Question

8. A 0.40-kg mass is attached to a spring with a force constant of k =...

8. A 0.40-kg mass is attached to a spring with a force constant of k = 387 N/m, and the mass–spring system is set into oscillation with an amplitude of A = 3.7 cm. Determine the following.

(a) mechanical energy of the system
J

(b) maximum speed of the oscillating mass
m/s

(c) magnitude of the maximum acceleration of the oscillating mass
m/s2

Homework Answers

Answer #1

(a) Mechanical energy of the system which will be given by -

From conservation of energy, we have

E = P.Espring

E = (1/2) k A2

where, k = spring constant = 387 N/m

A = amplitude of an oscillation = 0.037 m

then, we get

E = [(0.5) (387 N/m) (0.037 m)2]

E = 0.264 J

(b) Maximum speed of the oscillating mass which will be given by -

From conservation of energy, we have

E = K.Emax

(0.264 J) = (1/2) m v2

where, m = mass of an object = 0.4 kg

then, we get

v2 = [(0.264 J) / (0.2 kg)]

v = 1.32 m2/s2

v = 1.14 m/s

(c) Magnitude of the maximum acceleration of an oscillating mass which will be given by -

From obey's hooke law, we get

F = k A m amax = k A

amax = [(387 N/m) (0.037 m)] / (0.4 kg)

amax = [(14.319 N) / (0.4 kg)]

amax = 35.7 m/s2

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