Why does the value of cos(t) turn to 1 in the solution for v_c(0)=0?

[electron2]

\dot{v_c(0)}=2
v_c(0)=0

\ddot{v_c}+2\dot{v_c}+2v_c=\cos (t)

-\omega ^2V_c-2+2j\omega V_c +2V_c=cos(t)

in the solution i have +2 instead of the -2 that i got here(because we do -f'(0) when we find v_c double dot)

and why cos t turns to 1
??

 

[HallsofIvy]

The Fourier series for a function is a sine and cosine sum:
f(x)= A_0 + A_1 cos(x)+ B_1sin(x)+ A_2cos(2x)+ B_2sin(2x)+ A_3cos(3x)+ \cdot\cdot\cdot

If the function is simply a sine or cosine (or finite combination) the Fourier series is just that combination. In particular, if f(x)= cos(x)
f(x)= cos(x)= 0+ 1 cos(x)+ 0 sin(x)+ 0 cos(2x)+ 0 sin(2x)+ 0 cos(3x)+ \cdot\cdot\cdot
The only non-zero coefficient is A_1= 1.

 

[electron2]

whattt nooooooooooooooo

the forier transform of cos t is e^(i*f)
f-phase angle

and because its cos (t)
the phase angle is 0

so i get 1

!

 

[electron2]

is my formula correct
??

 

[Redbelly98]

whattt nooooooooooooooo

the forier transform of cos t is e^(i*f)

No, it is the sum of two delta functions located at f=(1/2π) Hz.

f-phase angle

and because its cos (t)
the phase angle is 0

so i get 1

!

What is 1?
And what was the question asked in the problem? Is it to solve the differential equation you wrote in post #1? Fourier transforms are not necessary for that. It is just not clear what is the problem that is to be solved here.

 

[electron2]

\ddot{v_c}+2\dot{v_c}+2v_c=\cos (t)

-\omega ^2V_c-2+2j\omega V_c +2V_c=1i was told that because the formula is e^(i*f)
f-is the phase ange
and we have cos t - phase angle =0
so e^(i*f)=1

ring a bell to you?

 

[Redbelly98]

Okay, the phase angle f=0.

So e^(i*f) = e^0 = ____?

 

[turin]

And what was the question asked in the problem? Is it to solve the differential equation you wrote in post #1? Fourier transforms are not necessary for that.

Nor appropriate. Since this is an initial value problem, use Laplace Transform. That way, your initial values will appear in the transformed equation. In that case, cos(t) certainly does not transform to 1.

Alternatively, assume an exponetial solution, and then solve for the parameters of the exponential. You can use superposition to get the cos(t) from two cleverly chosen exponentials. Then, match to the boundary (initial) conditions. (I would prefer the Laplace transform method - it is automatic.)