AQA Physics A - PHYA4 (11/06/12) - Exam thread

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  1. internet tough guy's Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    (Original post by m1a1tank)
    which question is this in the paper?
    I just came across a few summary questions in the textbook, so its not a specific past paper question - haven't done any yet :eek:
  2. number23's Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    (Original post by callmenighthawk)
    When the frequency of the external periodic force matches the natural frequency, there is efficient energy transfer from external to natural. The periodic frequency wave will be precisely 1/4 of a wavelength out of phase to the natural frequency and the external force will give the oscillating system extra force whenever the natural amplitude is not zero.

    not really sure about that one
    does anyone know why they are 90 degree out of phase?
  3. callmenighthawk's Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    (Original post by number23)
    does anyone know why they are 90 degree out of phase?
    because as soon as the displacement is anything other than 0, there will be an opposing force trying to get it back to equilibrium. the driving force (force = mass x velocity remember) has the same wave as velocity, so the external force must do the same not to oppose the natural frequency wave, and to boost the amplitude.

    hope that helps
  4. number23's Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    (Original post by callmenighthawk)
    because as soon as the displacement is anything other than 0, there will be an opposing force trying to get it back to equilibrium. the driving force (force = mass x velocity remember) has the same wave as velocity, so the external force must do the same not to oppose the natural frequency wave, and to boost the amplitude.

    hope that helps
    so its when there is minimal restoring force?
  5. callmenighthawk's Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    (Original post by number23)
    so its when there is minimal restoring force?
    well there is minimum external force at displacement maximum (when there is no movement, force = mass x velocity remember) so it's not just there is an external force suddenly at displacement 0, it's a wave the same as the velocity.

    but yes, there will be maximum external force whne there is 0 restoring force
  6. don'tTRIP.'s Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    What happens to gravitational field strength on a planet with density 2rho?
  7. m1a1tank's Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    (Original post by don'tTRIP.)
    What happens to gravitational field strength on a planet with density 2rho?
    instead of just giving you the answer to it, I'll give you a clue for how to get to the answer.

    rho (density) = Mass/ Volume
  8. don'tTRIP.'s Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    (Original post by m1a1tank)
    instead of just giving you the answer to it, I'll give you a clue for how to get to the answer.

    rho (density) = Mass/ Volume

    Need help with the full question; if you double radius as well as the density, why is the answer 4F? I know doubling the radius quadruples the force so, does the answer imply that density has no effect on field strength? :confused:
  9. m1a1tank's Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    (Original post by don'tTRIP.)
    Need help with the full question; if you double radius as well as the density, why is the answer 4F? I know doubling the radius quadruples the force so, does the answer imply that density has no effect on field strength? :confused:
    why would the radius have anything to do with the gravitational field strength. Gravitation field strength is g, which is equal to F/M right. or were you talking about the F=GmM/r^2 equation?
  10. don'tTRIP.'s Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    (Original post by m1a1tank)
    why would the radius have anything to do with the gravitational field strength. Gravitation field strength is g, which is equal to F/M right. or were you talking about the F=GmM/r^2 equation?
    Yeah, I was talking about the GMm/r^2 equation :holmes:
  11. jlnbello's Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    can some pls explain to what the principles of rate of change of flux linkage means? and how it relates to induced emf
    thanks in advance
  12. Mortson's Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    (Original post by don'tTRIP.)
    Need help with a question, but can't seem to find it It was something along of the lines of, what is the gravitational field strength of a planet with both, a density and radius twice that of Earth? The answer was 4F I think. Anyone?

    Thanks.
    I apologize in advance if this is unclear/patronizing.


    So, we know from the question that:

    density = 2x
    radius = 2x

    The equation for the magnitude of a gravitational field is-



    Straight away we can put it the fact that the radius is twice the value, getting-

    which is equal to

    In the equation, M is the mass of the planet. Seeing as we are given something to do with the density, we need to relate mass to density. So we can use the equation:



    Rearrange and substitute into the equation we had before to get:



    We know the density is twice the value of the earth's, so:



    Volume of a sphere is:



    Seeing as the radius is twice the earth's radius, we can subsitute this in and get:

    which is equal to

    So the volume is 8 times the earth's, the density is 2 times the earth and the radius is 2 times the earth. In the equation, we now have:



    Which equals 4. Done!
  13. FrightBright's Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    (Original post by callmenighthawk)
    well there is minimum external force at displacement maximum (when there is no movement, force = mass x velocity remember) so it's not just there is an external force suddenly at displacement 0, it's a wave the same as the velocity.

    but yes, there will be maximum external force whne there is 0 restoring force
    I maybe wrong.

    Isnt Maximum restoring force is at maximum displacement since Force = ma not mv
    Last edited by FrightBright; 02-06-2012 at 12:07.
  14. callmenighthawk's Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    (Original post by FrightBright)
    I maybe wrong.

    Isnt Maximum restoring force is at maximum displacement since Force = ma not mv
    **** I am a dumb****. all I know is that the external force is 1/4th out of phase to the natural displacement.

    there is maximum restoring force at maximum displacement though, which is why it is at 0 velocity then

    external force and restoring force are two different things, restoring forces are in everything but external forces increase/decrease amplitude, depending on how out of phase it is (or how close it is to the natural frequency)
    Last edited by callmenighthawk; 02-06-2012 at 12:16.
  15. FrightBright's Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    (Original post by callmenighthawk)
    **** I am a dumb****. all I know is that the external force is 1/4th out of phase to the natural displacement.

    there is maximum restoring force at maximum displacement though, which is why it is at 0 velocity then

    external force and restoring force are two different things, restoring forces are in everything but external forces increase/decrease amplitude, depending on how out of phase it is (or how close it is to the natural frequency)
    I've never used the term external force. However all I know is restoring force acts towards equilibruim

    So I guess when a body is moving away from equilibruim the resultant from (external-restoring) decreases. It reaches maximum displacement and then the restoring force is at max (negative accelaration)

    So Im guessing that restoring force is at maximum displacement and external force is maximum at equlibruim.

    I could be shatting gas though...
  16. callmenighthawk's Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    What tips have your teachers given you to use in the 6 marker written questions?

    Mine said to just write down everything you know related to the topic if you aren't sure what to write. Also write down some quick bullet points to structure your answer to make sure that you are actually answering the question - a lot of the time what people are writing down IS correct, but doesn't answer the question so gets zero marks.
  17. paul272's Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN11.PDF
    how do you do questions 11, 21 and 22
  18. don'tTRIP.'s Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    Is this a good definition of centripetal force?

    "The force needed for an object to execute circular motion"
  19. number23's Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    (Original post by don'tTRIP.)
    Is this a good definition of centripetal force?

    "The force needed for an object to execute circular motion"
    it always directed towards the centre of the circle, perpendicular to direction of motion
  20. callmenighthawk's Avatar
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    Re: AQA Physics A - PHYA4 (11/06/12) - Exam thread
    (Original post by paul272)
    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN11.PDF
    how do you do questions 11, 21 and 22
    11 - no clue - look up the mark scheme

    21 - decreases as the area the coil is cutting is 0 (it is perpendicular to the field) just do BAN cos 50 I believe.

    22 - you want to work out the area, as you're trying to work out BA

    you work out the area by 60 x (150 x 10) because it is traveling 1500m distance, and has a 60m width
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