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AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread] Watch

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    (Original post by CD223)
    Do you always assume the amplitude to be the horizontal distance from the equilibrium then?


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    Yeah, if the amplitude is small enough but i think the amplitude is the arc of the oscillation, isn't it?


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    (Original post by DCMed96)
    A quick question: is magnetic flux density a scalar or vector? I did a question and it said it is a vector, but I don't really get it, can someone explain why?
    It is a vector quantity but i really dint know why. It's magnetic flux per unit area of a magnetic field at right angles to the magnetic field. I think the magnetic field with which magnetic flux is defined makes flux density a vector

    EDIT: or maybe in terms of B=F/(IL), force makes flux density a vector. Not sure to be honest


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    Can someone help me how to do this question please? The answer is C and I do not see how!

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    (Original post by DCMed96)
    A quick question: is magnetic flux density a scalar or vector? I did a question and it said it is a vector, but I don't really get it, can someone explain why?
    Vector - B = F/IL

    It therefore depends on a vector, making it a vector quantity.


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    (Original post by Mehrdad jafari)
    Yeah, if the amplitude is small enough but i think the amplitude is the arc of the oscillation, isn't it?


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    Think so!


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    Hi can someone please explain the (i) and (iii) parts of this six marker from the specimen paper to me please?Name:  ImageUploadedByStudent Room1433679027.585974.jpg
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    The natural frequency of the spring system is 1.5Hz


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    (Original post by DCMed96)
    A quick question: is magnetic flux density a scalar or vector? I did a question and it said it is a vector, but I don't really get it, can someone explain why?
    It's defined as magnetic flux per unit area, which can be seen as number of field lines per unit area. Area is a scalar so we can ignore that and concentrate on number of field lines. Field lines are clearly vectors, so number of field lines = number*field lines = scalar*vector = vector. Thus magnetic flux per unit area = number of field lines per unit area = number of field lines / area = vector / scalar = vector.
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    (Original post by Disney0702)
    Can someone help me how to do this question please? The answer is C and I do not see how!

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    Does this help?

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    (Original post by TeddyKC)
    Hi can someone please explain the (i) and (iii) parts of this six marker from the specimen paper to me please?Name:  ImageUploadedByStudent Room1433679027.585974.jpg
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Size:  116.2 KBName:  ImageUploadedByStudent Room1433679082.207933.jpg
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    The natural frequency of the spring system is 1.5Hz


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    Which paper is this In please?

    It's all about forced oscillations and how a driving force from the rod makes the mass spring system either resonate or damp. I'll do it later unless someone eats me to it


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    Hi would it be possible for you to explain the equation used In part c final question


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    (Original post by CD223)
    Does this help?

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    Oh my! Thank you! I didn't think of it like that!
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    (Original post by Disney0702)
    Oh my! Thank you! I didn't think of it like that!
    No worries! Tricky question


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    (Original post by TeddyKC)
    Hi can someone please explain the (i) and (iii) parts of this six marker from the specimen paper to me please?Name:  ImageUploadedByStudent Room1433679027.585974.jpg
Views: 101
Size:  125.2 KBName:  ImageUploadedByStudent Room1433679066.093318.jpg
Views: 106
Size:  116.2 KBName:  ImageUploadedByStudent Room1433679082.207933.jpg
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Size:  121.4 KBThe natural frequency of the spring system is 1.5HzPosted from TSR Mobile
    i) Driving frequency 0.2Hz is much less than the natural frequency of the system 1.5Hz, so the two are roughly in phase and the system is forced to vibrate at an amplitude that is less than at natural frequency

    ii) Driving frequency equals the natural frequency of the system so the system oscillates with rapidly increasing amplitude as it gains more and more energy from the driving force, eventually reaching resonance where it oscillates at its maximum possible amplitude.

    iii) Driving frequency 10Hz is much more than the natural frequency of the system 1.5Hz, so the two are completely out of phase and the system is forced to vibrate at an amplitude that is less than at natural frequency.
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    Any idea what might come up in six marker this year?
    What's the hardest thing AQA can throw on us??
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    (Original post by DCMed96)
    Any idea what might come up in six marker this year?
    What's the hardest thing AQA can throw on us??
    SHM application question or a momentum question I think.


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    Can someone help me with this question please?

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    Thank you in advance
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    (Original post by NEWT0N)
    i) Driving frequency 0.2Hz is much less than the natural frequency of the system 1.5Hz, so the two are roughly in phase and the system is forced to vibrate at an amplitude that is less than at natural frequency

    ii) Driving frequency equals the natural frequency of the system so the system oscillates with rapidly increasing amplitude as it gains more and more energy from the driving force, eventually reaching resonance where it oscillates at its maximum possible amplitude.

    iii) Driving frequency 10Hz is much more than the natural frequency of the system 1.5Hz, so the two are completely out of phase and the system is forced to vibrate at an amplitude that is less than at natural frequency.
    Thank you!
    But I still don't quite understand, why are they roughly in phase if the frequency of the driver is much less than they natural frequency and vice versa for (iii) ?


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    Sometimes I can do these, but sometimes fail at them. Ended up with a quadratic which is totally wrong. Some help please.

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    (Original post by Disney0702)
    Can someone help me with this question please?

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    Thank you in advance
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    (Original post by Jimmy20002012)
    Sometimes I can do these, but sometimes fail at them. Ended up with a quadratic which is totally wrong. Some help please.

    Name:  ImageUploadedByStudent Room1433681576.471548.jpg
Views: 80
Size:  79.3 KB


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\dfrac{8}{2} = \dfrac{x^2}{(60-x)^2}

    Where x is the distance to the 8nC charge.

    

\therefore 4 = \dfrac{x^2}{(60-x)^2}

    

\therefore 2 = \dfrac{x}{(60-x)}

    

\therefore 2(60-x) = x

    

\therefore 120 - 2x = x

    

\therefore 120 = 3x

    

\therefore 40 = x

    Does that help?


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