A 2.15 kg lightly damped harmonic oscillator has an angular oscillation frequency of 0.261 rad/s. If the maximum displacement of 2.0 m occurs when t = 0.00 s, and the damping constant b is 0.74 kg/s what is the object's displacement when t = 4.01 s?
A) 0.50 m
B) 0.43 m
C) 0.58 m
D) 0.65 m

Question

Quizplus · Accepted Answer

The correct answer is A because the displacement of the object at time t = 4.01 s is 0.50 m. This can be calculated using the equation:x(t) = A * e^(-bt/2m) * cos(ωt + φ)where:x(t) is the displacement of the object at time tA is the maximum displacement of the objectb is the damping constantm is the mass of the objectω is the angular oscillation frequencyφ is the phase anglePlugging in the given values, we get:x(4.01 s) = 2.0 m * e^(-0.74 kg/s * 4.01 s / 2 * 2.15 kg) * cos(0.261 rad/s * 4.01 s + 0)= 0.50 m

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