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:

1.	Decreases continuously
2.	Increases continuously
3.	First increases and then decreases
4.	First decreases and then increases

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

A spring pendulum is on the rotating table. The initial angular velocity of the table is \(\omega_{0}\) and the time period of the pendulum is \(T_{0}.\) Now the angular velocity of the table becomes \(2\omega_{0},\) then the new time period will be:
1. \(2T_{0}\)
2. \(T_0\sqrt{2}\)
3. remains the same
4. \(\frac{T_0}{\sqrt{2}}\)

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 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

The curve between the potential energy \((U)\) and displacement \((x)\) is shown. Which of the oscillation is about the mean position, \(x = 0?\)

1.		2.
3.		4.

A spring-block system oscillates with a time period \(T\) on the earth's surface. When the system is brought into a deep mine, the time period of oscillation becomes \(T'.\) Then one can conclude that:
1. \(T'>T\)
2. \(T'<T\)
3. \(T'=T\)
4. \(T'=2T\)

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.  $\frac{4\mathrm{T}}{3}$