Chapter 16, Problem 82PQ

To determine

To prove the relation for system.

Answer to Problem 82PQ

The solution to the equation is $\underset{\_}{F_{\max }\cos \left({\omega }_{\text{drv}}t\right)=m\left(d^{2}y/d{t}^2\right)+b\left(dy/dt\right)+ky}$.

Explanation of Solution

Write the expression for the given solution of the equation.

  $y\left(t\right)=A_{\text{drv}}\cos \left({\omega }_{\text{drv}}t+{\varphi }_{\text{drv}}\right)$                                                                                  (I)

Here, $y\left(t\right)$ is the instantaneous displacement, $A_{\text{drv}}$ is the amplitude, ${\omega }_{\text{drv}}$ is the angular frequency, $t$ is the time and ${\varphi }_{\text{drv}}$ is the phase angle.

Write the expression for the equation.

  $F_{\max }\cos \left({\omega }_{\text{drv}}t\right)=m\left(d^{2}y/d{t}^2\right)+b\left(dy/dt\right)+ky$                                                 (II)

Here, $F_{\max }$ is the maximum force, $\left(d^{2}y/d{t}^2\right)$ is the second derivative and $\left(dy/dt\right)$ is the first derivative.

Write the expression for the angular frequency.

  ${\omega }_{\text{drv}}=\sqrt{\frac{k}{m}}$                                                                                                         (III)

Here, $k$ is wave vector and $m$ is mass.

Conclusion:

Taking first and second derivative from equation (I).

  $\frac{dy}{dt}=-{\omega }_{\text{drv}}{A}_{\text{drv}}\sin \left({\omega }_{\text{drv}}t+{\varphi }_{\text{drv}}\right)$                                                                        (IV)

  $\frac{d^{2}y}{d{t}^2}=-{\omega }_{\text{drv}}^2{A}_{\text{drv}}\cos \left({\omega }_{\text{drv}}t+{\varphi }_{\text{drv}}\right)$                                                                       (V)

Substitute $-{\omega }_{\text{drv}}{A}_{\text{drv}}\sin \left({\omega }_{\text{drv}}t+{\varphi }_{\text{drv}}\right)$ for $\frac{dy}{dt}$, $-{\omega }_{\text{drv}}^2{A}_{\text{drv}}\cos \left({\omega }_{\text{drv}}t+{\varphi }_{\text{drv}}\right)$ for $\frac{d^{2}y}{d{t}^2}$ and $A_{\text{drv}}\cos \left({\omega }_{\text{drv}}t+{\varphi }_{\text{drv}}\right)$ for $y\left(t\right)$ in Equation (II) to find $F_{\max }\cos \left({\omega }_{\text{drv}}t\right)$.

  $F_{\max }\cos \left({\omega }_{\text{drv}}t\right)=\left(\begin{array}{c}m\left[-{\omega }_{\text{drv}}^2{A}_{\text{drv}}\cos \left({\omega }_{\text{drv}}t+{\varphi }_{\text{drv}}\right)\right]+\\ b\left[-{\omega }_{\text{drv}}{A}_{\text{drv}}\sin \left({\omega }_{\text{drv}}t+{\varphi }_{\text{drv}}\right)\right]+\\ k\left[A_{\text{drv}}\cos \left({\omega }_{\text{drv}}t+{\varphi }_{\text{drv}}\right)\right]\end{array}\right)$                             (VI)

Using trigonometric identity and comparing both side the components of sine and cosine terms.

  $F_{\max }=A_{\text{drv}}\left[\left(k-m{\omega }_{\text{drv}}^2\right)\cos \left({\varphi }_{\text{drv}}\right)-b{\omega }_{\text{drv}}\sin {\varphi }_{\text{drv}}\right]$                                             (VII)

  $0=A_{\text{drv}}\left[-\left(k-m{\omega }_{\text{drv}}^2\right)\sin \left({\varphi }_{\text{drv}}\right)-b{\omega }_{\text{drv}}\cos {\varphi }_{\text{drv}}\right]$                                              (VIII)

Write the expression for the trigonometric identities by using equation (VIII).

  $\tan \left({\varphi }_{\text{drv}}\right)=\frac{-b{\omega }_{\text{drv}}}{k-m{\omega }_{\text{drv}}^2}$                                                                                       (IX)

  $\sin \left({\varphi }_{\text{drv}}\right)=\frac{-b{\omega }_{\text{drv}}}{\sqrt{{\left(b{\omega }_{\text{drv}}\right)}^2+{\left(k-m{\omega }_{\text{drv}}^2\right)}^2}}$                                                                 (X)

  $\cos \left({\varphi }_{\text{drv}}\right)=\frac{\left(k-m{\omega }_{\text{drv}}^2\right)}{\sqrt{{\left(b{\omega }_{\text{drv}}\right)}^2+{\left(k-m{\omega }_{\text{drv}}^2\right)}^2}}$                                                               (XI)

Substitute $\frac{-b{\omega }_{\text{drv}}}{\sqrt{{\left(b{\omega }_{\text{drv}}\right)}^2+{\left(k-m{\omega }_{\text{drv}}^2\right)}^2}}$ for $\sin \left({\varphi }_{\text{drv}}\right)$ and $\frac{\left(k-m{\omega }_{\text{drv}}^2\right)}{\sqrt{{\left(b{\omega }_{\text{drv}}\right)}^2+{\left(k-m{\omega }_{\text{drv}}^2\right)}^2}}$ for $\cos \left({\varphi }_{\text{drv}}\right)$ in equation (VII) to find $A_{\text{drv}}$.

  $\begin{array}{c}{F}_{\max }=A_{\text{drv}}\left[\left(k-m{\omega }_{\text{drv}}^2\right)\left[\frac{\left(k-m{\omega }_{\text{drv}}^2\right)}{\sqrt{{\left(b{\omega }_{\text{drv}}\right)}^2+{\left(k-m{\omega }_{\text{drv}}^2\right)}^2}}\right]-b{\omega }_{\text{drv}}\left[\frac{-b{\omega }_{\text{drv}}}{\sqrt{{\left(b{\omega }_{\text{drv}}\right)}^2+{\left(k-m{\omega }_{\text{drv}}^2\right)}^2}}\right]\right]\\ =A_{\text{drv}}\sqrt{{\left(b{\omega }_{\text{drv}}\right)}^2+m^2\left({\omega }^2-{\omega }_{\text{drv}}^2\right)}\\ A_{\text{drv}}=\frac{F_{\max }}{\sqrt{{\left(b{\omega }_{\text{drv}}\right)}^2+m^2\left({\omega }^2-{\omega }_{\text{drv}}^2\right)}}\end{array}$

Thus, the solution is satisfying both side of the equation.