If a particle in SHM has a time period of \(0.1\) s and an amplitude of \(6\) cm, then its maximum velocity will be:
1. \(120 \pi\)$\mathrm{}$ cm/s 
2. \(0.6 \pi\)$\mathrm{}$ cm/s 
3. \(\pi\)$\mathrm{}$ cm/s
4. \(6\) cm/s

                     
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. \(\dfrac{a\omega}{4}\)
2. \(\dfrac{a\omega}{2\pi}\)
3. \(\dfrac{2a\omega}{\pi}\)
4. \(\dfrac{a\omega}{\sqrt{3}\pi}\)

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:
1. decreases continuously
2. increases continuously
3. first increases and then decreases
4. first decreases and then increases