1. Use SHM relations

For a particle in simple harmonic motion:

$a = -\omega^2 y$

where $y$ is the displacement from mean position.

Given magnitudes at the instant:

• position $|y| = 4\,\text{m}$
• velocity $|v| = 2\,\text{m s}^{-1}$
• acceleration $|a| = 16\,\text{m s}^{-2}$

So,

$\omega^2 = \frac{|a|}{|y|} = \frac{16}{4} = 4$

Hence,

$\omega = 2\,\text{rad s}^{-1}$

2. Apply velocity-displacement relation in SHM

The standard relation is:

$v^2 = \omega^2(A^2 - y^2)$

Substitute the given values:

$2^2 = 4(A^2 - 4^2)$

$4 = 4(A^2 - 16)$

Divide by $4$:

$1 = A^2 - 16$

$A^2 = 17$

Therefore,

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

3. Match with the required form

Amplitude is given as $\sqrt{x}\,\text{m}$.

So,

$x = 17$