Edexcel Physics Unit 2 "Physics at work" June 2013

Now its difference to producing standing waves in a string. The harmonics are produced for odd multiples of half wavelength. Like this

Ive never understood why its lambda/4, why not lambda/2 for the fundamental?

But you are BRANNY101

Only kidding

Ive never understood why its lambda/4, why not lambda/2 for the fundamental?

I don't know. My teacher said that its about resonance and that will learn about it in A2. I guess this is the theory behind it


At the end of the tube the sound wave undergoes reflection and another wave is produced. Now at this point of incidence there should be node, a point with zero amplitude that undergoes no displacement at all. This is impossible if the wave has undergone half wavelength where it produces an antinode in the end which is impossible! Notice also that there should always be an antinode at the opening.
A hard reflection takes place. The sound wave tries to displace the particles of the tube at the end in one direction but its way to small a magnitude. According to Newton's Third law the tube then exerts a equal and opposite force, a force at the opposite direction, so the resultant displacement is zero because of complete destruction of anti phase points.

Its good to remember that at the opening an anti node is necessary and at the end a node is necessary for a standing wave. And if both are openings then both should be antinodes.

Thank you sooo much really!
one more question (from CIE) but who cares ...it's still electricity!
http://vvcap.net/db/rhNqORlDMtAxemgMkbWe.htp http://vvcap.net/db/DlCJFeu-fbrM6OA7lLmZ.htp
the second last one when it says : closed open open .... i thought it's 3 ?

Right, but in an open-ended tube , both sides are at anti-nodes, right? ( e.g. a tube )

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Yes that's what I said in the last sentence

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On part (c)(iii) How do i do it?

P is a fixed resistor, when the lamp is in parallel with the fixed resistor, the resistance of the parallel combination is smaller than the resistance of P. Therefore now that the lamp is removed you have a potential divider, the ratio of the resistance between P and R changes so that P has a higher proportion of the voltage (as the resistance of P > the resistance of P and the lamp in parallel), hence the voltmeter reading increases.

Now its difference to producing standing waves in a string. The harmonics are produced for odd multiples of half wavelength. Like this

Well. When the bulb is removed the total resistance of the circuit decreases. This is because now there's no parallel connection to the component P. As the total V is constant more R means less I (V=IR). So as current decreases and the resistance of P is assumed to be constant then V decreases.




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Please help with this question.



I don't know how to interpret the question to a diagram and then answer the question.please help.The answer is D according to mark scheme.


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Well phase difference is the fraction of difference of oscillation of one oscillator to another. It measured as an angle in radians or in degrees. It actually sees how the sine curve of one wave has been translated to get another along the x axis. A translation or phase difference of 0,2π,4π
...2nπ (multiples of 2π) gives the same curve back again so its thought to be in phase, while a phase angle difference of π,3π....2n+1π gives a sine curve that is totally opposite so its out of phase.
You can work with degrees also.

Path difference is the difference in the distance travelled between two waves from their source. It can be measured as distance or in wavelength. If the difference shows a multiple of one wavelength they are in phase while a odd multiple of half wavelengths(0.5lamda,1.5lambda) shows an exactly out of phase points in the two wave. The significance of this can be seen in the principle of superposition of waves, diffraction and interference patterns etc

The relationship between these two variables are simple. A phase angle difference of 2π or 360° corresponds to 1 lambda path difference so use this ratio method

Please help with this question.



I don't know how to interpret the question to a diagram and then answer the question.please help.The answer is D according to mark scheme.


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Sorry for it being messy. Basically there are 2 resistors in series, R2 and R3, with resistor R2 in parallel with both of them. Then there's a voltmeter across the whole set of 3 resistors. In parallel, components have the same pd, so pd across R1=PD across R2 and R3. All .resistors are identical, R2=R3=R1. Now let consider just the circuit's series part. PD across R2 and R3=PD across R2+PD across R3. So PD across say R2 or even just across R3 is HALF of PD across series set of R2 and R3, so therefore PD across R2 or even just across R3 is HALF of PD across whole parallel combination, which is HALF of PD across R1. By P=V^2/R, power dissipated by R2 alone or even R3 alone is equal to a quarter of power dissipated by R1, making D the answer

I was doing this yesterday... it was exploding my mind. Sorry if I'm wrong but I'll try my best to help you. This is the diagram I did:



Basically, let's call the top path, path 1, and the bottom path, path 2.
Path 1 has double the resistance of path 2, and since both paths are in parallel, they have the same potential difference across them. So:

$V=IR$

path 1 has half the current flowing through it than path 2 (current takes the path of least resistance).

Power dissipated is:

$P=I^2R$

So the power dissipated for ONE of the resistors in path 1 is:

$Current = \dfrac{1}{3}I; Resistance = R[br][br]Power = (\dfrac{1}{3}I)^2R[br] = \dfrac{1}{9}I^2R$

Power dissipated for the resistor in path 2 is:

$Current = \dfrac{2}{3}I; Resistance = R[br][br]Power = (\dfrac{2}{3}I)^2R[br]= \dfrac{4}{9}I^2R$

Hence, the power dissipated by the resistor in path 2 is four times the power dissipated by a resistor in path 1.

I hope this helped.