Question

Consider a 0.85 kg mass oscillating on a massless spring with spring constant of 45 N/m. This object reaches a maximum position of 12 cm from equilibrium. a) Determine the angular frequency of this mass. Then, determine the b) force, c) acceleration, d) elastic potential energy, e) kinetic energy, and f) velocity that it experiences at its maximum position. Determine the g) force, h) acceleration, i) elastic potential energy, j) kinetic energy, and k) velocity that it experiences at the equilibrium position. Show how you got each answer.

Answer #1

a)

m = mass oscillating = 0.85 kg

k = spring constant = 45 N/m

angular frequency is given as

w = sqrt(k/m)

w = sqrt(45/0.85)

w = 7.3 rad/s

b)

A = maximum stretch in the spring = 0.12 m

F_{max} = force = k A = 45 x 0.12 = 5.4 N

c)

acceleration is given as

a_{max} = Aw^{2} = (0.12) (7.3)^{2} =
6.4 m/s^{2}

d)

Elastic potential energy is given as

U_{max} = (0.5) k A^{2} = (0.5) (45)
(0.12)^{2} = 0.324 J

e)

kinetic energy at maximum position is zero.

KE_{max} = 0

f)

(0.5) m v^{2} = 0

hence v = 0

g)

at equilibrium position x = 0

F_{eq} = k x = 0

h)

a_{eq} = w^{2} x = 0

i)

Elastic potential energy is given as

U_{eq} = (0.5) k x^{2} = (0.5) (45)
(0)^{2} = 0

j)

using conservation of energy

kinetic energy = U_{max} = 0.324 J

k)

(0.5) m v_{eq}^{2} = 0.324

(0.5) (0.85) v_{eq}^{2} = 0.324

v_{eq} = 0.87 m/s

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