Question

i need help!

Answer 1

only b

Answer 2

a is 0.040

Answer 3

The first thing to do is simplify the function/expression for the masses position and write it as one trig function. Can you do that?

Answer 4

yeah

Answer 5

Then you have y(t) = R cos(8t - phi)
for some R and phi. Now you just need to find the values of t for which
8t - phi = pi/2 + k.pi
for any integer k, because this is when cos(8t - phi) = 0, the equilibrium position of the spring.

Answer 6

wow you lost me there

Answer 7

The spring is at the equilibrium position when y(t) = 0, yes?

Answer 8

So we need to find the values of t in the given range \( 0 \leq t \leq 1 \) where that is true.

Answer 9

Now how are you going to do that? How are you going to find when y(t) = 0? One strategy is to simplify that expression as one trig function. There are other strategies. You tell me where you want to go next with solving the equation y(t) = 0.

Answer 10

....

Answer 11

If you want to talk this through and find the answer, let me know.

Answer 12

well simply the equation

Answer 13

If y(t) = 0 then
cos 8t = 3 sin 8t
or
tan 8t = 1/3

Make sense? If so, can you now solve this?

Answer 14

i think tan would be easier to solve so i have to move the 8 to the other side

Answer 15

Yes ... so what are the solutions to that last equation
tan(8t) = 1/3 ?

Answer 16

tant=8/3?

Answer 17

then the inverse of tan?

Answer 18

No, you definitely can't do that.

If

tan(8t) = 1/3

then

8t = arctan(1/3) + k.pi

where k is any integer and arctan(1/3) is the one value of arctan between -pi/2 and pi/2. That is
arctan(1/3) = 0.3218 radians, or 18.4 degrees

Answer 19

and then divide 8?

Answer 20

Hence
\[ t = \frac{1}{8} (0.3218 + k\pi) \]

Now the first value of t in the range of the problem, \( 0 \leq t \leq 1 \) is just ...

Answer 21

\[ t = \frac{0.3218}{8} = 0.0402 \ seconds \]

Answer 22

Now find for what maximum value of k, the general form of t is still less than 1. That is the largest value of t and the answer to part b of your question.

Answer 23

so 0.0402=1/8(0.3218+kpi)?

Answer 24

No. We need to introduce k because arctan has infinitely many values. For example, for what values of x is
tan x = 0


Well x = 0, yes. But also x = pi. And x = 2pi. And x = -pi and x = -2 pi ... and so on.

Tan is a period function of period pi, and that is why when we take arctan of the equation
tan x = y
we find infinitely many values of y.

Answer 25

We use the parameter k to give us a way to label and find each of those answers.

Answer 26

So for your problem in particular, we have that
\[ t = 0.0402 + \frac{k\pi}{8} \]
That is for EVERY integer k, the corresponding value of t has the property that
tan(8t) = 1/3

So for k = 0, we have t = 0.0402
k = 1 ---> t = 0.0402 + 0.3927 = 0.4329
k = 2 ---> t = 0.0402 + 0.7854 = 0.8259
etc.

Answer 27

oh ok

Answer 28

This makes a lot of physical sense as well. It is saying that every pi/8 seconds, the bob on the pendulum goes through the equilibrium point.

Answer 29

0.040 would be the the smallest time, but how would you find the largest time? whats the equation?

Answer 30

JamesJ COME BACK

Answer 31

The largest time corresponds to the largest integer k for which
\[ t = \frac{1}{8} (\arctan(1/3) + k\pi) \ \leq 1 \]
So keep checking the values of k until you find one that is greater than 1. I began writing that table of values above.

Answer 32

k=3

Answer 33

Yes, for k = 3, t > 1.

Hence the maximum value of t is the value corresponding to k = 2, i.e.,
t = 0.8259 seconds

Answer 34

thank you!