One end of a spring of force constant \(k\) is fixed to a vertical wall and the other to a block of mass \(m\) resting on a smooth horizontal surface. There is another wall at a distance \(x_0\) from the block. The spring is then compressed by \(2x_0\) and then released. The time taken to strike the wall will be?

A spring having a spring constant of \(1200\) N/m is mounted on a horizontal table as shown in the figure. A mass of \(3\) kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of \(2.0\) cm and released. The frequency of oscillations will be:
    

The time period of a mass suspended from a spring is \(T\). If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be:
1. \(\frac{T}{4}\)
2. \(T\)
3. \(\frac{T}{2}\)
4. \(2T\)

A spring elongates by a length 'L' when a mass 'M' is suspended to it. Now a tiny mass 'm' is attached to the mass 'M' and then released. The new time period of oscillation will be:

1.  \(2 \pi \sqrt{\frac{\left(\right. M   +   m \left.\right) l}{Mg}}\)

2. \(2 \pi \sqrt{\frac{ml}{Mg}}\)

3. \(2 \pi \sqrt{L   /   g}\)

4. \(2 \pi \sqrt{\frac{Ml}{\left(\right. m   +   M \left.\right) g}}\)

An ideal spring with spring-constant K is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially un-stretched. Then the maximum extension in the spring will be:
1. 4 Mg/K 
2. 2 Mg/K
3. Mg/K 
4. Mg/2K