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.

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. \(T'=2T\)

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.  $\frac{4\mathrm{T}}{3}$

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

      
1. \(\sqrt{\frac{\alpha}{\beta}}\)
2. \(2\pi\sqrt{\frac{\alpha}{\beta}}\)
3. \(2\pi\left(\frac{\beta}{\alpha}\right)\)
4. \(2\pi\left(\frac{\alpha}{\beta}\right)\)