If a body is executing simple harmonic motion with frequency 'n', then the frequency of its potential energy is:

1.	3n	2.	4n
3.	n	4.	2n

A spring is stretched by \(5~\text{cm}\) by a force \(10~\text{N}\). The time period of the oscillations when a mass of \(2~\text{kg}\) is suspended by it is:
1. \(3.14~\text{s}\)
2. \(0.628~\text{s}\)
3. \(0.0628~\text{s}\)
4. \(6.28~\text{s}\)

The potential energy of a simple harmonic oscillator, when the particle is halfway to its endpoint, will be:
1. \(\frac{2E}{3}\)
2. \(\frac{E}{8}\)
3. \(\frac{E}{4}\)
4. \(\frac{E}{2}\)

A particle of mass m oscillates with simple harmonic motion between points x₁ and x₂, the equilibrium position being O. Its potential energy is plotted. It will be as given below in the graph:

1.		2.
3.		4.

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. T/4
2. T
3. T/2
4. 2T

The total energy of the particle performing SHM depends on: 
1. \(k,\) \(a,\) \(m\)
2. \(k,\) \(a\)
3. \(k,\) \(a\), \(x \)
4. \(k,\) \(x \)

The displacement between the maximum potential energy position and maximum kinetic energy position for a particle executing simple harmonic motion is:

1. $\pm \frac{\mathrm{a}}{2}$

2. $+\mathrm{a}$

3. $\pm \mathrm{a}$

4. $-1$

When a mass is suspended separately by two different springs, in successive order, then the time period of oscillations is \(t _1\) and \(t_2\) respectively. If it is connected by both springs as shown in the figure below, then the time period of oscillation becomes \(t_0.\) The correct relation between \(t_0,\) \(t_1\) & \(t_2\) is:

1. ${{\mathrm{t}}_0}^2={{\mathrm{t}}_1}^2+{{\mathrm{t}}_2}^2$

2. ${{\mathrm{t}}_0}^{-2}={{\mathrm{t}}_1}^{-2}+{{\mathrm{t}}_2}^{-2}$

3. ${{\mathrm{t}}_0}^{-1}={{\mathrm{t}}_1}^{-1}+{{\mathrm{t}}_2}^{-1}$

4. ${\mathrm{t}}_0={\mathrm{t}}_1+{\mathrm{t}}_2$

The frequency of a simple pendulum in a free-falling lift will be:
1. zero
2. infinite
3. can't say
4. finite

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