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?

          

1.	\(\frac{1}{6} \pi \sqrt{ \frac{k}{m}}\)	2.	\( \sqrt{\frac{k}{m}}\)
3.	\(\frac{2\pi}{3} \sqrt{ \frac{m}{k}}\)	4.	\(\frac{\pi}{4} \sqrt{ \frac{k}{m}}\)

A block is connected to a relaxed spring and kept on a smooth floor. The block is given a velocity towards the right. Just after this:
               

A mass m is suspended from two springs of spring constant $k_1$ $and$ $k_2$ as shown in the figure below. The time period of vertical oscillations of the mass will be
                

1. $2\mathrm{\pi }\sqrt{\left(\frac{{\mathrm{k}}_1+{\mathrm{k}}_2}{\mathrm{m}}\right)}$

2. $2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\left({\mathrm{k}}_1+{\mathrm{k}}_2\right)}}$

3. $2\mathrm{\pi }\sqrt{\frac{\mathrm{m}\left({\mathrm{k}}_1{\mathrm{k}}_2\right)}{\left({\mathrm{k}}_1+{\mathrm{k}}_2\right)}}$

4. $2\mathrm{\pi }\sqrt{\frac{\mathrm{m}\left({\mathrm{k}}_1+{\mathrm{k}}_2\right)}{\left({\mathrm{k}}_1{\mathrm{k}}_2\right)}}$

A body is executing simple harmonic motion. At a displacement \(x,\) its potential energy is \(E_1\) and at a displacement \(y\), its potential energy is \(E_2\)${\mathrm{}}_{}$. The potential energy \(E\) at displacement \(x+y\) will be?
1. \(E = \sqrt{E_1}+\sqrt{E_2}\)
2. \(\sqrt{E} = \sqrt{E_1}+\sqrt{E_2}\)
3. \(E =E_1 +E_2\)
4. None of the above