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 a 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 = 1.7e^(−0.22t) * cos(4.9t)


t = ??? seconds

Answer 1

Anyone? :(

Answer 2

substitute y by 3 and solve for t.
3 = 1.7e^(−0.22t) * cos(4.9t)

Answer 3

Hmm....

Answer 4

the maximum displacements will be achieved when the cos() term is equal to 1.
this happens when:\[4.9t=n\pi\]therefore, the maximum displacements occur when:\[t=\frac{n\pi}{4.9}\]

Answer 5

so in this case n = 3?

Answer 6

substitute that into your equation and set y=3 to get:\[3=1.73^{-0.22\times\frac{n\pi}{4.9}}\]

Answer 7

solve this for n, which will then give you a value for t

Answer 8

sorry - replace 1.73 by 1.7e above

Answer 9

I keep feeling like we need a log to bring down the n?

Answer 10

\[3=1.7e^{-0.22\times\frac{n\pi}{4.9}}\]

Answer 11

yes - you will need to use logs

Answer 12

sweet okay.

Answer 13

also, you can use wolframalpha to graph this. see here: http://www.wolframalpha.com/input/?i=y+%3D+1.7e^%28%E2%88%920.22t%29+*+cos%284.9t%29+for+t%3D0+to+10

Answer 14

ln 3 = ln -0.22 + ln(n*pi) +ln (1.7e) ?

Answer 15

not sure if my log rules are correct...

Answer 16

first divide both sides by 1.7 to get:\[\frac{3}{1.7}=e^{-0.22\times\frac{n\pi}{4.9}}\]then take log to base e to get:\[\log(\frac{3}{1.7})=-0.22\times\frac{n\pi}{4.9}\]

Answer 17

ah okay. :)

Answer 18

0.568 is what log(3/1.7) is so should I keep the orignal, or this?

Answer 19

then we would need to multiply by 4.9 and divide by 0.22 pi

Answer 20

I always leave the actual calculations to the end to avoid any rounding errors.

Answer 21

okies.

Answer 22

so then we should have\[\frac{ 4.9 \log(3/1.7)}{0.22\pi}\]

Answer 23

= n

Answer 24

you missed the negative sign.

Answer 25

oh didn't even see that...

Answer 26

which = -4.03

Answer 27

so would it be 4.03 seconds?

Answer 28

yes - if you look at its graph you will see that at t=0, its displacement from the equilibrium is about 1.6.
so this means t has to be negative for the displacement to be bigger than this.
are you sure the question is stated correctly?

Answer 29

yup

Answer 30

hang on - we missed something!

Answer 31

y is said to be in feet.
and we are supposed to find displacement of 3 /inches/!

Answer 32

so we should have set y = 3/12 = 1/4 = 0.25

Answer 33

so just replace the 3 in your final equation by 0.25

Answer 34

(4.9log(0.25/1.7))/(0.22pi) ?

Answer 35

yes

Answer 36

13.59

Answer 37

again - don't forget the minus sign!

Answer 38

yeah it was negative 13.59 :)

Answer 39

no the answer works out to be plus 13.59

Answer 40

hmm that's apparently still not right. Yeah it was negative 13.59 without the negative sign, so it's positive.

Answer 41

\[-\frac{4.9 \log(0.25/1.7)}{0.22\pi}\approx13.59\]

Answer 42

and remember this is the value for n

Answer 43

\[t=\frac{n\pi}{4.9}\]

Answer 44

t = npi/4.9

Answer 45

so t = 13.59pi/4.9

Answer 46

yes

Answer 47

8.71.

Answer 48

that looks right

Answer 49

still not right hmm.... FACEPALM.

Answer 50

I don't like this q :(

Answer 51

let me re-check everything above...

Answer 52

thanks :).

Answer 53

Is it because we forgot that cos?

Answer 54

But I guess that was the npi.

Answer 55

ok, I think when we found n, we needed to round it to the next WHOLE number as n is supposed to be an integer.
so we should have used n=14, which would then give us:\[t=\frac{14\pi}{4.9}\approx 8.98s\]

Answer 56

Nope that doesn't work either :(

Answer 57

:/

Answer 58

!!!!!!!!!!!!!!!!!!!!!!!!

Answer 59

the only other option I can think of is that we should have rounded n down to 13 instead which would give:\[t=\frac{13\pi}{4.9}\approx 8.33 s\]

Answer 60

according to this practice one this is the answer, maybe we will try to see if we can get the same 8.40? I'm not sure if it should be higher or lower because 1.66 is lower than 1.7..

Answer 61

maybe my assumption that the cos() term must equal 1 was flawed?

Answer 62

Idk I asked turning to come help :D. Maybe he would know... :P

Answer 63

hang on - from that image it looks like you can use the Newton-Raphson method

Answer 64

so just use that and iterate to the required level of accuracy

Answer 65

Never heard of that method..... I thought they were talking about one of newton's laws of physics :P

Answer 66

:)

Answer 67

I guess you are also allowed to approximate the answer by using your graphing utility of choice.
i.e. just visually check when the displacement becomes less that 3" from the equilibrium by inspecting the graph

Answer 68

Sadly no graphing calculator, only wolfman! :)

Answer 69

try this: https://www.desmos.com/calculator

Answer 70

shall I post what I think is the solution through wolfram?

Answer 71

TuringTest - I believe the solution is 8.40

Answer 72

8.40 is the solution to the practice :P.

Answer 73

oh I missed that the solution was known, sorry

Answer 74

I onyl have 1 more chance to get it right so lets hope we can do this, if not an hour wasted ftl :(

Answer 75

Na that's the solution to the practice question using 1.66 instead of 1.7... Idk what the solution to this ugly one is :(

Answer 76

unless that's what we are talking ABOOOT :P.

Answer 77

http://www.wolframalpha.com/input/?i=solve+0.25+%3D+1.7e%5E%28%E2%88%920.22t%29+*+cos%284.9t%29+for+0+to+10

this indicates to me that it is about 7.8

Answer 78

the effort here is not wasted @KonradZuse - it is more important to understand HOW to solve a problem than it is to get any particular answer correct.

Answer 79

but that's just me playing with a computer, exactly^

Answer 80

Yeah I know :P. This has been an interesting learning experience :P.

Answer 81

http://www.wolframalpha.com/input/?i=solve+0.25+%3D+1.66e%5E%28%E2%88%920.22t%29+*+cos%284.9t%29+for+0+to+10

This produces the same thing abouts.

Answer 82

My gut feeling is to use the Newton-Raphson method to solve this numerically.

Answer 83

lets doo it then :P. LEts try the practice one first :).

Answer 84

but as you haven't learnt that method yet I guess you will have to stay with the graphical method

Answer 85

oh - so you want to learn that method?

Answer 86

is it calc 1 stuff? I basically forgot everything in calc 1 :P. Calc 2 is rough :P.

Answer 87

have you learnt about calculus yet?

Answer 88

All this stuff seems to be a revamp of calc 1....

Answer 89

I took calculus 1 online about 3 years ago.

Answer 90

the methos itself is described here: http://en.wikipedia.org/wiki/Newton%27s_method

Answer 91

basically we use the iterative approach:\[x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\]where f'(x) is the differential of f(x)

Answer 92

in your case we are given:\[0.25=1.7e^{−0.22t} *\cos(4.9t)\]so this needs to be re-arranged to get:

Answer 93

\[f(t)=0.25-1.7e^{−0.22t} *\cos(4.9t)\]and we are trying to find a value for t that gives a zero of f(t)

Answer 94

making sense so far?

Answer 95

mhm

Answer 96

so you need to first find the derivative of f(t).
do you know how to do this?