The velocity v of a particle executing simple harmonic motion is...

The standard equation for the velocity of a particle in simple harmonic motion is:
$v = A \omega \cos(\omega t)$

Given equation:
$v = 50 \pi \cos(5 \pi t) \text{ cm/s}$

Comparing the given equation with the standard equation, we get:
$\omega = 5 \pi \text{ rad/s}$
$A \omega = 50 \pi \text{ cm/s}$

The time period $T$ is:
$T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{5\pi} = 0.4 \text{ s}$

The amplitude $A$ is:
$A = \dfrac{50 \pi}{\omega} = \dfrac{50 \pi}{5 \pi} = 10 \text{ cm}$

Thus, the amplitude is $10 \text{ cm}$ and the time period is $0.4 \text{ s}$.

Answer: $10 \text{ cm}, 0.4 \text{ s}$