Question

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/s^{2}

Answer #1

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

From conservation of energy, we have

E = P.E_{spring}

E = (1/2) k A^{2}

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.E_{max}

(0.264 J) = (1/2) m v^{2}

where, m = mass of an object = 0.4 kg

then, we get

v^{2} = [(0.264 J) / (0.2 kg)]

v = _{
}1.32 m^{2}/s^{2}

**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 a_{max} = k A

a_{max} = [(387 N/m) (0.037 m)] / (0.4 kg)

a_{max} = [(14.319 N) / (0.4 kg)]

**a _{max} = 35.7 m/s^{2}**

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