Chapter 4.6, Problem 73E

To determine

Find the damping factor and graph the given function and describe the behavior of the function as x increases without bound.

Answer to Problem 73E

The damping factor $e^{\frac{-x^2}{2}}$

Explanation of Solution

Given:

The period for $y=sinx$ is $2p$ .

The damping factor in $y=e^{\frac{-x^2}{2}}\sin x$ is $e^{\frac{-x^2}{2}}$

Let's plot the function $g(x)=e^{\frac{-x^2}{2}}\sin x$ and the damping factor $e^{\frac{-x^2}{2}}$ in the same viewing window from −2p to 2p, using the command

  ${}_{plots[multiple]}\left({}_{plot,}\left[e^{\left(\frac{-x·x}{2}\right)}.\sin (x),x=-2\pi \mathrm{..2}\pi \right],\left[e^{\left(\frac{-x·x}{2}\right)},x=-2\pi \mathrm{..2}\pi \right]\right)$

in the software Maple.

Thus the graph is

  

The above graph represents the equation $g(x)=e^{\frac{-x^2}{2}}\sin x$ and the damping factor $e^{\frac{-x^2}{2}}$

Further, from the above graph, x increases, the function approaches 0.

Therefore, $x\to \infty ,g(x)\to 0$