Let's begin by understanding the equations related to simple harmonic motion (SHM). For a particle executing SHM, the position $$x$$, velocity $$v$$, and acceleration $$a$$ are given by the following equations:

1. Position: $$x = A \cos(\omega t + \phi)$$

2. Velocity: $$v = -A \omega \sin(\omega t + \phi)$$

3. Acceleration: $$a = -A \omega^2 \cos(\omega t + \phi)$$

Here, $$A$$ is the amplitude of the motion, $$\omega$$ is the angular frequency, and $$\phi$$ is the phase constant.

Given the magnitudes at a certain instant:

$$x = 4 \, \mathrm{m}$$

$$v = 2 \, \mathrm{ms}^{-1}$$

$$a = 16 \, \mathrm{ms}^{-2}$$

Using the acceleration equation:

$$a = -A \omega^2 \cos(\omega t + \phi)$$

Since we’re given the magnitude of the acceleration, we remove the negative sign:

$$16 = A \omega^2 \cos(\omega t + \phi)$$

Using the position equation:

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

We already know $$x = 4 \, \mathrm{m}$$, so:

$$4 = A \cos(\omega t + \phi)$$

From these two equations, we know:

$$A \omega^2 \cos(\omega t + \phi) = 16$$

$$A \cos(\omega t + \phi) = 4$$

Therefore:

$$A \omega^2 \cdot 4/A = 16$$

$$4 \omega^2 = 16$$

$$\omega^2 = 4$$

$$\omega = 2 \, \mathrm{rad/s}$$

Next, using the velocity equation:

$$v = -A \omega \sin(\omega t + \phi)$$

Again, we consider the magnitude:

$$2 = A \cdot 2 \sin(\omega t + \phi)$$

$$2 = 2A \sin(\omega t + \phi)$$

$$\sin(\omega t + \phi) = \dfrac{1}{A}$$

We know from the position equation that:

$$\cos(\omega t + \phi) = \dfrac{4}{A}$$

Using the identity $$\sin^2(\theta) + \cos^2(\theta) = 1$$, we get:

$$\left(\dfrac{1}{A}\right)^2 + \left(\dfrac{4}{A}\right)^2 = 1$$

$$\dfrac{1}{A^2} + \dfrac{16}{A^2} = 1$$

$$\dfrac{17}{A^2} = 1$$

$$A^2 = 17$$

$$A = \sqrt{17} \, \mathrm{m}$$

Therefore, the amplitude of the motion is $$\sqrt{17} \, \mathrm{m}$$, meaning $$x$$ is 17.