Tricky Simple harmonic motion question

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rub em out
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This question is from the unit 4b specimen paper.

We find out in previous parts of the question that the natural frequency of oscillation is 1.5Hz and that the spring constant is 61.3N/m.

The answer is here:

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I understand about the resonant frequency being hit at 1.5Hz causing large oscillations but I can't see how you can tell the size of oscillations or the phase difference of oscillations!?

any help would be great, thanks!
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The Mr Z
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(Original post by rub em out)
I understand about the resonant frequency being hit at 1.5Hz causing large oscillations but I can't see how you can tell the size of oscillations or the phase difference of oscillations!?

any help would be great, thanks!
Well, the answer doesn't ask for the size of the resonant oscillations, just for you to mention that they're >> the non-resonant ones.
However you can work out the magnitude by considering a balance of energy; in steady state the work done by the force is going to be equal to the energy dissipated by dampening over 1 cycle.

For the phase, either just know this, or solve the equation for SHM using the obvious solutions.

so if F=Asin(\omega t)

Then you have  m\frac{d^2x}{dt^2} - b\frac{dx}{dt} + kx = Asin(\omega t)

And the obvious solutions are B sin(wt) + C cos (wt). Just put these possible solutions in and you'll see that at resonance (where w2=k/m) B = 0, so x = C cos(wt). As cos(y) = sin (y+90o), then x is 90o out of phase with F.

(Fully solving it properly requires complex exponentials which you probably haven't done in maths yet)
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(Original post by The Mr Z)
Well, the answer doesn't ask for the size of the resonant oscillations, just for you to mention that they're >> the non-resonant ones.
However you can work out the magnitude by considering a balance of energy; in steady state the work done by the force is going to be equal to the energy dissipated by dampening over 1 cycle.

For the phase, either just know this, or solve the equation for SHM using the obvious solutions.

so if F=Asin(\omega t)

Then you have  m\frac{d^2x}{dt^2} - b\frac{dx}{dt} + kx = Asin(\omega t)

And the obvious solutions are B sin(wt) + C cos (wt). Just put these possible solutions in and you'll see that at resonance (where w2=k/m) B = 0, so x = C cos(wt). As cos(y) = sin (y+90o), then x is 90o out of phase with F.

(Fully solving it properly requires complex exponentials which you probably haven't done in maths yet)
Thanks for the help but im struggling to understand those equations :L I guess what I dont get is how they can say the oscillations are in phase if they have completely different time periods... This is how i'm imagining it:

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Stonebridge
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Question discussed here
http://www.thestudentroom.co.uk/show....php?t=2006770
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(Original post by Stonebridge)
Question discussed here
http://www.thestudentroom.co.uk/show....php?t=2006770
oh great, I think that actually settles most of my questions
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The Mr Z
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(Original post by rub em out)
Thanks for the help but im struggling to understand those equations :L I guess what I dont get is how they can say the oscillations are in phase if they have completely different time periods... This is how i'm imagining it:

Image
Basically you get two responses at different frequencies;

1. The natural response, at the harmonic frequency of the system. This decays over time due to dampening. It's the solution of m\frac{d^2x}{dt^2} - b\frac{dx}{dt} + kx = 0, ie the complimentary solution.

2. The forced response, at the frequency of the force term. This does not decay over time because the force adds energy to the system at the same rate at which it is lost due to dampening. It is the specific solution of the equation.

They have different frequencies (in general) so don't interact - the natural response is unaffected by the force driving the forced response and so decays away. At longish times after you start the force the natural response has decayed to zero and only the forced response remains.

So in steady state, ie at large times where the response is unchanging, only the forced term remains. Hence x has the same frequency as F, so it is possible to talk about a phase difference between them.
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2013zislam
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(Original post by The Mr Z)
Basically you get two responses at different frequencies;

1. The natural response, at the harmonic frequency of the system. This decays over time due to dampening. It's the solution of m\frac{d^2x}{dt^2} - b\frac{dx}{dt} + kx = 0, ie the complimentary solution.

2. The forced response, at the frequency of the force term. This does not decay over time because the force adds energy to the system at the same rate at which it is lost due to dampening. It is the specific solution of the equation.

They have different frequencies (in general) so don't interact - the natural response is unaffected by the force driving the forced response and so decays away. At longish times after you start the force the natural response has decayed to zero and only the forced response remains.

So in steady state, ie at large times where the response is unchanging, only the forced term remains. Hence x has the same frequency as F, so it is possible to talk about a phase difference between them.
i thought frequency is independant of the amplitude so wont the amplitude be the same throughout
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