A particle starts executing simple harmonic motion (SHM) of amplitude $$a$$ and total energy $$E$$. At any instant, its kinetic energy is $$\frac{3E}{4}$$, then its displacement $$y$$ is given by:

For a particle in simple harmonic motion, the total mechanical energy is the sum of its kinetic and potential energies. The standard results are:

• Total energy  $$E = \dfrac{1}{2}\,k\,a^{2}$$  where $$k$$ is the spring (force) constant and $$a$$ is the amplitude.
• Potential energy at displacement $$y$$ from the mean position  $$U = \dfrac{1}{2}\,k\,y^{2}$$.
• Kinetic energy at the same instant  $$K = E - U$$.

We are told that at some instant the kinetic energy is

$$K = \dfrac{3E}{4}.$$

Using the relation $$K = E - U$$, we substitute the given value:

$$E - U = \dfrac{3E}{4}.$$

Rearranging, we find the potential energy at that instant:

$$U = E - \dfrac{3E}{4} = \dfrac{E}{4}.$$

Now we equate this to the expression for potential energy in SHM:

$$\dfrac{1}{2}\,k\,y^{2} = \dfrac{E}{4}.$$

But the total energy is $$E = \dfrac{1}{2}\,k\,a^{2}$$, so we substitute $$E$$ on the right-hand side:

$$\dfrac{1}{2}\,k\,y^{2} = \dfrac{1}{4}\left(\dfrac{1}{2}\,k\,a^{2}\right).$$

We observe that the factor $$\dfrac{1}{2}\,k$$ appears on both sides, allowing us to cancel it out completely:

$$y^{2} = \dfrac{a^{2}}{4}.$$

Taking the principal value of the square root (the question asks for the magnitude of displacement), we obtain

$$y = \dfrac{a}{2}.$$

Hence, the correct answer is Option D.