Question

The displacement from equilibrium of a mass oscillating on the end of a spring suspended from a ceiling is given below, where y is the displacement (in feet) and t is the time in seconds. Use a graphing utility to graph the displacement function on the interval [0, 10]. Find the value of t past which the displacement is less than 3 inches from equilibrium. (Round your answer to two decimal places.) y = [1.54e^(−0.22t)]*cos(4.9t)

Answer 1

\[y = 1.54e^{-0.22t}\cos(4.9t)\]

You want the value of \(t\) where \(y < \frac{3}{12}\)

\[\frac{3}{12} > 1.54e^{-0.22t}\cos(4.9t)\]
Key here is to realize that \(\cos(4.9t)\) oscillates between 1 and -1. We can replace that part with 1 to find the point at which the damping exponential function is a small enough value that we'll never be more than 3 inches from equilibrium.

Answer 2

Here's a graph of a similar equation (numbers changed, so don't just copy and paste this into your answer!) Two gridlines show the amplitude of 3 inches.

Answer 3

Where do you get\[\frac{ 3 }{ 12 }\]from?

Answer 4

" y is the displacement (in feet)"
"displacement is less than 3 inches from equilibrium"

Answer 5

3 inches = 3/12 of a foot, right?

Answer 6

So then what does t equal?

Answer 7

t=nπ4.9

Answer 8

t=nPi/4.9

Answer 9

oops lol

Answer 10

What is n then?

Answer 11

The answer is suppose to be a number

Answer 12

Yes. You need to find the value of \(t\) so that the "envelope" function (the exponential) is going to be less than the cutoff value for the displacement. Remember, the cos function ranges between -1 and 1.

Answer 13

The dashed line is the exponential "envelope" function that is multiplied by the cos function. When the envelope gets down to 3/12, no matter what the value of the cos function is, the resulting displacement will be less than 3 inches.

Answer 14

So do I put \[\int\limits_{0}^{3}1.54e ^{-.22t}\cos(4.9t) dt\]?

Answer 15

No...Solve the following:

\[\frac{3}{12} = 1.54e^{-0.22t}\]That's all there is to it...

Answer 16

All you are trying to do is find the value of \(t\) where the dashed line crosses the horizontal line in this figure:

Answer 17

I got 8.26 but still got it wrong

Answer 18

If you've copied the problem correctly, that should be correct. Beyond that value of \(t\), it is impossible for the displacement to be greater than 3 inches.

Answer 19

I already put it on my calculator: 1.54e^(−0.22*8.26399) = .25 = 3/12

Answer 20

Why does Cos(4.9t) get left out?

Answer 21

Because it can't be larger than 1. At its maximum value, the displacement is just the value of that exponential.