Chapter 8.5, Problem 42AYU

To determine

To find:

a. An equation that relates the displacement $\text{d}$ of the object from its rest position after $\text{t}$ seconds and to graph the equation found in part (a) for 5 oscillations using a graphing utility.

Answer to Problem 42AYU

$d\left(t\right) = - 16e^{ - \frac{0.65t}{30}}\text{cos}\sqrt{4\pi ²/25 - \frac{.4225}{900}}t$

Explanation of Solution

Given:

$\text{m} = 15, \text{a} = 16, \text{b} = 0.65, \text{T} = 5$

Formula used:

The displacement $\text{d}$ of an oscillating object from its at–rest position at time $\text{t}$ is given by $d\left(t\right) = a{e}^{ - \frac{bt}{2m}}\text{cos}\sqrt{{\omega }^2 - \frac{b^2}{4m^2}}t$, where $\text{b}$ is the damping factor or damping coefficient and m is the mass of the oscillating object. Here |a| is the displacement at $\text{t} = 0$, and $\frac{2\text{π}}{\text{w}}$ is the period under simple harmonic motion (no damping).

Calculation:

$\text{d}\left(\text{t}\right) = {\text{ ae}}^{ - \frac{\text{bt}}{2\text{m}}}\text{cos}\sqrt{{\text{ω}}^2 - \frac{{\text{b}}^2}{4{\text{m}}^2}}\text{t}$

$\text{m} = 15, \text{a} = 16, \text{b} = 0.65, \text{T} = 5$

$d\left(t\right) = - 16e^{ - \frac{0.65t}{2\left(15\right)}}\text{cos}\sqrt{{\left(\frac{2\pi }{5}\right)}^2 - \frac{{0.65}^2}{4 * {15}^2}}t$

$d\left(t\right) = - 16e^{ - \frac{0.65t}{30}}\text{cos}\sqrt{4\pi ²/25 - \frac{.4225}{900}}t$

Graph of $d\left(t\right) = - 16e^{ - \frac{0.65t}{30}}\text{cos}\sqrt{4\pi ²/25 - \frac{.4225}{900}}t$

The graph for $d\left(t\right) = - 16e^{ - \frac{0.65t}{30}}\text{cos}\sqrt{4\pi ²/25 - \frac{.4225}{900}}t$ is plotted above using the graphing tool.