1. Write the energy expressions for SHM

For a particle executing SHM with amplitude $A$ and angular frequency $\omega$:

$x = A\cos(\omega t + \phi)$

The total mechanical energy is

$E = \frac{1}{2}m\omega^2A^2$

Potential energy at displacement $x$ is

$U = \frac{1}{2}m\omega^2x^2$

Kinetic energy is

$K = E - U = \frac{1}{2}m\omega^2(A^2 - x^2)$

2. Check statement (1): "Potential energy is always equal to its K.E."

This is false.

In SHM, $U=K$ only at special positions:

$\frac{1}{2}m\omega^2x^2 = \frac{1}{2}m\omega^2(A^2-x^2)$

$x^2 = A^2-x^2$

$2x^2=A^2 \Rightarrow x=\pm \frac{A}{\sqrt{2}}$

So they are equal only at particular instants, not always.

3. Check statement (2): "Average potential and kinetic energy over any given time interval are always equal."

This is false.

They are equal only when averaged over a complete time period (or an integer multiple of the period), not over any arbitrary interval.

For example, over a very short interval near the extreme position, potential energy is large and kinetic energy is small, so their averages are not equal.

4. Check statement (3): "Sum of the kinetic and potential energy at any point of time is constant."

This is true.

Since

$K + U = \left(\frac{1}{2}m\omega^2(A^2-x^2)\right) + \left(\frac{1}{2}m\omega^2x^2\right)$

$K+U = \frac{1}{2}m\omega^2A^2 = \text{constant}$

So total mechanical energy remains constant in SHM.

5. Check statement (4): "Average K.E. in one time period is equal to average potential energy in one time period."

This is true.

Over one complete time period,

$\langle K \rangle = \langle U \rangle = \frac{E}{2}$

This is a standard result for SHM.

6. Select the correct option

• (1) False
• (2) False
• (3) True
• (4) True

Hence the correct option is:

$\boxed{\text{A: (3) and (4)}}$

7. Comparison with stored answer

Stored correct answer: A

Derived answer: A

So, the derived answer agrees with the stored answer.