Question

Whoever can answer this the best I will send you 10 owlbucks.
Can someone please explain how we could use this differential equation to model an un-damped oscillator.
(d^2 x)/(dt^2 )-10 dx/dt+34x=0

Then if we were to extend this differential equation into a forced damped oscillator, explain how you could model such an oscillator using a forced second order differential equation.

The description must be commensurate to a level 5 qualification.

Answer 1

Need help @carlj0nes

Answer 2

I believe the damping term is associated with the dx/dt term

find the roots to the "characteristic equation", and examine the discriminant to classify the behavior of the system

Answer 3

Your question is not clear to me. You can start with the physics of , for example, a spring that you pull and then release. Making some simple assumptions, you can derive a differential equation that models the behavior of the spring. You can introduce damping to model how the spring slows down. You will get an equation that looks like the one you posted.

Answer 4

The original question was. A Damped oscillation with no external forces can be modelled by the equation I previously posted.
Where x(in mm) is the amplitude of the oscillation at time t seconds. The initial amplitude of the oscillation is 2mm (i.e. when t=0) and the velocity is 0mm/s.

By using the information provided above, determine the particular solution of the differential equation.

Just to confirm I have already completed this part of the question but I need to do the question posted for a higher grade

Answer 5

an undamped harmonic oscillator would have this equation
\[ \frac{ d^2x }{ dt^2 }+ \cancel{-10\frac{ dx }{ dt }}+34x = 0 \]

Answer 6

btw, are you sure it is -10 dx/dt ?
that -10 leads to e^5t term which grows (as opposed to being damped) with time.

Answer 7

This is what I have so far
(d^2x)/(dt^2)-10 dx/dt +34x=0
r ^2 −10r+34=0
(r−5)^ 2 −25+34=0
(r−5) ^2 =−9
(r−5)=±√−9 =±√9 √ −1
r−5=±3i
r=5±3i
so
x(t)=ae ^(5+3i)t +be ^(5−3i)t
x(t)=e ^5t (asin(3t)+bcos(3t))=
e^ 5t (ksin(3t+θ))

−e 5t ≤e 5t sin(3t+θ)≤e 5t

Answer 8

not sure if it is correct but I am after a explanation as well

Answer 9

I would stop at
x(t)=e ^5t (asin(3t)+bcos(3t))

and use the initial conditions to find a and b
at t=0, x= 2
so
\[ 2 = e^{5 \cdot 0} ( a \sin(3\cdot 0) + b \cos(3\cdot 0)\\
2= 1 \cdot(0 + b\cdot 1) \\ b=2
\]

Answer 10

now find dx/dt (which is the velocity), so we can use the second initial condition
v=0 at t=0
\[ \frac{d}{dt} x(t)= \frac{d}{dt} \left(e^{5t} (a \sin(3t)+b\cos(3t))\right) \]

Answer 11

should replace b with 2 and dx/dt with 0 (i.e. v=0)
in that, and solve for a

Answer 12

you should get
\[ \frac{d}{dt} \left(e^{5t} (a \sin(3t)+b\cos(3t))\right) = 0 \\
\left(\frac{d}{dt} e^{5t}\right) (a \sin(3t)+b\cos(3t)) + e^{5t} \frac{d}{dt} (a \sin(3t)+b\cos(3t)) = 0 \]

Answer 13

or
\[ 5 e^{5t} (a \sin(3t) + 2 \cos(3t)) + e^{5t} ( 3a \cos(3t) -6\sin(3t) ) = 0 \]
now plug in t=0

Answer 14

to "force" the oscillator, set the equation equal to some function of t
\[ \frac{ d^2x }{ dt^2 }+ -10\frac{ dx }{ dt }+34x = f(t) \]
for example, let f(t)= sin(t)

Answer 15

***A Damped oscillation with no external forces can be modelled by the equation I previously posted.***

Again, I would double check the -10 coefficient. Real oscillations grow smaller, not grow larger with time.

Answer 16

@phi thanks for your help but can someone explain how we could use this differential equation to model an un-damped oscillator.

Answer 17

http://openstudy.com/study#/updates/558c1099e4b0c2c04901546b

Answer 18

if it was +10dx/dt
was damping
\[x(t)=e^{-5t}(asin(3t)+bcos(3t))\]