Consider a 0.50 kg mass on a spring with a constant of 600 N/m. The spring and m

Question

Consider a 0.50 kg mass on a spring with a constant of 600 N/m. The spring and mass system are originally at equilibrium when the mass is at y = 0. The mass is puled down and it is 40 0 cm below its equilibrium position and then released (y = -40.0 cm). What is the period of oscillation when the mass is released? What is the frequency of the mass as it oscillates? What is the net force acting on the mass when y = 20 0 cm? What is the acceleration of the mass when y = 20.0 on? What is the velocity of the mass when y = 20.0 cm'

Explanation / Answer

from the given

mass m = 0.50 kg

force constant or spring constant k = 600 N / m

displacement y = 40.0 cm = 0.40 m

(a)

the timeperiod is given by

T = 2 pi sqrt(m / k)

= 2 (3.14) sqrt(0.50 / 600)

= 0.181 s

(b)

frequency f = (1 / T)

= (1 / 0.181)

= 5.52 Hertz

(c)

force acting is

F = k y

= (600) (0.20 m)

= 120 N

(d)

acceleration a = (k / m) y

= (600 / 0.50) (0.20)

= 240 m / s2

(e)

velocity

v = sqrt(k / m) y

= sqrt(600 / 0.50) (0.20)

= (34.64) (0.20)

= 6.928 m / s

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