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Chapter 14: Problem 48

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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Most popular questions from this chapter

Object \(A\) is four times heavier than object B. Fach object is attached to a spring, and the springs have equal spring constants. The two objects are then pulled from their equilibrium positions and released from rest. What is the ratio of the periods of the two oscillators if the amplitude of \(A\) is half that of \(B ?\) a) \(T_{A}: T_{\mathrm{g}}=1: 4\) c) \(T_{A}: T_{\mathrm{B}}=2\) : b) \(T_{A}: T_{B}=4: 1\) d) \(T_{A}: T_{B}=1: 2\)

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