A mass of \(0.404 \mathrm{~kg}\) is attached to a spring with a spring constant of \(206.9 \mathrm{~N} / \mathrm{m}\). Its oscillation is damped. with damping constant \(b=14.5 \mathrm{~kg} / \mathrm{s}\). What is the frequency of this damped oscillation?

Short Answer

Expert verified
Answer: The frequency of the damped oscillation is approximately \(3.687 \mathrm{~Hz}\).

Step by step solution

01

Identify the given values

As given in the problem, the mass \(m = 0.404 \mathrm{~kg}\), the spring constant \(k = 206.9 \mathrm{~N} / \mathrm{m}\), and the damping constant \(b = 14.5 \mathrm{~kg} / \mathrm{s}\).
02

Calculate intermediate values

Before calculating the damped frequency, we first need to find \((b/2m)^2\). To do this, divide the damping constant by two times the mass: $$ damping\_term = \frac{b}{2m} = \frac{14.5}{2 \times 0.404} = 17.945 \mathrm{~s}^{-1} $$ Now square this value: $$ damping\_term^2 = (17.945)^2 = 322.05 \mathrm{~s}^{-2} $$
03

Calculate the damped frequency

Use the formula for damped frequency to find the solution: $$ f_{damped} = \frac{1}{2 \pi} \sqrt{k/m - (b/2m)^2} $$ Substituting the given values in the equation: $$ f_{damped} = \frac{1}{2 \pi} \sqrt{\frac{206.9}{0.404} - 322.05} $$ Simplify the equation to get the damped frequency: $$ f_{damped} = \frac{1}{2 \pi} \sqrt{512.621 - 322.05} = \frac{1}{2 \pi} \sqrt{190.571} $$ $$ f_{damped} \approx 3.687 \mathrm{~Hz} $$ The frequency of the damped oscillation is approximately \(3.687 \mathrm{~Hz}\).

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