The kinetic energy \((K)\) of a simple harmonic oscillator varies with displacement \((x)\) as shown. The period of the oscillation will be: (mass of oscillator is \(1\) kg)

                     
1. \(\frac{\pi}{2}~\text{s}\)
2. \(\frac{1}{2}~\text{s}\)
3. \(\pi~\text{s}\)
4. \(1~\text{s}\)

A particle is executing linear simple harmonic motion with an amplitude \(a\) and an angular frequency \(\omega\). Its average speed for its motion from extreme to mean position will be:
1. \(\frac{a\omega}{4}\)
2. \(\frac{a\omega}{2\pi}\)
3. \(\frac{2a\omega}{\pi}\)
4. \(\frac{a\omega}{\sqrt{3}\pi}\)

All the surfaces are smooth and springs are ideal. If a block of mass \(m\) is given the velocity \(v_0\) in the right direction, then the time period of the block shown in the figure will be:

                       
1. \(\frac{12l}{v_0}\)
2. \(\frac{2l}{v_0}+ \frac{3\pi}{2}\sqrt{\frac{m}{k}}\)
3. \(\frac{4l}{v_0}+ \frac{3\pi}{2}\sqrt{\frac{m}{k}}\)
4. \( \frac{\pi}{2}\sqrt{\frac{m}{k}}\)

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:

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. \(\dfrac{T_0}{\sqrt{2}}\)