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    (Original post by rbnphlp)
    Yes.

    Any system that undergoes simple harmonic motion exhibits two key features.

    1. When the system is displaced from equilibrium there must exist a restoring force that tends to restore it to equilibrium.
    2. The restoring force must be proportional to the displacement, or approximately so.

    it does not have to be always a spring , it can be a string , a boat tied to harbour as long as they satisfy the above two conditions they will perform or can be modelled as performing SHM
    Thanks for your reply!!! could you take a look at query 6 as well please, would be much appreciated
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    (Original post by boromir9111)

    Query 6 : Lenses : if the focal length of an lens is 30 cm would it be right to say that image distance will be at 30cm too?
    I wont be able to answer that without know where the object is placed from the lens..i.e you need to know what v(object distance) is .. to find u(image distance)
    Also you ned to know if its concave or convex lenses as stonebridge pointed out
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    (Original post by boromir9111)
    are you talking about 1/u + 1/v = 1/f.... that's for a thin lens though. But is it for a converging for diverging?
    Yes, the formula applies to a so-called thin lens. Thick lenses are a bit more complicated and you won't ever need to worry about them at A Level.
    It works for both types of lens, and gives you the method for calculating, for example, the image distance (from the lens) given the object distance and the focal length.
    There's not much more to it really, except that you can also solve the problem by drawing a ray diagram to scale and measuring the values.
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    (Original post by Stonebridge)
    Yes, the formula applies to a so-called thin lens. Thick lenses are a bit more complicated and you won't ever need to worry about them at A Level.
    It works for both types of lens, and gives you the method for calculating, for example, the image distance (from the lens) given the object distance and the focal length.
    There's not much more to it really, except that you can also solve the problem by drawing a ray diagram to scale and measuring the values.
    Hey there mate, thanks for your reply..... oh ok, so there's not much depth to it from what i already know on this. That's good and thanks once again!!!
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    if a convex lens has a focal length of 30cm then the image of a "distant" object will be 30cm from the lens.

    if u = infinity then 1/v = 1/f so v = f
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    (Original post by Drummy)
    if a convex lens has a focal length of 30cm then the image of a "distant" object will be 30cm from the lens.

    if u = infinity then 1/v = 1/f so v = f
    oh if u = infinity only then v = f. Thanks for that mate
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    Can someone please explain this wave-particle duality to me please and how this relates to de Broglie wavelegnth???

    Thanks in advance.
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    anyone?
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    You need to look at the topic "Electron Diffraction". This is evidence that particles (electrons) can behave as waves. The idea is used in an electron microscope.
    The de Broglie wavelength is what the electron's wavelength is when it's behaving like a wave. There is a formula to calculate it.
    Wavelength= h/p where h is Planck's constant and p is the momentum of the electron.
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    (Original post by Stonebridge)
    You need to look at the topic "Electron Diffraction". This is evidence that particles (electrons) can behave as waves. The idea is used in an electron microscope.
    The de Broglie wavelength is what the electron's wavelength is when it's behaving like a wave. There is a formula to calculate it.
    Wavelength= h/p where h is Planck's constant and p is the momentum of the electron.
    Yeah i know that, is there anymore to that i need to know?
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    Just combine it with the photoelectric effect as an example of waves (light) behaving like particles, and you have the two sides of the wave-particle duality topic at A Level.
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    http://www.edexcel.com/migrationdocu...e_20080611.pdf

    question 3. Just so i know, for the field lines i would draw arrows going from north to south yes?
    For 3 (c) not sure on the reason to make really.
    For 3 (d) - the poles have been reversed, the wire would still levitate wouldn't it?

    edit - oh wait for 3(c) could i say that flux = BA and therefore B = flux/A-----> so as distance increases (area) the flux decreases?
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    (Original post by Stonebridge)
    Just combine it with the photoelectric effect as an example of waves (light) behaving like particles, and you have the two sides of the wave-particle duality topic at A Level.
    Ahh yes, the photoelectric effect, that will help me in my latter exam for physics. Thanks once again mate and if you get time, please take a look at my next question please
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    For 3 (c) not sure on the reason to make really.
    oh wait for 3(c) could i say that flux = BA and therefore B = flux/A-----> so as distance increases (area) the flux decreases?
    This is true, but not sure that it will be the answer they're looking for given that the question is about ampere's law. More likely they want this:

    An electric current in a wire creates a magnetic field via Ampere's law, or more specifically in this situation, the Biot-Savart law. The direction of this field is given by  l \times r , equivalently found using the right hand rule. In this case, with the configuration shown, we see that the field generated by the current (electron motion from the negative to positive terminal) will be in the opposite sense to that of the magnet. Net result - the two magnetic fields partially cancel, resulting in the flux measured for the magnet being less than that of the manufacturers rating.

    For 3 (d) - the poles have been reversed, the wire would still levitate wouldn't it?
    Now the poles are reversed, the electrons are flowing in the opposite direction so using the right hand rule shows that the force is now in the opposite direction, i.e. the wire is forced downward and won't levitate.
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    When can you not SUVAT equations and when can you use them? please explain to an A level standard. Thanks in advance.
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    (Original post by spread_logic_not_hate)
    This is true, but not sure that it will be the answer they're looking for given that the question is about ampere's law. More likely they want this:

    An electric current in a wire creates a magnetic field via Ampere's law, or more specifically in this situation, the Biot-Savart law. The direction of this field is given by  l \times r , equivalently found using the right hand rule. In this case, with the configuration shown, we see that the field generated by the current (electron motion from the negative to positive terminal) will be in the opposite sense to that of the magnet. Net result - the two magnetic fields partially cancel, resulting in the flux measured for the magnet being less than that of the manufacturers rating.



    Now the poles are reversed, the electrons are flowing in the opposite direction so using the right hand rule shows that the force is now in the opposite direction, i.e. the wire is forced downward and won't levitate.
    Thank you very much for your reply!!!! that has really helped me alot mate thanks once again.
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    When can you not SUVAT equations and when can you use them? please explain to an A level standard.
    SUVAT eqautions apply only to bodies moving with constant accelleration, i.e. accelleration that does not change in time.

    eg if a car accellerates from 1ms-1 to 10ms-1 with accelleration a = 2ms-2, then SUVAT is ok.

    If the car accellerates from 1ms-1 to 10ms-1 with accelleration a = (1 + t)ms-2, where t is time, then SUVAT is not ok as the accelleration is no longer constant in time.

    Also SUVAT is only valid for 1d motion, as all of the quantities are scalars. It may be possible to use it to describe 2 or 3d motion, by assigning each direction its own set of SUVAT equations, but im not sure on this and anyway a vector based approach would be far easier there.
 
 
 
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