Q. 84 - Calculus

Question

In the previous exercise we gave an equation describing spring motion without air resistance. If we take into account friction due to air resistance, the mass will oscillate up and down according to the equation

$s\left(t\right)=e^{(-f/2m)t}\left(A\sin \left(\frac{\sqrt{4km-f^2}}{2m}t\right)\right.\left.+B\cos \left(\frac{\sqrt{4km-f^2}}{2m}t\right)\right)$

where m, k, A, and B are the constants described in Problem $83$ and f is a positive "friction coefficient" that measures the amount of friction due to air resistance.

1. Find the limit of $s\left(t\right)$ as $t\to \infty$. What does this say about the long-term behavior of the mass on the end of the spring?
2. Explain how this limit relates to the fact that the new equation for $s\left(t\right)$ does take friction due to air resistance into account.
3. Suppose the bob at the end of the spring has a mass of $2$ grams, the coefficient for the spring is $k=9$, and the friction coefficient is $f=6$. Suppose also that the spring is released in such a way that $A=4$ and $B=2$. Use a graphing utility to graph the function $s\left(t\right)$ that describes the distance of the mass from its equilibrium position. Use your graph to support your answer to part (a).

Step-by-Step Solution

Answer

Part(a). The limit is equal to zero.

Part(b). Since the limit is zero, there is some friction to damp the oscillations of the spring.

Part(c). The graph is as follows,

Part(a) Step 1. Given Information

We are given a function,

$s\left(t\right)=e^{(-f/2m)t}\left(A\sin \left(\frac{\sqrt{4km-f^2}}{2m}t\right)\right.\left.+B\cos \left(\frac{\sqrt{4km-f^2}}{2m}t\right)\right)$

where m, k, A, and B are the constants.

Part(a) Step 2. Finding the limit

The limit is given by, 

$$\begin{gathered} \underset{t\to \infty }{\lim }s\left(t\right)=\underset{t\to \infty }{\lim }\left[e^{(-f/2m)t}\left[A\sin \left(\frac{\sqrt{4km-f^2}}{2m}t\right)+B\cos \left(\frac{\sqrt{4km-f^2}}{2m}t\right)\right]\right. \\ =e^{(-f/2m)\infty }\left[A\sin \left(\frac{\sqrt{4km-f^2}}{2m}\infty \right)+B\cos \left(\frac{\sqrt{4km-f^2}}{2m}\infty \right)\right] \\ =0\left[A\sin \right(\infty )+B\cos (\infty \left)\right] \\ =0 \end{gathered}$$

Part(b) Step 1. About the friction due to air resistance

As the limit is equal to zero, there is some friction to damp the oscillations of the spring. 

Part(c) Step 1. Graphing the function

Putting

 $$\begin{gathered} m=2, \\ k=9, \\ f=6, \\ A=4, \\ B=2 \end{gathered}$$ 

in the equation, we get

$$\begin{gathered} s\left(t\right)=e^{(-6/2·2)t}\left[4\sin \left(\frac{\sqrt{4·9·2-6^2}}{2·2}t\right)+2\cos \left(\frac{\sqrt{4·9·2-6^2}}{2·2}t\right)\right] \\ =e^{(-3/2)t}\left[4\sin \left(\frac{\sqrt{72-36}}{4}t\right)+2\cos \left(\frac{\sqrt{72-36}}{4}t\right)\right] \\ =e^{(-3/2)t}\left[4\sin \left(\frac{\sqrt{36}}{4}t\right)+2\cos \left(\frac{\sqrt{36}}{4}t\right)\right] \\ =e^{(-3/2)t}\left[4\sin \left(\frac{6}{4}t\right)+2\cos \left(\frac{6}{4}t\right)\right] \\ =e^{(-3/2)t}\left[4\sin \left(\frac{3}{2}t\right)+2\cos \left(\frac{3}{2}t\right)\right] \end{gathered}$$

The graph is as follows,