Chapter 3, Problem 69RE

To determine

To find: The velocity and the acceleration of the object.

Explanation of Solution

Given: The equation of motion is $s=A{e}^{-ct}\cos \left(\omega t+\delta \right)$

Velocity is the derivation of the first position.

So, the expression for velocity is calculated as:

  $\begin{array}{c}v=y^{\prime }\\ =A{e}^{-ct}\left(-c\right)\cos \left(\omega t+\delta \right)+A{e}^{-ct}\left(-\sin \left(\omega t+\delta \right)\right)\left(\omega \right)\\ =-A{e}^{-ct}\left(c\cos \left(\omega t+\delta \right)+\omega \sin \left(\omega t+\delta \right)\right)\end{array}$

Acceleration is the second derivative of position. So,

  $\begin{array}{c}a=y^{″}\\ =\left(-A{e}^{-ct}\left(-c\right)\right)\left(c\cos \left(\omega t+\delta \right)+\omega \sin \left(\omega t+\delta \right)\right)+\left(A{e}^{-ct}\right)\left(c^2\cos \left(\omega t+\delta \right)+{\omega }^2\cos \left(\omega t+\delta \right)\right)\\ =A{e}^{-ct}\left[c^2\cos \left(\omega t+\delta \right)+c\omega \sin \left(\omega t+\delta \right)+c\omega \left(\omega t+\delta \right)-{\omega }^2\cos \left(\omega t+\delta \right)\right]\omega t+\delta \\ =A{e}^{-ct}\left[\left(c^2-{\omega }^2\right)\cos \left(\omega t+\delta \right)+2c\omega \sin \left(\omega t+\delta \right)\right]\end{array}$

Therefore, the velocity of the object is $-A{e}^{-ct}\left(c\cos \left(\omega t+\delta \right)+\omega \sin \left(\omega t+\delta \right)\right)$ and the acceleration of the objectis $A{e}^{-ct}\left[\left(c^2-{\omega }^2\right)\cos \left(\omega t+\delta \right)+2c\omega \sin \left(\omega t+\delta \right)\right]$ .