Answer:
a
  [tex]m = 0.169 \ kg[/tex]
b
 [tex]|v_{max} |= 0.5653 \ m/s[/tex]
Explanation:
From the question we are told that
  The  spring constant is  [tex]k = 14 \ N/m[/tex]
   The  maximum extension of the spring is  [tex]A = 6.0 \ cm = 0.06 \ m[/tex]
   The number of oscillation is  [tex]n = 30[/tex]
   The  time taken is  [tex]t = 20 \ s[/tex]
Generally the the angular speed of this oscillations is mathematically represented as
      [tex]w = \frac{2 \pi}{T}[/tex]
where T is the period which is mathematically represented as
   [tex]T = \frac{t}{n}[/tex]
substituting values
   [tex]T = \frac{20}{30 }[/tex]
   [tex]T = 0.667 \ s[/tex]
Thus Â
    [tex]w = \frac{2 * 3.142 }{ 0.667}[/tex]
    [tex]w = 9.421 \ rad/s[/tex]
this angular speed can also be represented mathematically as
    [tex]w = \sqrt{\frac{k}{m} }[/tex]
=> Â [tex]m =\frac{k }{w^2}[/tex]
substituting values
   [tex]m =\frac{ 15 }{(9.421)^2}[/tex]
   [tex]m = 0.169 \ kg[/tex]
In SHM (simple harmonic motion )the equation for velocity is  mathematically represented as
    [tex]v = - Awsin (wt)[/tex]
The  velocity is maximum when  [tex]wt = \(90^o) \ or \ 1.5708\ rad[/tex]
   [tex]v_{max} = - A* w[/tex]
=> Â [tex]|v_{max} |= A* w[/tex]
=> Â Â [tex]|v_{max} |= 0.06 * 9.421[/tex]
=> Â [tex]|v_{max} |= 0.5653 \ m/s[/tex]
