For the damped oscillator system shown in Figure 2, the block has a mass of 1.50 kg and thespring constant is 8.00 N/m. The damping force is given by -b(dx/dt), where b = 230 g/s. Theblock is pulled down 12.0 cm and released.(a) Calculate the time required for the amplitude of the resulting oscillations to fall to one-thirdof its initial value.(b) How many oscillations are made by the block in this time?
Question
For the damped oscillator system shown in Figure 2, the block has a mass of 1.50 kg and thespring constant is 8.00 N/m. The damping force is given by -b(dx/dt), where b = 230 g/s. Theblock is pulled down 12.0 cm and released.(a) Calculate the time required for the amplitude of the resulting oscillations to fall to one-thirdof its initial value.(b) How many oscillations are made by the block in this time?
Solution
(a) The equation for the amplitude of a damped oscillator as a function of time is given by:
A(t) = A0 * e^(-bt/2m)
where A0 is the initial amplitude, b is the damping constant, m is the mass, and t is time. We want to find the time t when the amplitude A(t) is one third of the initial amplitude A0. So we set up the equation:
A0/3 = A0 * e^(-bt/2m)
Solving for t gives:
t = -2m/b * ln(1/3)
Substituting the given values m = 1.50 kg, b = 230 g/s = 0.230 kg/s, we get:
t = -2*1.50/0.230 * ln(1/3) ≈ 6.49 s
(b) The frequency of oscillation of a damped oscillator is approximately the same as the frequency of the corresponding undamped oscillator, which is given by:
f = 1/2π * sqrt(k/m)
where k is the spring constant. Substituting the given values k = 8.00 N/m, m = 1.50 kg, we get:
f = 1/2π * sqrt(8.00/1.50) ≈ 0.47 Hz
The number of oscillations in time t is then f*t, so:
N = f*t = 0.47 Hz * 6.49 s ≈ 3.05
So the block makes approximately 3 oscillations in this time.
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