Answer:
a) T = 1.69 s, b)  k = 0.825 N / m, c)  v = 1.46 feet/s, d) a = 5.41 ft / s²,
e)   v = - 1,319 ft / s,    a = - 2.70 ft / s², f) K = 4.8 10⁻³ J, U = 1.49 10⁻³ J
Explanation:
In a mass-spring system with simple harmonic motion, the angular velocity is
          w =  
a) find the period
 angular velocity, frequency, and period are related
          w = 2π f = 2π / T
           f = 1 / T
           T = 1 / f
            T = 1 / 0.59
            T = 1.69 s
b) the spring constant
          w = 2π f
          w = 2π 0.59
          w = 3.70 rad / s
          w² = k / m
           k = w² m
           k = 3.70² 0.060
           k = 0.825 N / m
c) the maximum speed
simple harmonic movement is described by the expression
           x = A cos (wt + Ф)
speed is defined by
          v = 
           v = -A w sin (wt + fi)
the speed is maximum when the cosine is ± 1
           v = A w
           v = 0.394 3.70
           v = 1.46 feet/s
d) maximum acceleration
             a =  
             a = - A w² cos wt + fi
the acceleration is maximum when the cosine is ±1
             a = A w²
             a = 0.394 3.70²
             a = 5.41 ft / s²
e) velocity and acceleration for x = 6 cm
let's reduce the cm to feet
             x = 6 cm (1 foot / 30.48 cm) = 0.1969 foot
Before doing this part we must find the phase angle (Ф), the most common way to start the movement is to move the spring a small distance and release it, so its initial speed is zero for t = 0 s
let's use the expression for the velocity
            v = -A w sin (0 + Фi)
            0 = - A w sin Ф
so sin Ф = 0 which implies that Фi = 0
 the equation of motion is
             x = A cos wt
             x = 0.394 cos 3.70t
we substitute
            0.1969 = 0.394 cos 370t
            3.70 t = cos⁻¹ (0.1969 / 0.394)
let's not forget that the angle is in radians
            3.70, t = 1.047
            t = 1.047 / 3.70
            t = 0.2826 s
we substitute this time in the equation for velocity and acceleration
            v = - Aw sin wt
            v = - 0.394 3.70 sin 3.70 0.2826
            v = - 1,319 ft / s
            a = - A w² cos wt
            a = - 0.394 3.70² cos 3.70 0.2826
            a = - 2.70 ft / s²
f) the kinetic and potential energy at this point
            K = ½ m v²
let's slow down to the SI system
            v = 1.319 ft / s (1 m / 3.28 ft) = 0.402 m / s
            
            K = ½ 0.060 0.402²
            K = 4.8 10⁻³ J
            U = ½ k x²
            U = ½ 0.825 0.06²
            U = 1.49 10⁻³ J