domm1
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https://isaacphysics.org/questions/c...a-cf7875531231 Part B.

I was able to find the expression for z, but it says that the solution for x is the Re(z). I converted my expression for z from exponential form using Euler's formula, which I thought would gave me the real and imaginary parts of z, but I was wrong (kinda guessed it as it would've meant delta =0). I'm confused how else I can express z to find it's real part? I can't find the entire solution (i.e. complementary function also) as I'd end up with unknown constants due to no boundary conditions being given. Where have I gone wrong?
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DFranklin
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(Original post by domm1)
https://isaacphysics.org/questions/c...a-cf7875531231

I was able to find the expression for z, but it says that the solution for x is the Re(z). I converted my expression for z from exponential form using Euler's formula, which I thought would gave me the real and imaginary parts of z, but I was wrong (kinda guessed it as it would've meant delta =0). I'm confused how else I can express z to find it's real part? I can't find the entire solution (i.e. complementary function also) as I'd end up with unknown constants due to no boundary conditions being given. Where have I gone wrong?
As ever, it would help if you posted your working.

It's hard to know what you mean by "I converted my expression for z from exponential form using Euler's formula, which I thought would gave me the real and imaginary parts of z", because (obviously), if you'd done it correctly, it would have to give you a valid expression for z, and if you'd extracted the real and imaginary parts correctly, they would have to be correct (tautologically).

But as something you may be getting wrong: you need to be careful about "real and imaginary parts", because if z = Ae^{i\omega t}, then \dfrac{dz}{dt} = i\omega A e^{i\omega t} = i\omega z, and so the real part of this comes from the imaginary part of z.

Regarding the initial conditions: the question is asking for a particular integral - this will be the solution you find by assuming z is of the given form.
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domm1
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(Original post by DFranklin)
As ever, it would help if you posted your working.

It's hard to know what you mean by "I converted my expression for z from exponential form using Euler's formula, which I thought would gave me the real and imaginary parts of z", because (obviously), if you'd done it correctly, it would have to give you a valid expression for z, and if you'd extracted the real and imaginary parts correctly, they would have to be correct (tautologically).

But as something you may be getting wrong: you need to be careful about "real and imaginary parts", because if z = Ae^{i\omega t}, then \dfrac{dz}{dt} = i\omega A e^{i\omega t} = i\omega z, and so the real part of this comes from the imaginary part of z.

Regarding the initial conditions: the question is asking for a particular integral - this will be the solution you find by assuming z is of the given form.
I would've posted my workings, but there's not really much to it apart from a conversion of e^{iwt} to cos(wt) +isin(wt), and then expanding the term, giving the Real{z} = X_0cos(wt), where X_0 equalled the coefficient of e^{iwt} from part A. However as I mentioned I realise this must've been incorrect as \delta would have to equal zero.

Have I re-wrote e^{iwt} incorrectly?
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mqb2766
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For part a) you need to assume the steady state solution is
z = Ae^(iwt)
and solve for A using the complex ode, then reason that the real part is the solution of the real ode. Complex A will scale and phase shift
cos(wt).
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domm1
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(Original post by mqb2766)
For part a) you need to assume the steady state solution is
z = Ae^(iwt)
and solve for A using the complex ode, then reason that the real part is the solution of the real ode. Complex A will scale and phase shift
cos(wt).
I've done part A it's the latter parts I am stuck on. I have my expression for A and I just thought that to convert z(t) into its real and imaginary parts I could use e^{i\theta} = cos(\theta) + isin(\theta), and multiply by A (using the cosine term as my Real{z}) however this does not give an expression where a phase shift is apparent, i.e. \delta =0.

I can't see how I could end up getting a \delta in my cosine term?
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DFranklin
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(Original post by domm1)
I've done part A it's the latter parts I am stuck on. I have my expression for A and I just thought that to convert z(t) into its real and imaginary parts I could use e^{i\theta} = cos(\theta) + isin(\theta), and multiply by A (using the cosine term as my Real{z}) however this does not give an expression where a phase shift is apparent, i.e. \delta =0.

I can't see how I could end up getting a \delta in my cosine term?
Since you haven't posted it, I'm just going to assume you've found A correctly.

If you write A in the form re^{i\alpha}, then \text{Re}(A e^{i\theta}) = r \text{Re}(e^{i(\theta + \alpha)}) (hence the phase shift).
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domm1
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(Original post by DFranklin)
Since you haven't posted it, I'm just going to assume you've found A correctly.

If you write A in the form re^{i\alpha}, then \text{Re}(A e^{i\theta}) = r \text{Re}(e^{i(\theta + \alpha)}) (hence the phase shift).
I originally thought that my expression for A was correct (due to the answer for part A being Ae^{iwt} where A is the expression found from substitutions into the complex DE), however I am now very confused, as my expression for A could not be converted into complex exponential form (or just that I'm missing something very obvious). For reference, my expression for A is,

A = \dfrac{F_0}{k-m\omega^{2}+ib\omega}
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mqb2766
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(Original post by domm1)
I originally thought that my expression for A was correct (due to the answer for part A being Ae^{iwt} where A is the expression found from substitutions into the complex DE), however I am now very confused, as my expression for A could not be converted into complex exponential form (or just that I'm missing something very obvious). For reference, my expression for A is,

A = \dfrac{F_0}{k-m\omega^{2}+ib\omega}
That shouldn't be too hard to get the modulus / argument? In its basic form it's
z =1/(a+ib)
Which gives
|z| = 1/sqrt(a^2 + b^2)
arg(z) = atan(-b/a)
Which gives the scaling and shifting values.
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domm1
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(Original post by mqb2766)
That shouldn't be too hard to get the modulus / argument? In its basic form it's
z =1/(a+ib)
Which gives
|z| = 1/sqrt(a^2 + b^2)
arg(z) = atan(-b/a)
Which gives the scaling and shifting values.
Oh I did do that, but I got confused when looking to translate it in to complex exponential form A=re^{i\alpha} as I thought it would mean \alpha = tan^{-1}(\dfrac{m\omega^{2}-k}{b\omega}) which didn't seem very pleasant so thought I was going in the wrong direction. Would this be the case or no?
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mqb2766
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(Original post by domm1)
Oh I did do that, but I got confused when looking to translate it in to complex exponential form A=re^{i\alpha} as I thought it would mean \alpha = tan^{-1}(\dfrac{F_0k - F_0m\omega^{2}}{-F_0b\omega}) which didn't seem very pleasant so thought I was going in the wrong direction. Would this be the case or no?
Cancel the F0, but the phase shift will be something like that. It has to depend on the parameters of the ode.
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domm1
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(Original post by mqb2766)
Cancel the F0, but the phase shift will be something like that. It has to depend on the parameters of the ode.
I tried both \alpha = tan^{-1}(\dfrac{m\omega^{2}-k}{b\omega}) and \alpha = tan^{-1}(\dfrac{k-m\omega^{2}}{b\omega}) (as from what you recommended the arg came out to be the reciprocal of what I had calculated), but both incorrect? I'm confused why this is as surely \delta depends on nothing but the arg of A? I.e. \delta = \alpha.
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DFranklin
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(Original post by domm1)
I tried both \alpha = tan^{-1}(\dfrac{m\omega^{2}-k}{b\omega}) and \alpha = tan^{-1}(\dfrac{k-m\omega^{2}}{b\omega}) (as from what you recommended the arg came out to be the reciprocal of what I had calculated), but both incorrect? I'm confused why this is as surely \delta depends on nothing but the arg of A? I.e. \delta = \alpha.
mqb has told you that if z = 1/(a+ib), then arg z = -b/a.

In this case, you want the argument of A = \dfrac{F_0}{k-m\omega^{2}+ib\omega}

(and so \text{arg }A = \text{arg }z, where z = \dfrac{1}{k-m\omega^{2}+ib\omega}).
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domm1
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(Original post by DFranklin)
mqb has told you that if z = 1/(a+ib), then arg z = -b/a.

In this case, you want the argument of A = \dfrac{F_0}{k-m\omega^{2}+ib\omega}

(and so \text{arg }A = \text{arg }z, where z = \dfrac{1}{k-m\omega^{2}+ib\omega}).
arg(z)=tan^{-1}(\dfrac{-b\omega}{k-m\omega^{2}}) no?
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DFranklin
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(Original post by domm1)
arg(z)=tan^{-1}(\dfrac{-b\omega}{k-m\omega^{2}}) no?
Yes. Note this is not what you had posted before...
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domm1
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(Original post by DFranklin)
Yes. Note this is not what you had posted before...
But my point is that this is still incorrect. Is it not multiplied by F_0 or something? I cannot see why this should still be incorrect.
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(Original post by domm1)
But my point is that this is still incorrect. Is it not multiplied by F_0 or something? I cannot see why this should still be incorrect.
The calculation is correct. It's not the *answer*, because you have z = |A| e^{i(\alpha + \omega t)}, and the question is looking for an expression of form z = X_0 e^{i(\omega t - \delta)}, so it isn't quite true to say that \delta = \alpha. Since you're considering the ratio of the imaginary and real parts of A, multiplying by a real constant such as F_0 isn't going to change anything.
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(Original post by DFranklin)
The calculation is correct. It's not the *answer*, because you have z = |A| e^{i(\alpha + \omega t)}, and the question is looking for an expression of form z = X_0 e^{i(\omega t - \delta)}, so it isn't quite true to say that \delta = \alpha. Since you're considering the ratio of the imaginary and real parts of A, multiplying by a real constant such as F_0 isn't going to change anything.
I used the thing you mentioned earlier Re(A e^{i\theta}) = r \text{Re}(e^{i(\theta + \alpha)}) As the \text{Re}(Ae^{i\omega t})=\text{Re}(z), but I didn't seem to get very far in success with it all. Surely if it is written in the form given above, r \text{Re}(e^{i(\theta + \alpha)}) that should give the answer, or is there still need for adaption to account that \alpha \neq \delta?
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mqb2766
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Can you post your working? It really does help.
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(Original post by domm1)
I used the thing you mentioned earlier Re(A e^{i\theta}) = r \text{Re}(e^{i(\theta + \alpha)}) As the \text{Re}(Ae^{i\omega t})=\text{Re}(z), but I didn't seem to get very far in success with it all. Surely if it is written in the form given above, r \text{Re}(e^{i(\theta + \alpha)}) that should give the answer, or is there still need for adaption to account that \alpha \neq \delta?
What is the real part of e^{i(\omega t + \alpha)}?
And what is the form of the answer expected?

Do you not see that this is not (quite) compatible with taking \delta = \alpha?
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domm1
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(Original post by mqb2766)
Can you post your working? It really does help.
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I have very minimal workings for this part, but here they are. (They are also a bit all over the place, not a very nice flow to it all).
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