } \mathrm{unit}$

2.  $\frac{3\mathrm{\pi }}{2} \mathrm{unit}$

3.  $\frac{\mathrm{\pi }}{2} \mathrm{unit}$

4.  $\frac{\mathrm{\pi }}{4} \mathrm{unit}$

A block of mass m is attached to a massless spring having a spring constant k. The other end of the spring is fixed from the wall of a trolley, as shown in the figure. Spring is initially unstretched and the trolley starts moving toward the direction shown. Its velocity-time graph is also shown.

The energy of oscillation, as seen from the trolley is:

1.  $\frac{32{\mathrm{m}}^2}{6\mathrm{k}}$

2.  $\frac{9{\mathrm{m}}^2}{8\mathrm{k}}$

3.  $\frac{9{\mathrm{m}}^2}{32\mathrm{k}}$

4.  $\frac{8{\mathrm{m}}^2}{9\mathrm{k}}$

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