A 50.0-g hardboiled egg moves on the end of a spring with force constant k = 20.

Question

A 50.0-g hardboiled egg moves on the end of a spring with force constant k = 20.0 N/w It is released with amplitude 0.60 m at t = 0 s. A damping force F+s = -by acts on the egg. After it oscillates for 4.00 s. the amplitude of the motion has decreased to 0.150 m a) Calculate the magnitude of the damping coefficient b. b) Calculate the angular frequency w c) Verify that the phase angle phi = 0. d) Find the displacement times at time t = 4.00.s e) How long it takes to reduce the amplitude less than 1%.

Explanation / Answer

frequency w0 = sqrt(k/m) = 20
period T = 2pi/w0 = 0.314
N (no. of cycles) = elapsed time/T = 4/0.314 = 12.74
is the "logarithmic decrement", the proportionate damping loss over one oscillation cycle. AR is the amplitude ratio = 4.
= ln(AR)/N = 0.1088
= /sqrt(4pi^2+^2) = 0.0173
B = 2*w0*m = 2*k/w0 = 2/sqrt(km) = 0. 0346 N-s/m

a) damping coefficient = 0.0346 N-s/m

b) Angular frequency = 20

c) Phase angle = 0

d) displacement = 0.15*cos(20*4) = 0.026 m = 2.6 cm

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