A spring is stretched 220 mm by a 16 kg block. If the block is displaced 90 mm d

Question

A spring is stretched 220 mm by a 16 kg block. If the block is displaced 90 mm downward from its equilibrium position and given a downward velocity of 0 80 m/s, determine the equation which describes the motion. Assume that positive displacement is downward y = 0.120 sin 6.68t - 0.220cos 6.68t y = 0.120 sin 6.68t + 9.00 times 10^-2 cos6.68t y = 0.120 cos 6.68t - 9.00 times 10^-2 sin 6.68t y = -0.120 sin 6.68t + 9.00 times 10^-2 cos6.68t What is the phase angle? Express your answer with the appropriate units. Phi =

Explanation / Answer

F = kx - mg = 0 (for equilibrium position)

k (0.220) = 16 x 9.8

k = 712.7 N/m

w = sqrt(k/m) = sqrt(712.7 / 16) = 6.68 rad/s

v^2 = w^2 ( A^2 - x^2)

0.80^2 = 6.68^2 ( A^2 - 0.09^2)

A = 0.150 mm

x = A cos(wt + phi) = 0.150 cos(6.68t + phi)

putting t = 0 , x = 0.09 m

0.09 = 0.150 cos( phi)

phi = 0.926 rad

(A) x = 0.150 cos(6.68t + phi)

cos(A+ B) = cosA cosB - sinA sinB

x = 0.150[ cos6.68t cos0.926 - sin6.68t sin0.926]

x = 0.090 cos6.68t - 0.120 sin6.68 t

Ans: x = - 0.120 sin6.68t + 9x 10^-3 cos6.68t

Last option

(B) phase angle = 0.926 rad

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