The displacement of a body undergoing simple harmonic motion is...

The standard equation for simple harmonic motion is given by $y = A \cos(\omega t + \phi)$.

Comparing the given equation $y = 12 \cos\left(4\pi t - \dfrac{\pi}{4}\right) \mathrm{cm}$ with the standard equation, we get:

Amplitude $A = 12 \mathrm{~cm}$

Angular frequency $\omega = 4\pi \mathrm{~rad/s}$

The time period $T$ is given by $T = \dfrac{2\pi}{\omega}$.

Substituting the value of $\omega$, we get:

$T = \dfrac{2\pi}{4\pi} = 0.5 \mathrm{~s}$

Therefore, the amplitude is $12 \mathrm{~cm}$ and the time period is $0.5 \mathrm{~s}$.

Answer: $12 \mathrm{~cm}, 0.5 \mathrm{~s}$