ALL CIE PHYSICS DOUBTS HERE !!!

[Deadly_king]

W07 Q 5(a) P2
phi/360=AB/lambda
(60/360) * lambda = AB
AB = 1/6 * 6
     = 1
So it leads or lags behind (question didnt state which) by 1
see attachment

S06 Q6 P2
What part didnt u understand exactly?

Good work

He said he need the whole question. Do you mind to take care of it?

I've to help others.

[$!$RatJumper$!$]

W07 Q 5(a) P2
phi/360=AB/lambda
(60/360) * lambda = AB
AB = 1/6 * 6
     = 1
So it leads or lags behind (question didnt state which) by 1
see attachment

S06 Q6 P2
What part didnt u understand exactly?

Thankx that makes sense about the phase diff. But dont we need to do anything about the amplitude since it mentions its intensity is half? even the mark scheme says something about amplitude.

And if you could do all parts on q6 it would be wonderful

thankx

[TJ-56]

Good work

He said he need the whole question. Do you mind to take care of it?

I've to help others.

(b) 32.4 cm is the distance between 1 nodes, i.e. half the distance of a complete wavelength. v=f * lambda
frequencey is stated, 512 Hz
Lambda= 32.4*2 *10^-2 (im meters)
so 512*64.8 * 10^-2 = 330 m\s
(c) the exact distance of the antinode is half 32.4
So 32.4/2 = 16.2
So the antinode is 16.2 cm above the surface of water in the tube on the left.
As 15.7 cm is the length of the column of air INSIDE the tube, then 16.2-15.7 is the length of air above the where the antinode is located.
Hope that was helpful

[TJ-56]

Thankx that makes sense about the phase diff. But dont we need to do anything about the amplitude since it mentions its intensity is half? even the mark scheme says something about amplitude.

And if you could do all parts on q6 it would be wonderful

thankx

oh sorry my bad, i will post the correct answer with the amplitude part in a few min

[TJ-56]

Thankx that makes sense about the phase diff. But dont we need to do anything about the amplitude since it mentions its intensity is half? even the mark scheme says something about amplitude.

And if you could do all parts on q6 it would be wonderful

thankx

say the maximum amp of the given wave is 2 (taking each 5 small blocks as 1)
then we have:
I(new)/I = A(new)²/A²
0.5I/I = A(new)²/(2A)²
Making A(new) ² the subject gives us = sqr root(2A²)
                                                      = sqr root(2) * A
which is 1.4 (the new amplitude)

[$!$RatJumper$!$]

say the maximum amp of the given wave is 2 (taking each 5 small blocks as 1)
then we have:
I(new)/I = A(new)²/A²
0.5I/I = A(new)²/(2A)²
Making A(new) ² the subject gives us = sqr root(2A²)
                                                      = sqr root(2) * A
which is 1.4 (the new amplitude)

Thank you so much i finally understand how to do it just to make sure can u please draw it out again? thank you. and im cheking out the other answers for my other question now. thankx again! +rep

[$!$RatJumper$!$]

(b) 32.4 cm is the distance between 1 nodes, i.e. half the distance of a complete wavelength. v=f * lambda
frequencey is stated, 512 Hz
Lambda= 32.4*2 *10^-2 (im meters)
so 512*64.8 * 10^-2 = 330 m\s
(c) the exact distance of the antinode is half 32.4
So 32.4/2 = 16.2
So the antinode is 16.2 cm above the surface of water in the tube on the left.
As 15.7 cm is the length of the column of air INSIDE the tube, then 16.2-15.7 is the length of air above the where the antinode is located.
Hope that was helpful

that was very helpful thankx! though just a question.. why did you only draw half a wavelength in the tube? are there certain rules as in how many wavelengths to draw when drawing stationary waves

[TJ-56]

Thank you so much i finally understand how to do it just to make sure can u please draw it out again? thank you. and im cheking out the other answers for my other question now. thankx again! +rep

There u go

[$!$RatJumper$!$]

gr8 stuff thankx.. could you please answer my other question aswell about the stationary waves

[TJ-56]

gr8 stuff thankx.. could you please answer my other question aswell about the stationary waves

I didnt understand ur question, the qp though implies that only half a wavelength should be drawn, (difference between 2 loud sounds)

[$!$RatJumper$!$]

I didnt understand ur question, the qp though implies that only half a wavelength should be drawn, (difference between 2 loud sounds)

it just simply says sketch the form of the stationery wave set up in the tube. no mention of only half a wavelength

[TJ-56]

w08 q6 (b)
can someone state 2 acceptable experiments?
for the first one i drew a light source and an obstacle, where light is diffracted and an eye at the other end, which observes the diffracted light (and cannot see the source)
for the second one i drew a person shouting, with an obstacle in front of him, and an observer behind the obstacle , where the observer can hear the shouting after being diffracted
Are these correct responses?
thanx in advance

[TJ-56]

it just simply says sketch the form of the stationery wave set up in the tube. no mention of only half a wavelength

Ok what do u mean by certain rules?
The tube on the left show an anti node at the top,
To find the NEXT antinode, half a wavelength is drawn,
if we were to draw a complete wavelength instead, 3 anti nodes will result, but they want the NEXT antinode, which is formed by drawing half a wavelength
sorry but i dont know how to explain it more than that,
did that cover ur question?
maybe i didnt fully understand ur que

[$!$RatJumper$!$]

oh right yeas that makes sense now. thankx a lot man

[$!$RatJumper$!$]

w08 q6 (b)
can someone state 2 acceptable experiments?
for the first one i drew a light source and an obstacle, where light is diffracted and an eye at the other end, which observes the diffracted light (and cannot see the source)
for the second one i drew a person shouting, with an obstacle in front of him, and an observer behind the obstacle , where the observer can hear the shouting after being diffracted
Are these correct responses?
thanx in advance

I drew a light source with a cardboard with a slit in it to act as the place where the diffraction will occur. Then a screen where the diffracted light will be seen.
In your case I dont think we can use an eye as a suitable detection media. We will need to use some sort of equipment in order for us to interpret/observe the diffracted light.

For the second one i simply drew a sound source with a small obstacle and had a microphone and cathode ray oscilloscope on the other end to see the wave pattern on the CRO.
I think your second one is fine.

You may want to check the mark scheme to see what cambridge are looking for exactly in your answers