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    I want to check if my answer to part (iv) is correct. It seems too simple.

    2pi/0.5 = w = 4pi But w = √(k/m)

    16pi2 * 80 = k = 12633.

    Since the two springs are in parallel we have the spring constant for an individual spring = 6317 N/m

    Also, could I have a hint to part (v) ?
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    (Original post by Stonebridge)
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    Could you help me, please ? Its been quite a long time...
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    I think I've worked out the answer to part (v) as well, but could you check my method, please ?

    At 11 km/h we can also imagine the bumps as having a forcing frequency that is very nearly equal to that of the natural frequency of the springs i.e. the forcing frequency is 2pi/0.49 whilst the natural frequency is 2pi/0.5


    However, at 40 km/h I'm facing a dilemma. I still see a good increase in amplitude above the original amplitude of A1 but how does this translate into the person not being ejected ?
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    I agree with your reasoning for part iv with one small point to make.
    The resonant frequency for free oscillations of a damped oscillator is not equal to that of an undamped one.
    The damped oscillator's natural frequency will always be less.
    So the formula for k, using the undamped value of w is not strictly true.
    However, as the damping is very small (1/20 times the critical value of damping) it means that the natural frequency of the damped motion is very very close to the undamped value. (You may have a formula in your notes that relates the damped and undamped natural frequencies in terms of the damping constant or some other similar quantity (See below*). I think you'll find that the difference is about one part in 1000. It's insignificant.)

    For the last part of the question, you now have forced damped oscillations as you correctly point out.
    Again, the natural frequency for forced damped oscillations is not equal to the undamped natural frequency (and neither is it equal to the unforced damped natural frequency.) But again, for this low level of damping, the values are very very close such that the difference is insignificant.
    The frequency of the bumps as you ride them gives the driving frequency. (There is an old A-Level question with this idea in it.) If, as you say, this is very close to the natural frequency, the amplitude will build up and could cause the rider to fall off.
    With such a low level of damping, the curve of amplitude against frequency will have a very sharp maximum. This means that for driving frequencies only slightly above or below the natural frequency, the amplitude increase will be small.
    The question seems to be asking for this. THose other two speeds and corresponding driving frequencies are actually a long way from the peak resonant value.

    * As I don't know exactly what you have studied on this topic it's difficult to advise more, but you will get some useful information and equations on
    http://en.wikipedia.org/wiki/Damping

    and some of the links from there, especially Damping factor and Damping ratio
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    im curious. is this uni physicswht year?
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    (Original post by Stonebridge)
    I agree with your reasoning for part iv with one small point to make.
    The resonant frequency for free oscillations of a damped oscillator is not equal to that of an undamped one.
    The damped oscillator's natural frequency will always be less.
    So the formula for k, using the undamped value of w is not strictly true.
    However, as the damping is very small (1/20 times the critical value of damping) it means that the natural frequency of the damped motion is very very close to the undamped value. (You may have a formula in your notes that relates the damped and undamped natural frequencies in terms of the damping constant or some other similar quantity (See below*). I think you'll find that the difference is about one part in 1000. It's insignificant.)
    I think you mean : http://latex.codecogs.com/gif.latex?\omega^2=\omega_0^2-\frac{\gamma^2}{4}

    Through a manner of substitutions I calculate w(damped frequency) = 12.56244 (12.56637 is the UNdamped frequency).

    You see, the problem is how do I relate this to the spring constant ? I am only aware of one formula
    Wo = (k/m)^0.5

    For the last bit of the question are my workings in the attachments in the third post correct ? My method seems to break down for the case of 40 km/h...

    I've been taught it all. Everything from quality factor, the general solutions of each case, the energy loss per cycle.... My knowledge is deep.
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    That formula for spring constant with undamped w is always correct.
    You have been given w for damped oscillations and you have the damping factor.
    Can't you, from this, find the undamped w? Hence k?

    I've been taught it all. Everything from quality factor, the general solutions of each case, the energy loss per cycle.... My knowledge is deep.
    In which case you know as much as I do.
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    (Original post by Stonebridge)
    That formula for spring constant with undamped w is always correct.
    You have been given w for damped oscillations and you have the damping factor.
    Can't you, from this, find the undamped w? Hence k?
    In which case you know as much as I do.
    Fine, I got it. I have another question which is bugging me :

    http://www.mediafire.com/view/?x7nqqpkl6xfzff6

    In (ii) one can use the reduced mass formula to make the system appear as if only one mass is involved. This singular mass will have a mass very close to that of the 0.2 kg ball.The large cannon ball is unlikely to experience much SHM and can be approximated to be still.

    The trouble is having is how to answer this question for the 5 marks on offer ? What else can I say ?

    In (iii) calculating w we get 2pi. Using w^2 = (k/m)^0.5 we can get k. However, isn't this a little too simple for 6 marks ? Should I be taking into account the drag due to the air in the system ? Hence, should I be considering damped SHM ?

    My workings to part (iv) are below :

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    (Original post by Stonebridge)
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    Please try and help me with what's in the 9th post.

    Thank you
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    My interpretation of question ii is that it is asking you to define reduced mass and derive the formula from 1st principles.
    That ought to be worth the marks. As I'm not familiar with these questions or what you have been taught etc, and how mark schemes work here, (unlike A-Level) it's difficult to answer reliably.
    In iii as far as I can see, you consider the motion as undamped as the air resistance is negligible. It's the reduced mass you use.
    Part iv is then the case of damped oscillations.
 
 
 
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