The displacement y of a particle executing simple harmonic motion is given by the equation:

From the equation, the angular frequency is directly given.

$$\omega = 0.2 \text{ rad/s}$$

The angular frequency is $\boxed{0.2 \text{ rad/s}}$.

The frequency is related to the angular frequency by the formula $f = \frac{\omega}{2\pi}$.

$$f = \frac{0.2 \text{ rad/s}}{2\pi \text{ rad}}$$ $$f = \frac{0.1}{\pi} \text{ Hz}$$ $$f \approx \frac{0.1}{3.14159} \text{ Hz}$$ $$f \approx 0.0318 \text{ Hz}$$

The frequency is $\boxed{0.0318 \text{ Hz}}$.

The period is the reciprocal of the frequency, $T = \frac{1}{f}$, or can be calculated as $T = \frac{2\pi}{\omega}$.

$$T = \frac{2\pi \text{ rad}}{0.2 \text{ rad/s}}$$ $$T = 10\pi \text{ s}$$ $$T \approx 10 \times 3.14159 \text{ s}$$ $$T \approx 31.4 \text{ s}$$

The period is $\boxed{31.4 \text{ s}}$.

The phase constant (or initial phase) is directly given in the equation.

$$\phi = 0.3 \text{ rad}$$

The phase is $\boxed{0.3 \text{ rad}}$.

Velocity is the first derivative of displacement with respect to time.

$$v = \frac{dy}{dt} = \frac{d}{dt} [A \sin(\omega t + \phi)]$$ $$v = A\omega \cos(\omega t + \phi)$$

Substitute the values:

$$v = (4 \text{ m})(0.2 \text{ rad/s}) \cos(0.2t + 0.3)$$ $$v = 0.8 \cos(0.2t + 0.3) \text{ m/s}$$

$$v_0 = 0.8 \cos(0.2(0) + 0.3)$$ $$v_0 = 0.8 \cos(0.3) \text{ m/s}$$

Using $\cos(0.3 \text{ rad}) \approx 0.9553$:

$$v_0 \approx 0.8 \times 0.9553 \text{ m/s}$$ $$v_0 \approx 0.764 \text{ m/s}$$

The initial velocity is $\boxed{0.764 \text{ m/s}}$.

Acceleration is the first derivative of velocity with respect to time.

$$a = \frac{dv}{dt} = \frac{d}{dt} [A\omega \cos(\omega t + \phi)]$$ $$a = -A\omega^2 \sin(\omega t + \phi)$$

Substitute the values:

$$a = -(4 \text{ m})(0.2 \text{ rad/s})^2 \sin(0.2t + 0.3)$$ $$a = -4 \times 0.04 \sin(0.2t + 0.3) \text{ m/s}^2$$ $$a = -0.16 \sin(0.2t + 0.3) \text{ m/s}^2$$

$$a_0 = -0.16 \sin(0.2(0) + 0.3)$$ $$a_0 = -0.16 \sin(0.3) \text{ m/s}^2$$

Using $\sin(0.3 \text{ rad}) \approx 0.2955$:

$$a_0 \approx -0.16 \times 0.2955 \text{ m/s}^2$$ $$a_0 \approx -0.0473 \text{ m/s}^2$$

The initial acceleration is $\boxed{-0.0473 \text{ m/s}^2}$.

The maximum velocity occurs when $\cos(\omega t + \phi) = \pm 1$.

$$v_{max} = A\omega$$ $$v_{max} = (4 \text{ m})(0.2 \text{ rad/s})$$ $$v_{max} = 0.8 \text{ m/s}$$

The maximum velocity is $\boxed{0.8 \text{ m/s}}$.

The maximum acceleration occurs when $\sin(\omega t + \phi) = \pm 1$.

$$a_{max} = A\omega^2$$ $$a_{max} = (4 \text{ m})(0.2 \text{ rad/s})^2$$ $$a_{max} = 4 \times 0.04 \text{ m/s}^2$$ $$a_{max} = 0.16 \text{ m/s}^2$$

The maximum acceleration is $\boxed{0.16 \text{ m/s}^2}$.