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.

Equation of a simple harmonic motion is  given by x = asin$\mathrm{\omega }$t. For which value of x, kinetic energy is equal to the potential energy?

1.  $\mathrm{x}$ $=$ $\pm$ $\mathrm{a}$

2.  $\mathrm{x}$ $=$ $\pm$ $\frac{\mathrm{a}}{2}$

3.  $\mathrm{x}$ $=$ $\pm$ $\frac{\mathrm{a}}{\sqrt{2}}$

4.  $\mathrm{x}$ $=$ $\pm$ $\frac{\sqrt{3}\mathrm{a}}{2}$

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. \(\frac{E}{2}\)

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)

A block of mass \(4~\text{kg}\) hangs from a spring of spring constant \(k = 400~\text{N/m}\). The block is pulled down through \(15~\text{cm}\) below the equilibrium position and released. What is its kinetic energy when the block is \(10~\text{cm}\) below the equilibrium position? [Ignore gravity]
1. \(5~\text{J}\)
2. \(2.5~\text{J}\)
3. \(1~\text{J}\)
4. \(1.9~\text{J}\)

The kinetic energy of a particle executing SHM is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particle is 5.12 kg, the time period of its oscillation will be:
1. $\frac{\mathrm{\pi }}{5}sec$ 
2. $2\mathrm{\pi }$ $\sec$
3. $20\mathrm{\pi }$ $\sec$   
4. $5\mathrm{\pi }$ $\sec$

A particle executing simple harmonic motion has a kinetic energy of \(K_0 cos^2(\omega t)\). The values of the maximum potential energy and the total energy are, respectively:
1. \(0\) and \(2K_0\)
2. \(\frac{K_0}{2}\) and \(K_0\)
3. \(K_0\) and \(2K_0\)
4. \(K_0\) and \(K_0\)