Physics: Discussion, Questions & Potential Solutions

[izam25]

Wasn't that a completely different question?
I meant the one with the two diffraction patterns

Idk if this was the right question but I think I used Ek=p^2/2m, found p from the de Broglie wavelength (which was the same as he wavelength of the x ray)

[jpulvirenti]

think a + cutoff will  be lower

[aussiboi]

Guys what was the answer to the uncertainity principle mc qs pls tell by whole 20/20 depends on it :c

[Richard Feynman 101]

think a + cutoff will  be lower

Defeniately. I think uncertainity was B (cannot really remember only did it a couple of hours ago).

[KiNSKi01]

Yo why for the two speakers question r people using path difference. That’s what I initially thought but then I went with the fact that a standing wave would have formed and the quiet regions would be where nodes exist

[jpulvirenti]

Yo why for the two speakers question r people using path difference. That’s what I initially thought but then I went with the fact that a standing wave would have formed and the quiet regions would be where nodes exist

Same Did you get 1 meter form centre?!

[KiNSKi01]

Lol I said 2 because when it said quiet region I wasn’t sure whether you counted where they were initially standing as the first quiet region. I think the answer is one metre and not two (according to my teachers)

[izam25]

Lol I said 2 because when it said quiet region I wasn’t sure whether you counted where they were initially standing as the first quiet region. I think the answer is one metre and not two (according to my teachers)

Oh shoot that's what I did (I got 1m) but I read this thread and thought I was wrong. Because it's not diffraction right??? I might be completely off though

[S200]

Felt really good to have done it, but I think I did some janky methods to get to things...

What did ppl get for the radius of that zero gravity question?

Oh shoot that's what I did (I got 1m) but I read this thread and thought I was wrong. Because it's not diffraction right??? I might be completely off though

I said 2 as well, but I think 1 is the correct answer...
How does diffraction work in with path difference?

[aussiboi]

the 0 gravity was honestly a troll question.

I guessed that and just used v = root(rg) and solved for r using g = 9.8 idk why it says 0 gravity when g = 9.8

[mzhao]

Lol I said 2 because when it said quiet region I wasn’t sure whether you counted where they were initially standing as the first quiet region. I think the answer is one metre and not two (according to my teachers)

I said 2 as well, but I think 1 is the correct answer...

0.75m is the correct answer.
https://www.vcaa.vic.edu.au/Documents/exams/physics/2016/2016physics-amd-w.pdf
Take a look at Q7 from Detailed Study 6 for an identical question

You can get find solution using either path difference or standing wave theory.

Using standing waves:
Note that adjacent nodes (or adjacent antinodes) are spaced λ/2 apart. Median point is a maximum, so go λ/4 to the right to reach first minimum.
Therefore, the second minimum is at λ/4 + λ/2 = 0.75m to the right.

Using path difference:
P.D. = 3λ/2, half of it is 0.75m (as you move one unit either side, you are changing P.D. by two units)

[Theodric_Ironfist]

FWIW standing waves are special cases of constructive/destructive interference patterns that you generally see during diffraction. Antinodes are where constructive interference occurs, just like the antinodal lines in diffraction interference patterns. Which is why you can get the same answer assuming a standing wave or by using the path difference equation.

[Theodric_Ironfist]

What did people say about the "must the spaceship be an inertial reference frame" question?

The question mentioned that the ship was travelling at a constant speed, not velocity, so the ship may be an inertial reference frame if travelling in a straight line, but it doesn't have to be in the case of travelling in a circular path, which still has constant speed.

[KiNSKi01]

Yeah had the same as you

[Richard Feynman 101]

What did people say about the "must the spaceship be an inertial reference frame" question?

The question mentioned that the ship was travelling at a constant speed, not velocity, so the ship may be an inertial reference frame if travelling in a straight line, but it doesn't have to be in the case of travelling in a circular path, which still has constant speed.

It was not accelerating