The Physics PHYA2 thread! 5th June 2013

for a single diffraction,

is saying the 'fringe width would be wider', the same as saying 'there will be wider separation between the fringes'?

are they basically the same thing? :s-smilie:

this is for question 3d in jan 2011 paper, it asks me to state two ways that the appearance of the single slit diffraction fringes would change if the slit was made narrower. I put down 'the width of each fringe would be wider', I'm not sure if that gets the mark.

Reply 761

PrinceyJ

12

Original post by fuzzybear

for a single diffraction,

is saying the 'fringe width would be wider', the same as saying 'there will be wider separation between the fringes'?

are they basically the same thing? :s-smilie:

this is for question 3d in jan 2011 paper, it asks me to state two ways that the appearance of the single slit diffraction fringes would change if the slit was made narrower. I put down 'the width of each fringe would be wider', I'm not sure if that gets the mark.

No the mark scheme references the separation and not the size of the fringes.

Reply 762

Raimonduo

Original post by Zozzy

Could somebody help explain how to answer this question for me please because I don't really understand how you calculate the answer. It is from the Nelson Thornes AS physics textbook and is on page 107, question 4:

A small toy of weight 2.8N is suspended from a horizontal beam by means of two cords that are attached to the same point on the toy. One cord makes an angle of 60' to the beam and the other cord makes an angle of 40' to the vertical. Calculate the tension in each cord.

Apparently the cord at 40' has a tension of 1.5 and the other cord has a tension of 1.9, I've tried many ways to work back from the answer but I can just not seem to figure it out :/.

Despite the use of sine rule not being needed in the specification, by using the sine rule, I actually managed to get the right answer. To use the sine rule though, you need to arrange the forces into a force vector triangle. I made the assumption that because the toy was not in motion, all the forces must be in equilibrium, so then it was possible to use a force triangle.

Reply 763

fuzzybear

12

Original post by PrinceyJ

No the mark scheme references the separation and not the size of the fringes.

but the question asks for the appearance of the fringes if the slit was made narrower

so is it not true that the width of the fringes would get bigger?

Reply 764

Jack93o

15

Original post by ladynova

you would lose a mark so stick to 2 or 3 SF

what has significant figures got to do with this?

the significant figures would be the same whether you write the answer in SI or not

I put 500000, converted into SI (which is what the mark scheme put), is 5*10^5 which is still in the same number of significant figures as 500000

Reply 765

Jack93o

15

Original post by Raimonduo

For 7c, I initially thought that a minimum would be generated as well, because, as you've said, they both reach their equilibrium point at the point they reach the dector (displacement of both waves = 0 at the detector).

However, upon looking at it closer, I realised that, actually, the wave on the left being received must travel a longer distance, than the wave on the right. So through actually counting the amount of peaks and troughs in both the waves being received, I saw that there were 6 whole wave periods on the wave on the left, and 5 whole wave periods on the wave on the right. Which meant that, considering path difference, if the path difference is a whole integer number of λ, a maximum would occur. Which, in this case, the path difference was 1λ, which meant a maximum would be detected.

yeah I saw that as well

actually thinking about it, it makes sense, the intensity doesn't actually stay constant

like the anti-node of a stationary wave, the intensity would osciallate from peak to equilibrium to trough, and back again

so at that split instant in the diagram, the intensity would be zero, but 1/4 cyle later, you'd have the peaks of both waves reinforcing each other

the appearance of 'constant' bright fringes is an illusion, the bright fringes just flicker very fast from max intensity to zero intensity and back again to make it seem like the bright fringes are always at that same brightness/intensity

(edited 10 years ago)

Reply 766

Jack93o

15

Original post by NedStark

For jan 13 7e, can you explain the answer? it's a bit confusing.

Original post by Raimonduo

Mmm, not too sure about that dude, I got that wrong as well. But I wouldn't worry about it, since it came up in Jan13, it's unlikely that it'll come up tomorrow.

frequency doubled, would mean that the wavelength would be halved

because the speed of the microwave stays the same (at speed of light), remember this equation:

speed of light = frequency * wavelength, if frequency goes up, wavelength must go down

the detector would still detect a maximum, because the path difference between the two waves would still be a whole number of wavelengths, so the waves would be in phase, their peaks would meet each other at the same time, their troughs would meet each other at the same time as well

what I didn't understand about that question is why the mark scheme placed so much emphasis on you stating 'even number of wavelength difference', surely there would still be a maximum even if it was an odd number of wavelengths? all that matters for reinforcement of waves is that there is a whole number of wavelengths in their path difference

Reply 767

Jack93o

15

Original post by Zozzy

Could somebody help explain how to answer this question for me please because I don't really understand how you calculate the answer. It is from the Nelson Thornes AS physics textbook and is on page 107, question 4:

A small toy of weight 2.8N is suspended from a horizontal beam by means of two cords that are attached to the same point on the toy. One cord makes an angle of 60' to the beam and the other cord makes an angle of 40' to the vertical. Calculate the tension in each cord.

Apparently the cord at 40' has a tension of 1.5 and the other cord has a tension of 1.9, I've tried many ways to work back from the answer but I can just not seem to figure it out :/.

tension in both strings would be different, call tension in one of them T1 and the other T2 (or some other name)

equate the horizontal components of forces, and then do the same with the vertical components

so for example, equating the horizontal components would get you:

(T1)cos60 = (T2)sin40

(T1)/2 = (T2)sin40 (because cos60 = 1/2)

T1 = 2(T2)sin40

sub this into the vertical components equation, now you'll be able to solve T2

Reply 768

username1201943

19

Original post by wallaby

This is my 3rd time sitting it! 2nd resit :P So I'm really not either haha. 3rd time lucky :wink:

Oh wow. Yea good luck! :smile:

I'm resitting it too lol so we're kind of the same

Reply 769

masryboy94

8

quick question, how would you know whether point C would move up or down in the first 1/4 cycle?

wave diagram.gif2.1KB

Reply 770

Ben LeBlanc

Hello Physicians!

Sorry if this has already been mentioned/answered, but the 6-marker question in the Jan11 paper is about the formation of stationary waves.

How would I go about answering this 6-marker? Thanks :smile:

Reply 771

Im Digital

Original post by masryboy94

quick question, how would you know whether point C would move up or down in the first 1/4 cycle?

Does the question mention anything about what direction the wave travels? Because assuming it travels to the right, C would go down in its first 1/4 cycle

Unfortunately, I can't nail the description as to why, as I'm not revising the waves section yet

Posted from TSR Mobile

Reply 772

taliaa

please help! i seem to be losing marks on all of the equations that involve trigonometry rules. I remember s=o/h c=a/h t=o/a from maths and am confident with this but seem to be using the wrong one (mixing up between sin and cos)
e.g. jan 2013 - q 2 b i, i am conviced you have to use cos but the mark scheme says sin?! why is it sin and not cos?
thaaaanks!!

Reply 773

wallaby

4

Hi, sorry if anyone's asked this before :smile:

Just marking January 2009 paper, Q4a the markscheme is not at all clear to me which direction the velocity vector is meant to go in... If anyone could help :smile:

It says "velocity vector tangential to path and drawn from the ball, arrow in correct direction". I wish they'd draw a diagram to show you :P

Reply 774

OliverG

2

Original post by Ben LeBlanc

Hello Physicians!

Sorry if this has already been mentioned/answered, but the 6-marker question in the Jan11 paper is about the formation of stationary waves.

How would I go about answering this 6-marker? Thanks :smile:

If that is the one on formation of a stationary wave:

Wave travels down from A to B.
Wave reflects at B, new wave formed.
Two waves interfere constructively to form antinode, interfere destructively to form nodes.
State wavelength
State number of nodes/antinodes.
Energy does not travel along stationary waves.

Hope that helps!

Reply 775

wallaby

4

Original post by UnknownRoyalist

Oh wow. Yea good luck! :smile:

I'm resitting it too lol so we're kind of the same

Good luck to you too :smile:

Hopefully we will nail it this time :smile:

Reply 776

masryboy94

8

Original post by wallaby

Hi, sorry if anyone's asked this before :smile:

Just marking January 2009 paper, Q4a the markscheme is not at all clear to me which direction the velocity vector is meant to go in... If anyone could help :smile:

It says "velocity vector tangential to path and drawn from the ball, arrow in correct direction". I wish they'd draw a diagram to show you :P

basically draw a tangent line to the curve (path of ball) at the point where the ball is. and then draw an arrow pointing downwards on that tangent line.

edit: its brought forward by this concept http://mathinsight.org/image/parametrization_circle_4

(edited 10 years ago)

Reply 777

wallaby

4

Original post by masryboy94

basically draw a tangent line to the curve (path of ball) at the point where the ball is. and then draw an arrow pointing downwards on that tangent line.