A 0.500 kg mass is attached to a spring of constant 150 N/m. A driving force F(t) = (12.0 N) $\cos (ω t)$ is applied to the mass, and the damping coefficient b is 6.00 Ns/m. What is the amplitude (in cm) of the steady-state motion of is equal to half of the natural frequency $ω_{0}$ of the system?

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answer is 9.7.

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Detailed Solution

$D = (F_{0} / m) / {[{(ω_{0}^{2} - ω^{2})}^{2} + {(b ω / m)}^{2}]}^{1 / 2}$

$ω_{0}^{2} = k / m = 300, ω = ω_{0} / 2, ω^{2} = 300 / 4 = 75$

$F_{0} / m = 12 / 0.5 = 24 .$

$D = 24 / {[{(300 - 75)}^{2} + 36 * 75 / 0.25]}^{1 / 2} = 0.0968$

m = 9.7 cm is the amplitude of the steady

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