Question

Each of the following IVPs corresponds to a harmonic oscillator. Find the solution and specify its amplitude, phase angle and period of motion.
x¨ + 9x = 0, x(0) = 1, x˙(0) = 1

Answer 1

@myininaya do you happen to know how to do this?

Answer 2

is that a differential equation?

Answer 3

just want to make sure i'm seeing right
because we were just talking about trig and it seems like we went to diff equations

Answer 4

first of all do you know how to solve r^2+9=0

Answer 5

r=3

Answer 6

well now exactly
r^2-9=0
would have given r=3 or r=-3

Answer 7

not exactly*
r^2+9=0
would give r=3i or r=-3i

Answer 8

so the solution to x''+9x=0 is
\[x(t)=A \cos(3t)+B \sin(3t)\]

Answer 9

use your conditions to find A and B

Answer 10

x(0)=1
x'(0)=1

Answer 11

what do you mean?

Answer 12

x(0)=1
means x is 1 when t is 0
x'(0)=1
means x' is 1 when t is 0

Answer 13

You will have to find x' give x=Acos(3t)+Bsin(3t)

Answer 14

to apply the other condition there

Answer 15

you should end up with a system of equations to solve

Answer 16

x(0)=1 means
x=1 when t=0
so we have \[1=A \cos(3(0))+B \sin(3(0))\]
1=A+B(0)
1=A
This should make your life easier when finding B now

Answer 17

and it should also show you better how to apply x'(0)=1

Answer 18

can you find x' given x=cos(3t)+Bsin(3t) ?

Answer 19

1

Answer 20

\[x' \text{ or also called } \frac{dx}{dt}=(\cos(3t))'+(B \sin(3t))'\]

Answer 21

to find the derivative of either cos(3t) or sin(3t) you will need chain rule

Answer 22

hey @mony01 I'm so sorry I have to leave you

Answer 23

@PaxPolaris when you get chance will you help mony

Answer 24

@myininaya Aye Captain my captain! I'll try!

Answer 25

@mony01 ... where are we stuck ?

Answer 26

well i think i did the chain rule right i got -sin(3t)(3)+Bcos(3t)(3)

Answer 27

i feel the need to start over ...by guessing
\[x(t)=A \cos \left( 3t+\phi \right)\]

Answer 28

\[x(0)=A \cos(\phi)=1\]\[x'(0)=-3A \sin(\phi)=1\]

\[\implies \tan (\phi)=-1/3\]

Answer 29

how did you get -3?

Answer 30

\[x'(t)=A \left[ \cos \left( 3t+\phi \right) \right]'\\=A \left[ -\sin \left( 3t+\phi \right) \right]\cdot3\]

Answer 31

the derivative of cos is negative sin...

Answer 32

oh ok

Answer 33

so phase \(=\phi=-\tan^{-1}(-1/3)=-0.32\)radians

Answer 34

and from that you can derive Amplitude \(=A=\sqrt{10}/3\)

Answer 35

how is tan -1/3 and where did the 3t go from the derivative?

Answer 36

at t=0 we are given the initial conditions x(0)=1 x'(0)=1

because t=0, 3t disappears

Answer 37

And period we already know to be \(2\pi/3\)

Answer 38

how you know tan=-1/3 and where you get 10 and 2pi/3....sorry but im so lost

Answer 39

\[x(0)=A \cos(\phi)=1\]\[x'(0)=-3A \sin(\phi)=1\]
do you get these 2 equations?

Answer 40

how did you get those equations?

Answer 41

x(0)=1 and x'(0)=1 ... is given, correct?

if \(x(t)=A\cos(3t+\phi)\)... just plug in t=0

Answer 42

ok got that

Answer 43

same thing for x'(0) .

... then divide the second equation from the first to get rid of A\[{-3A \sin(\phi) \over A \cos(\phi)}=\frac11\]

Answer 44

whats the equation for x'(0)?

Answer 45

ok never mind now i see why tan is -1/3

Answer 46

\[x(t)=A \cos(3t+\phi)\\ x'(t)=-3A \sin(3t+\phi)\]

\(x'(0)=-3A \sin(\phi)\)...... and \(x'(0)=1\)

Answer 47

ok. so we get the phase angle \(\phi\) ....


and \(cos\ \phi=3/\sqrt{10}\)

Answer 48

why is it sqrt10

Answer 49

its a right triangle (pythagorean theorem)
\[\sqrt{(-1)^2+3^2}\]

Answer 50

ok

Answer 51

since \(A\cos(\phi)=1, \implies A=1/\cos(\phi),\implies A=\sqrt{10}/3\)

Answer 52

earlier you said the phase angle was -tan^-1 (-1/3) why?

Answer 53

sorry an extra negative sign.

\(\phi\) is the phase angle, and \(\tan(\phi)=-1/3\) .... so \(\phi=...\)

Answer 54

tan^-1 (-1/3)

Answer 55

right so we have the equation:
\[x(t)={\sqrt{10}\over3}\cos \left( 3t-0.32.... \right)\]

we have the phase angle, and amplitude, ...it fits all the original requirements

Answer 56

what else do we need?

Answer 57

period of motion

Answer 58

for the radians i got -0.32

Answer 59

cos(t) has period 2pi ... correct?

Answer 60

yes

Answer 61

the general function:\[A \cos(\omega\cdot t+\phi)\]

has the period: \(2\pi/\omega\)

Answer 62

so the period is 2pi/3

Answer 63

right ....

another way to think of it is
cos(t) completes 1 period in 2pi
cos(3t) completes 3 periods in 2pi

Answer 64

the radians has to be positive right?

Answer 65

idk about that.... guess technically the phase angle can be \(\tan^{-1}(-1/3)+2\pi \cdot n\)

Answer 66

for any integer n

Answer 67

so its always positive?

Answer 68

no i think you can leave it as it is ...

Answer 69

so -.32?

Answer 70

yeah

Answer 71

if it asks to find the simple harmonic motion for the equation, what would it be?

Answer 72

gahhhhhhhhh...........

Answer 73

i've been leading you completely wrong for the last hour. Sorry

Answer 74

what do you mean?

Answer 75

forget everything i said ... go back to what myininaya was doing

Answer 76

i think you were doing it right, i was asking how would i do it if it asked for the harmonic motion

Answer 77

sorry for Panicking ... was checking the answer on wolframalpha ... but initially my equation looked nothing like what they had.

it is on correct form

Answer 78

The equation is correct ... one of many ways to write it
We got:\[x(t)={\sqrt{10}\over3}\cos \left( 3t-\tan^{-1} \left( \frac13 \right)\right)\]

it can also be written using sin instead of cos [BUT the phase angle will be different]
\[x(t)={\sqrt{10}\over3}\color{blue}\sin\left( 3t\color{blue}{+\tan^{-1} \left( 3 \right)}\right)\]

i don't know which you are supposed to use ... but i was feeling partial to cosine

Answer 79

the equation @myininaya had almost arrived at was also correct: \[x(t)=\frac13\sin(3t)+\cos(3t)\]

but you 'd still have to do some work to arrive at Amplitude and phase.

Answer 80

wait how did she arrived at that?