Physics Question

nobodythere

http://www.examinations.ie/archive/exampapers/2002/LC021ALP1EV.pdf

Question 7, holy god tell me that it's easier than I thought! Any help would be appreciated

Doctor D-Man

Whats wrong with Q7. The first 3 bits are grand. Is it the calculations..

(i) use c = f(lambda) => wavelength = c/f = 10km

(II) 1500 + 5 = 1505km (1500 + 1/2 wavelength)

(iii)pythagoras

dudara

Is it the definitions and explanations, or the maths?

passive

I think the average mark on that question the year it was out was 38%.

I've done it a few times since my first experience... none nearby though. it's really not that awkward. I'd like something similar tomorrow

Richard_Fonzie

In part (ii), are they looking for the distance from transmitter to receiver, or just the distance from point of reflection to receiver?

nobodythere

Doctor D- I get the first part, but don't understand the rest, at all...

Not so much the maths as it is the concepts...

Richard_Fonzie

(ii) Well for destructive interference to happen a trough needs to meet a crest. So if you add an extra half wavelength on, the trough/crest will be meeting (or two other points on each wave that are oscillating differently).

dudara

The minimum distance that the waves can travel is 1500 km. However if two waves left the transmitter together, they would arrive in phase at the receiver causing constructive interference.

Here, they want the minimum distance travelled for destructive interference. This occurs when two waves mix which are anti-phase, or separated in space by 1/2 wavelength. ie one wave travels either a 1/2 wavelength behind or ahead of the other.

Here a 1/2 wavelength is 5 km. Thus the minimum distance is 1505 km

nobodythere

But isn't there already destructive interference at the 1500m because it said that the signal "fades"

Richard_Fonzie

Do you have an answer for part (iii)? 61.28m ?

dudara

The loss of signal strength due to fading does not cause destructive interference. Destructive interference has a very precise definition in physics, either (A) the mixing of coherent waves separated in space by 1/2 wavelength or (B) the meeting of a crest and trough when two coherent waves mix.

Nothing else can give you destructive interference,

Anything in the region of 61 m is correct for the last part

nobodythere

(i) Calculatel for radiowaves
l f v = 3
104 m / 10 km 3

(ii) What is the minimum distance
half wavelength / 5 km 3
1500 km + 5 km 3
1505 km 3

(iii) Calculate minimum h
Pythagoras theorem (any implication) 3
substitution 3
61 km……………… // 61000 m 5
(-1 for lack of units or incorrect units)

Doctor D-Man

Ok so part (i) is fine

(ii) min distance for relected wave: we know the wavelength of the wave is 10km. The distance the wave has to tavel to get to the receiver is 1500km. Destructive interference occurs when the wave is out of phase (i.e half wavelength out of phase) => must travel 1500 + half wavelenth => 1505km

Thats the best I can explain It. I find it hard even to explain it to myself but it kindof makes sense on some level

(iii) just need to find min value of h. use the right angled triangle and apply pythagoras. The base is 750km, height h and the hypotenuse.


If you make a mistake by subbing in the wrong value they only deducted 1m according to marking scheme.

I doubt they will ask anything that complicated though.

dudara

^
straight out of the marking scheme but correct.

Richard_Fonzie

grasshopa wrote:

(-1 for lack of units or incorrect units)

aha :eek:

nobodythere

I was just posting it cos he asked for the answer for question 3. It specifically says the signal fades due to "destructive interference":

The radio reception temporarily “fades” due to destructive interference between the waves travelling parallel to the ground and the waves reflected from a layer (ionosphere) of the earth’s atmosphere, as indicated in the diagram.

My problem with seeing this is that one of the waves travels parallel to the ground while the other is hitting off the reflected layer, and is hence travelling a different distance.

nobodythere

Doctor D-Man wrote:

Ok so part (i) is fine

(iii) just need to find min value of h. use the right angled triangle and apply pythagoras. The base is 750km, height h and the hypotenuse.

Is the hypotenuse 1505 divided by 2?

*Angel*

grasshopa wrote:

Is the hypotenuse 1505 divided by 2?

Correct

nobodythere

Ok, I'm just going to assume it's not coming up since it came up in 2002. Thanks everyone.

sixdraw

there's actually an error in that question aswell.The reflected wave would actually just be half the 1505km cause its not the entire length.It wasn't brought up at the marking conference and the marking scheme has it wrong.You probably would get the marks anyways but it just makes it all more confusing

Doctor D-Man

ka-ching.. now that we've solve this mind-bogglingly difficult physics question I think we're well prepared for anything those examiners try to throw at us.

Let's all celebrate by indulging in a cool glass of Duff beer....mmmm Duff...

Don't worry we'll all do grand. Physics is not a difficult subject when you think about it though. know the experiments and the graphs...30% in the bag..probly know em already after doing em all. Then definitions and formulae. The rest of the paper is just apply these formulae and maybe a few general questions thrown in also. Alot of the questions just test your problem soving abilities. Putting the formulae into action.

Btw.. formulae are fine for definitions arent they. like the Volt V=W/Q or capacitance C=Q/V. if i just write down the formula with what the letters stand for do I still get full marks

sixdraw

ya you will for most questions but there's a few that it wont work. I know the ampere is one but you'll have to look at the syallabus to be sure

*Angel*

Doctor D-Man wrote:

Btw.. formulae are fine for definitions arent they. like the Volt V=W/Q or capacitance C=Q/V. if i just write down the formula with what the letters stand for do I still get full marks

It depends on the marking scheme, sometimes you need the full definition written out, like sixdraw said you would need the definition for the ampere, sometimes the marks given will tell you.