At a specific instant, the magnitudes of the position, velocity,...

Given $x = 2 \text{ cm} = 0.02 \text{ m}$, $v = 1 \text{ m/s}$, and $a = 10 \text{ m/s}^2$.

In simple harmonic motion, the magnitude of acceleration is given by $a = \omega^2 x$.

Substituting the given values:

$10 = \omega^2 \times 0.02$

$\omega^2 = \dfrac{10}{0.02} = 500 \text{ rad}^2/\text{s}^2$

$\omega = \sqrt{500} = 10\sqrt{5} \text{ rad/s}$

The time period $T$ is given by:

$T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{10\sqrt{5}} \approx \dfrac{2 \times 3.14}{22.36} \approx 0.28 \text{ s}$

The velocity in SHM is given by $v = \omega \sqrt{A^2 - x^2}$.

Squaring both sides:

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

Substituting the known values:

$1^2 = 500 (A^2 - 0.02^2)$

$1 = 500 (A^2 - 0.0004)$

$A^2 - 0.0004 = \dfrac{1}{500} = 0.002$

$A^2 = 0.002 + 0.0004 = 0.0024 \text{ m}^2$

$A = \sqrt{0.0024} \text{ m} = \sqrt{24} \times 10^{-2} \text{ m} = \sqrt{24} \text{ cm}$

Since $\sqrt{24} \approx 4.899$, the amplitude is $A \approx 4.9 \text{ cm}$.