A 3.60-kg object is attached to a spring and placed on frictionless, horizontal

Question

A 3.60-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 15.0 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.

Explanation / Answer

A) force constant k = F/x = 15/0.2 = 75 N/m

B) f = (1/2pi)*sqrt(k/m) = (1/(2*3.142))*sqrt(75/3.6) = 0.726 Hz

C) Vmax = A*w = 0.2*sqrt(75/3.6) = 0.91 m/s

D) x = 0 m at equilibrium position

E) amax = A*w^2 = 0.2*(k/m) = 0.2*(75/3.6) = 4.17 m/s^2

F) x = + or - 0.2 m

g) Total energy TE = 0.5*k*A^2 = 0.5*75*0.2*0.2 = 1.5 J

h) using law of conservation of energy

1.5 = 0.5*k*x^2 + 0.5*m*v^2

1.5 = 0.5*75*(0.2/3)^2 + 0.5*3.6*v^2 = 0.86 m/s

i) F = k*x = m*a

a = (k*x)/m = (75*0.2/3)/(3.6) = 1.4 m/s^2

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