A spring pendulum is placed on a rotating table. The initial angular velocity of the table is \(\omega_{0}\) and the time period of the pendulum is \(T_{0}.\) If the the angular velocity of the table becomes \(2\omega_{0},\) then the new time period of the pendulum will be:

1.	\(2T_{0}\)	2.	\(T_0\sqrt{2}\)
3.	the same	4.	\(\dfrac{T_0}{\sqrt{2}}\)

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 graph of potential energy \((U)\) versus displacement \((x)\) is shown. Which of the following describes the oscillation about the mean position, \(x = 0\text{?}\)

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. \(T'=2T\)

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