Post Your AS LEVEL PHYSICS PAPER 2 DOUBTS

[applepie1996]

Thanx man but u should knw that i also did it till this part couldn't do the rest of it

If n is an odd integer, there will be destructive interference.
If n is an even integer, there will be constructive interference.
(It's the destructive interference that is being asked for in the question).


Using Pythagoras Theorem (right-angled triangles, etc.),
we can calculate that the difference in the distance from the most distant sound source to the microphone to that of the nearest sound source is 0.28 meter.

So, n.(lambda/2) = the path difference = 0.28 (meters) . . . . [1]
for constructive or destructive interference (depending on whether n is odd or even).

Now since speed of sound in air, v = f x lambda
we get: lambda = speed / frequency,

Now we know v; it is 330 m/s.
From [1] above, we can now say:

n.(330/2f) = the path difference = 0.28.

This can easily be re-arranged to give:
330n = 0.56f, which simplifies to:
f = 589n . . . . . .[2]

We know that if n = an even number, there will be constructive interference

but the question is asking for minima which is destructive
We know that if n = an odd number, there will be destructive interference.
If you put successive odd numbers into [2], you get the following frequencies:
n = 1; f = 589
n = 3; f = 1767
n = 5; f = 2945
n = 7; f = 4123
the frequency range in the question is 1000hz to 4000hz
Looking at the destructive interference table above, we can see that there are TWO frequencies that lie in this range
SO TWO MINIMA

 

[applepie1996]

s09 variant 1 qs 5b....plz help
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_2.pdf

Using Pythagoras Theorem (right-angled triangles, etc.),
we can calculate that the difference in the distance from the most distant sound source to the microphone to that of the nearest sound source is 0.28 meter.

So, n.(lambda/2) = the path difference = 0.28 (meters) . . . . [1]
for constructive or destructive interference (depending on whether n is odd or even).

Now since speed of sound in air, v = f x lambda
we get: lambda = speed / frequency,

Now we know v; it is 330 m/s.
From [1] above, we can now say:

n.(330/2f) = the path difference = 0.28.

This can easily be re-arranged to give:
330n = 0.56f, which simplifies to:
f = 589n . . . . . .[2]

We know that if n = an even number, there will be constructive interference

but the question is asking for minima which is destructive
We know that if n = an odd number, there will be destructive interference.
If you put successive odd numbers into [2], you get the following frequencies:
n = 1; f = 589
n = 3; f = 1767
n = 5; f = 2945
n = 7; f = 4123
the frequency range in the question is 1000hz to 4000hz
Looking at the destructive interference table above, we can see that there are TWO frequencies that lie in this range
SO TWO MINIMA

 

[applepie1996]

http://xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s08_qp_2.pdf

PLEAAASE 6B how do you calculate the POWER IN EACH PARTTTTTTTT.

for da first one S1 is opened and it is the main switch so if it is open the power in the whole circuit will also be zero
for da second one current always takes the path with the least resistance in dis case it goes through the loop with th B heater so power is 240^2/38.4=1.5kw
u'll take sum of the resistance as i said current always takes the path with the least resistance in dis case it goes through the loop with th B heater and the C heater which are parallel 2 each other so resistance on parallel is 38.4 x 38.4/38.4 + 38.4=19.2 so power = voltage^2/resistance=240^2/19.2=3kw
for da third onewell this pretty easy as it is a normal series circuit so u add the resistance 38.4+38.4=76.8 then u find the power=240 squared /76.8=0.75kw
u find total resistance first which is 76.8 x 38.4/76.8 + 38.4=25.6 now find power 240 squared /25.6 =2.25kw

 

[USMAN Sheikh]

Please mere bhe koi reply krdo plssss :"( :"( :"( on previos page 2 ques one ov graph and other of stress-strain pls !!

 

[freakybandi]

If n is an odd integer, there will be destructive interference.
If n is an even integer, there will be constructive interference.
(It's the destructive interference that is being asked for in the question).


Using Pythagoras Theorem (right-angled triangles, etc.),
we can calculate that the difference in the distance from the most distant sound source to the microphone to that of the nearest sound source is 0.28 meter.

So, n.(lambda/2) = the path difference = 0.28 (meters) . . . . [1]
for constructive or destructive interference (depending on whether n is odd or even).

Now since speed of sound in air, v = f x lambda
we get: lambda = speed / frequency,

Now we know v; it is 330 m/s.
From [1] above, we can now say:

n.(330/2f) = the path difference = 0.28.

This can easily be re-arranged to give:
330n = 0.56f, which simplifies to:
f = 589n . . . . . .[2]

We know that if n = an even number, there will be constructive interference

but the question is asking for minima which is destructive
We know that if n = an odd number, there will be destructive interference.
If you put successive odd numbers into [2], you get the following frequencies:
n = 1; f = 589
n = 3; f = 1767
n = 5; f = 2945
n = 7; f = 4123
the frequency range in the question is 1000hz to 4000hz
Looking at the destructive interference table above, we can see that there are TWO frequencies that lie in this range
SO TWO MINIMA

thankyou!!!!

 

[freakybandi]

listen,
First u calculate S2M (Pythagoras)
√(80²+100²)= 128 cm
then path difference is 28.0 cm (128-100)
Calculate lambda at 1.0 KHz which is (330/[1.0 x 10³]) = 0.330 m = 33 cm
then claculate lambda at 4.0 KHz which is (330/[4.0 x 10³]) = 0.0825 m = 8.25 cm
so lambda (wavelength) has decreased from 33.0 cm to 8.25 cm
u know if the path difference is half a wavelength then it is destructive interference [zero intensity].
if u calculated the path difference in terms of lambda at both extremes u'll get
28/33= .850 lambda & 28/8.25= 3.93 lambda
so the path difference (p) ,when frequency increased might be; .850<p<3.93
within this range how many X.5 are there (where X is any number) ? Ans: three
p= 1.50,2.50 or 3.50 Lambda

but the ans is two!!

 

[applepie1996]

thankyou!!!!

ur welcum

 

[applepie1996]

and one more question pls Q4 part b plsss be quick mate

Es the shaded region is between the two lines because the line with the arrow goin upwards means loading the line wid da arrow downwards means unloading
i'll solve da nxt part and tell u

 

[applepie1996]

and one more question pls Q4 part b plsss be quick mate

here is the answer of how 2 find shaded region
area of the bigger triangle (just make it on the graph) - area of smaller triangle
=1/2 x 0.90 x 80(EXTEND THE LINE TILL 80 TO FORM A TRIANGLE) - 1/2 x (0.9-0.1) x 74(I'M SUBTRACTING 0.9 FROM 0.1 BECAUSE THE TRIANGLE STARTS FROM 0.1 AND NOT THE ORIGIN) = 6.4mJ

 

[applepie1996]

Please mere bhe koi reply krdo plssss :"( :"( :"( on previos page 2 ques one ov graph and other of stress-strain pls !!

DERE I SOLVED UR DOUBTS U CAN CHECK THEM OUT

 

[freakybandi]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w09_qp_21.pdf
qs 5c....

 

[applepie1996]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_21.pdf
qs 5c....

ok for diffraction
use a laser screen and a slit and pass the laser through the slits as the light passes through the slit it is diffracted so it spreads out onto the screen showing diffraction
interference
a coherent light source such as a laser beam and let it pass through two parallel slits, and the light passing through the slits is observed on a screen behind the plate. The wave nature of light causes the light waves passing through the two slits to interfere, producing bright and dark fringes on the screen
I HOPE U CAN DRAW THEM URSELF

 

[freakybandi]

ok for diffraction
use a laser screen and a slit and pass the laser through the slits as the light passes through the slit it is diffracted so it spreads out onto the screen showing diffraction
interference
a coherent light source such as a laser beam and let it pass through two parallel slits, and the light passing through the slits is observed on a screen behind the plate. The wave nature of light causes the light waves passing through the two slits to interfere, producing bright and dark fringes on the screen
I HOPE U CAN DRAW THEM URSELF

yep i can draw them....thnkx!!

 

[applepie1996]

yep i can draw them....thnkx!!

ur welcum
anytym
GOOD LUK FOR UR EXAM

 

[freakybandi]

thnkx dat good luck is needed!!!....same to u!!

 

[applepie1996]

thnx

 

[leosco1995]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_22.pdf
Q4 (c) please.

 

[applepie1996]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_22.pdf
Q4 (c) please.

i think it's done like this
lowest frequecy = 1st harmonic (first harmonic is the first sound wave produced) which over here is lambda/4
so it will lambda/4=0.45m
=1.8m
f=v/lambda
=183Hz

 

[leosco1995]

i think it's done like this
lowest frequecy = 1st harmonic (first harmonic is the first sound wave produced) which over here is lambda/4
so it will lambda/4=0.45m
=1.8m
f=v/lambda
=183Hz

Wait.. can you explain this harmonic and frequency stuff a bit more? I don't get it.

 

[applepie1996]

Wait.. can you explain this harmonic and frequency stuff a bit more? I don't get it.

Harmonic is the first ever wave thing produced so over here it is an antinode and node which lambda/4 and this in other words is also called the fundamental frequency meaning lowest frequency