Answer
given,
mass of the block, m = 0.5 Kg
displacement, x = 30 cm = 0.3 m
Spring constant, k = 2 N/m
a) Angular frequency
  [tex]\omega = \sqrt{\dfrac{k}{m}}[/tex]
  [tex]\omega = \sqrt{\dfrac{2}{0.5}}[/tex]
 [tex]\omega = 2\ rad/s[/tex]
b) Period of oscillation
  [tex]T=\dfrac{2\pi}{\omega}[/tex]
  [tex]T=\dfrac{2\pi}{2}[/tex]
      T = 3.14 s
c) Position at  t = 2
 x = -A cos ω t
A = 0.3  ω = 2 rad/s
x = -0.3 cos (2 x 2)
x = 0.196 m
d) velocity at t= 2
 v = A ω sin ω t
 v = 0.3 x 2 x sin 4
 v = -0.454 m/s
e) acceleration at t= 2
  a = A ω² cos ω t
  a = 0.3 x 2² cos 4
  a = -0.784 m/s²
