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

Do you always assume the amplitude to be the horizontal distance from the equilibrium then?


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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?

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?

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



Thank you in advance

Hi can someone please explain the (i) and (iii) parts of this six marker from the specimen paper to me please?

The natural frequency of the spring system is 1.5Hz


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Does this help?



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Oh my! Thank you! I didn't think of it like that!

Any idea what might come up in six marker this year?
What's the hardest thing AQA can throw on us??

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.

Can someone help me with this question please?



Thank you in advance

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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$[br]\dfrac{8}{2} = \dfrac{x^2}{(60-x)^2}[br]$

Where x is the distance to the 8nC charge.

$[br]\therefore 4 = \dfrac{x^2}{(60-x)^2}[br]$

$[br]\therefore 2 = \dfrac{x}{(60-x)}[br]$

$[br]\therefore 2(60-x) = x[br]$

$[br]\therefore 120 - 2x = x[br]$

$[br]\therefore 120 = 3x[br]$

$[br]\therefore 40 = x[br]$

Does that help?


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