Solving Resonance Questions Homework

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Erbil
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Homework Statement



1) The equation of the particle on one dimension given by;

d^2y/dt^2 + 2 dy/dt + 6y = 10sinωt

a) Find the displacement as a function of time.
b) find T= ? A=? (period,amplitude)

2) Show that the amplitude of the vibration damping halved on time 1,39/γ.
3)On one LC circuit;
C = 100mikroF and resonance frequency is 2MHz.Find the value of coil. (L)

Homework Equations

General form = d^2y/dt+γdy/dt+y=F0cosωt
And I don't know can we use this formula but maybe can help;
A=-F0/m /(ω^4+γ^2ω^2)^1/2

d^2θ/dt^2 = -W0^2q

The Attempt at a Solution

I just tried to use the formula for amplitude but there's no mass.And I don't have an idea about 2 and 3.
 
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Do you know how to solve the homogeneous equation y'' + ay' + by = 0?

y=c1e^r1x+c2er^2x (distinct) if we have two root,
y= c1e^rx+c2xe^2x (repeated) if we have one root,
y= e^θx*(c1cos(Bx)+c2sin(Bx)
 
d^2y/dt^2 + 2 dy/dt + 6y = 10sinωt

I found imaginer roots ;

r1 = (-1-10i)
r2 = (-1+10i)

But MATLAB says that roots are ;

-1.0000 + 2.2361i
-1.0000 - 2.2361i

So, what? is it a non-lineer homogenous dif eq.?

IF we use the general solution formula ;

y= e^θx*(c1cos(Bx)+c2sin(Bx)

where can we use our roots?Edit : -1.0000 + 2.2361i
-1.0000 - 2.2361i are roots.I have forgot to take the square root of ∇
 
Last edited:
y'' + 2y' + 6y = 10sinωt

r1=(-1 + 2,236)i
r2=(-1-2,2236)I

Yp = Asinωt + Bcosωt
Y' = ωAcosωt -ωBsinωt
Y'' = -ω^2Asinωt - ω^2Bcosωt

(-ω^2Asinωt-ω^2Bcosωt)+2ωAcosωt-2ωBsinωt+6Asinω+6Bcosωt=10sinωt

-ω^2A-2ωB+6A=10
-ω^2B+2ωA+6B=0

A = 10/(-ω^2)+6b
B=0

Yp=10/(-ω^2)+6bsinωt

y= e^θx*(c1cos(Bx)+c2sin(Bx)+10/(-ω^2)+6bsinωt
 
Last edited:
No.Because my expression is so complex :D
 
haruspex said:
Did you verify that it satisfies the equation?

No? How can I do it? Where can I use roots.
 
Your solution for the roots of the homogeneous equation was correct and leads to
y = Ae(-1+i√5)t+Be(-1-i√5)t
You can rewrite that as
y = e-t(C cos(αt) + D sin(αt)) where α=√5.
Don't confuse that α with the given ω.
Now we just have to find a particular solution for the inhomogeneous equation. Clearly this will be of the form A cos(ωt) + B sin(ωt) (different A and B from before).
you correctly obtained
2A-2ωB+6A=10
2B+2ωA+6B=0
but I don't understand where you went from there. How did you deduce B = 0?
Wrt q 2, what is γ in this context?
 
haruspex said:
Your solution for the roots of the homogeneous equation was correct and leads to
y = Ae(-1+i√5)t+Be(-1-i√5)t
You can rewrite that as
y = e-t(C cos(αt) + D sin(αt)) where α=√5.
Don't confuse that α with the given ω.
Now we just have to find a particular solution for the inhomogeneous equation. Clearly this will be of the form A cos(ωt) + B sin(ωt) (different A and B from before).
you correctly obtained
2A-2ωB+6A=10
2B+2ωA+6B=0
but I don't understand where you went from there. How did you deduce B = 0?
Wrt q 2, what is γ in this context?

2A-2ωB+6A=10
2B+2ωA+6B=0 I can't solve it.I had a mistakes.I just figure now.There's no γ in my context.
 
Erbil said:
2A-2ωB+6A=10
2B+2ωA+6B=0 I can't solve it.
You still can't solve it? ω is a given constant here. It's just a pair of linear simultaneous equations in A and B.
There's no γ in my context.
Q2 reads:
2) Show that the amplitude of the vibration damping halved on time 1,39/γ.​