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. \(\frac{E}{2}\)

When the displacement is half the amplitude in an SHM, the ratio of potential energy to the total energy is:
1. 1 / 2

2. 1 / 4

3. 1

4. 1 / 8

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. \(1.9~\text{J}\)

Kinetic energy of a particle executing simple harmonic motion in straight line is \(pv^2\) and potential energy is \(qx^2,\) where \(v\) is speed at distance \(x\) from the mean position. The time period of the SHM is given by the expression:

1. $2\mathrm{\pi }\sqrt{\frac{\mathrm{q}}{\mathrm{p}}}$

2. $2\mathrm{\pi }\sqrt{\frac{p}{q}}$

3. $2\mathrm{\pi }\sqrt{\frac{\mathrm{q}}{\mathrm{p}+\mathrm{q}}}$

4. $2\mathrm{\pi }\sqrt{\frac{p}{p+q}}$

The total energy of a particle, executing simple harmonic motion is:

1. $\propto$ $x$                 

2. $\propto$ $x^2$

3.  Independent of x 

4.  $\propto$ $x^{1/2}$

If the potential energy U (in J) of a body executing SHM is given by U = 20 + 10 (${\sin }^2$100$\mathrm{\pi }$t), then the minimum potential energy of the body will be:

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. $-1$