#### Answer

(a) $k = 24.5~N/m$
(b) $T = 0.898~s$
(c) $v_{max} = 0.700~m/s$
The maximum speed occurs as the mass passes through the equilibrium position.

#### Work Step by Step

(a) We can find the spring constant as:
$kx = mg$
$k = \frac{mg}{x}$
$k = \frac{(0.500~kg)(9.80~m/s^2)}{0.20~m}$
$k = 24.5~N/m$
(b) We can find the period of oscillation as:
$T = 2\pi~\sqrt{\frac{m}{k}}$
$T = 2\pi~\sqrt{\frac{0.500~kg}{24.5~N/m}}$
$T = 0.898~s$
(c) We can find the maximum speed as:
$v_{max} = A~\omega$
$v_{max} = A~\sqrt{\frac{k}{m}}$
$v_{max} = (0.100~m)~\sqrt{\frac{24.5~N/m}{0.500~kg}}$
$v_{max} = 0.700~m/s$
The maximum speed occurs as the mass passes through the equilibrium position.