A mass, \(M=1.6 \mathrm{~kg}\), is attached to a wall by a spring with \(k=578 \mathrm{~N} / \mathrm{m}\). The mass slides on a frictionless floor. The spring and mass are immersed in a fluid with a damping constant of \(6.4 \mathrm{~kg} / \mathrm{s}\). A horizontal force, \(\mathrm{F}(\mathrm{t})=\mathrm{F}_{\mathrm{d}} \cos \left(\omega_{\mathrm{d}} \mathrm{t}\right)\) where \(F_{d}=52 \mathrm{~N},\) is applied to the mass through a knob, caus. ing the mass to oscillate back and forth. Neglect the mass of the spring and of the knob and rod. At approximately what frequency will the amplitude of the mass's oscillation be greatest, and what is the maximum amplitude? If the driving frequency is reduced slightly (but the driving amplitude remains the same), at what frecuency will the amplitude of the mass's ascillation be half of the maximum amplitude?

Step by step solution

01

Find the natural frequency of the system

The natural frequency, \(\omega_0\), can be found using the spring constant (k) and the mass (M) according to the following equation: \(\omega_0 = \sqrt{\frac{k}{M}}\) Now, plug in the values of k and M. \(\omega_0 = \sqrt{\frac{578\,\text{N/m}}{1.6\,\text{kg}}}\) \(\omega_0 = \sqrt{361.25\,\text{s}^{-2}}\) \(\omega_0 = 19\,\text{rad/s}\) The natural frequency of the system is 19 rad/s.

02

Find the damping factor and quality factor

First, we find the damping factor, \(\gamma\), using the damping constant (b): \(\gamma = \frac{b}{2M}\) \(\gamma = \frac{6.4\,\text{kg/s}}{2 \cdot 1.6\,\text{kg}}\) \(\gamma = 2\,\text{rad/s}\) Now, we find the quality factor (Q), which is the ratio of the natural frequency to the damping factor: \(Q=\frac{\omega_0}{\gamma}\) \(Q=\frac{19\,\text{rad/s}}{2\,\text{rad/s}}\) \(Q=9.5\) The damping factor is 2 rad/s and the quality factor is 9.5.

03

Find the maximum amplitude and the frequency at which it occurs

The amplitude of the mass's oscillation will be greatest when the driving frequency is equal to the resonant frequency. The resonant frequency, \(\omega_{max}\), can be found using the following equation: \(\omega_{max}=\sqrt{\omega_0^2-\gamma^2}\) \(\omega_{max}=\sqrt{(19\,\text{rad/s})^2-(2\,\text{rad/s})^2}\) \(\omega_{max}=\sqrt{357\,\text{s}^{-2}}\) \(\omega_{max}=18.87\,\text{rad/s}\) Now, we can find the maximum amplitude, A_max, using the quality factor and the driving force (F_d): \(A_{max}=\frac{Q \cdot F_d}{k}\) \(A_{max}=\frac{9.5 \cdot 52\,\text{N}}{578\,\text{N/m}}\) \(A_{max}=0.864\,\text{m}\) The maximum amplitude is 0.864 m and occurs at a driving frequency of 18.87 rad/s.

04

Find the frequency at which amplitude is half of the maximum amplitude

When the amplitude is half of the maximum amplitude, the power (P) absorbed by the system is also half of the maximum power, thus: \(\frac{P_{max}}{2}=\frac{F_d^2}{4k} \cdot Q \cdot \left(\frac{\gamma^2}{(\omega_{max}^2-\omega^2)^2+\gamma^2\omega^2}\right)\) Where \(\omega\) is the frequency at which the amplitude is half of the maximum amplitude. Simplifying it yields the following equation: \(\frac{2}{Q}=\frac{\gamma^2}{(\omega_{max}^2-\omega^2)^2+\gamma^2\omega^2}\) Now, substitute the known values and solve for \(\omega\): \(\frac{2}{9.5}=\frac{(2\,\text{rad/s})^2}{(18.87\,\text{rad/s})^2-\omega^2)^2+(2\,\text{rad/s})^2\omega^2}\) This equation can be solved numerically to find the value of \(\omega\): \(\omega \approx 18.48\,\text{rad/s}\) The frequency at which the amplitude of the mass's oscillation is half of the maximum amplitude is approximately 18.48 rad/s.

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