A mass of .5 kg hangs on a spring. The stiffness of the spring is 100 N/m, and t
ID: 3278353 • Letter: A
Question
A mass of .5 kg hangs on a spring. The stiffness of the spring is 100 N/m, and the mechanical resistance is 1.4 kg/s. The force (N) driving the system f=2cos(5t). (a) What will be the steady state amplitude, speed amplitude, and average power dissipation? (.0228 m , 0.114 m/s , 9.08mW) (b) What is the phase angle between speed and force? (-85.4 degrees) (c) What is the resonance frequency and what would be the displacement amplitude, speed amplitude at this frequency and for the same force magnitude as in (a)? ( 2.25 Hz, .101 m, 1.43 m/s, 1.43 W) (d) What is the Q of the system, and over what range of frequencies will the power loss be at least 50% of its resonance value? (5.05, between 2.04 and 2.48 Hz)
Given answers given inside parenthesis
Explanation / Answer
(A) a = Fo / m sqrt[(w0^2 - w^2)^2 + 4 b^2 w^2]
Fo = 2 N
m = 0.5 kg
w0 = sqrt(k/m) = sqrt(100/0.5) = 14.14 rad/s
w0^2 = 200
w = 5 rad/s
b = 1.4/ (2 x0.5) = 1.4
putting in values,
a = 0.0228 m .......Ans(Amplitude)
Speed apmlitude = a w = 0.0228 x 5 =0.114 m/s ........Ans
average power dissipation = (b Fo^2 / m) ( w^2 / [(w0^2 - w^2)^2 + 4 b^2 w^2] )
= (1.4 x 2^2 / 0.5) [ 5^2 / [ (200 - 5^2)^2 + 4(1.4^2)(5^2)]]
= 9.08 x 10^-3 W Or 9.08 mW
(b) average P = Fo vo cos(phi) / 2
9.08 x 10^-3 = (2)(0.114) cos(phi) / 2
cos(phi) = 0.0796
phi = 85.4 deg
(c) At resonance frequency wr = sqrt[ wo^2 - 2b^2]
= sqrt[200 - 2(1.4^2)]
= 14 rad/s
so resonance frequency = wr / 2pi = 2.23 Hz ........Ans
a = Fo / 2 b m sqrt(w0^2 - b^2)
a = 0.102 m .........Ans
Speed amplitude = a w = 0.102 x 14 = 1.43 m/s .......Ans
(d)
Q = w0 / 2 b
= sqrt(200) / (2 x 1.4)
= 5.05 .........Ans
w1 = - w0/2Q + w0 sqrt[1 + 1/(4 Q^2)]
w1 = - 1.40 + 14.21
w1 = 12.81 rad/s
f1 = w1 /2pi = 2.04 Hz ......Ans
w2 = w0/2Q + w0 sqrt[1 + 1/(4 Q^2)]
w2 = 1.40 + 14.21
w2 = 15.61 rad/s
f2 = w2 / 2pi
= 2.48 Hz ....Ans
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