A 2.20-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 29.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position, and it subsequently undergoes simple harmonic oscillations. (a) Find the force constant of the spring. N/m (b) Find the frequency of the oscillations. Hz (c) Find the maximum speed of the object. m/s (d) Where does this maximum speed occur? x = ± m (e) Find the maximum acceleration of the object. m/s2 (f) Where does the maximum acceleration occur? x = ± m (g) Find the total energy of the oscillating system. J (h) Find the speed of the object when its position is equal to one-third of the maximum value. m/s (i) Find the magnitude of the acceleration of the object when its position is equal to one-third of the maximum value. m/s2

a. 145 N/m b. 1.29 Hz c. 1.62 m/s d.  0 m e. 13.2 m/s² f. ± 0.2 m g. 2.9 J h. 0.54 m/s i. 4.39 m/s²

Explanation:

a. The force constant of the spring

The spring force F = kx and k = F/x where k is the spring constant. F = 29.0 N and x = 0.200 m

k = 29.0 N/0.200 m = 145 N/m

b. The frequency of oscillations, f

f = 1/2π√(k/m)    m = mass = 2.20 kg

f = 1/2π√(145 N/m/2.20 kg) = 1.29 Hz

c. maximum speed of the object

The maximum elastic potential energy of the spring = maximum kinetic  energy of the object

1/2kx² = 1/2mv²

v = (√k/m)x where v is the maximum speed of the object

v = (√145/2.2)0.2 = 1.62 m/s

d Where does the maximum speed occur?

The maximum speed occurs at  0 m

e. The maximum acceleration

a = kx/m = 145 × 0.2/2.2 = 13.2 m/s²

f. The maximum acceleration occurs at x = ± 0.2 m

g. The total energy of the system is the maximum elestic potential energy of the system

E = 1/2kx² = 1/2 × 145 × 0.2² = 2.9 J

h. When x = x₀/3

1/2k(x₀/3)² = 1/2mv²

kx₀²/9 = mv²

v = 1/3(√k/m)x₀ = 1/3(√145/2.2)0.2 = 0.54 m/s

i When x = x₀/3

a = kx₀/3m =  145 × 0.2/(2.2 × 3)= 4.39 m/s²