MULTISTEP In this problem, derive the expression for the

Chapter , Problem 105

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MULTISTEP In this problem, derive the expression for the average power delivered by a driving force to a driven oscillator (Figure 14-39).

                                                 

(a) Show that the instantaneous power input of the driving force is given by \(P=F v=-A \omega F_0 \cos \omega t \sin (\omega t-\delta)\).

(b) Use the identity \(\sin \left(\theta_1-\theta_2\right)=\sin \theta_1 \cos \theta_2-\cos \theta_1 \sin \theta_2\) to show that the equation in Part (a) can be written as \(P=A \omega F_0 \sin \delta \cos ^2 \omega t \sin -A \omega F_0 \cos \delta \cos \omega t \sin \omega t\)

(c) Show that the average value of the second term in your result for Part (b) over one or more periods is zero, and that therefore \(P_{\text {av }}=\frac{1}{2} A \omega F_0 \sin \delta\).

(d) From Equation 14-56 for \(\tan \delta\), construct a right triangle in which the side opposite the angle \(\delta\) is \)b \omega\) and the side adjacent is \(m\left(\omega_0^2-\omega^2\right)\), and use this triangle to show that

                                   \(\sin \delta=\frac{b \omega}{\sqrt{m^2\left(\omega_0^2-\omega^2\right)^2+b^2 \omega^2}}=\frac{b \omega A}{F_0} \text {.ssm }\)

(e) Use your result for Part (d) to eliminate \(\omega A\) from your result for Part (c), so that the average power input can be written as

                                   \(P_{\mathrm{av}}=\frac{1}{2} \frac{F_0^2}{b} \sin ^2 \delta=\frac{1}{2}\left[\frac{b \omega^2 F_0^2}{m^2\left(\omega_0^2-\omega^2\right)^2+b^2 \omega^2}\right]\)