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Phase difference past paper question

Hi,
I was doing an AQA physics past paper (linked below) about phase differences.
For question 6 a ii:
I got 3/4 pi and 7/4 pi
However, the mark scheme says pi and 2pi. Surely for it to be pi and 2pi point X must be at the peak of the wave, right? Or am I misunderstanding it. Thanks!

Paper: http://filestore.aqa.org.uk/subjects/AQA-PHYA2-QP-JUN12.PDF
Mark Scheme: http://filestore.aqa.org.uk/subjects/AQA-PHYA2-W-MS-JUN12.PDF
Original post by gigglej
Hi,
I was doing an AQA physics past paper (linked below) about phase differences.
For question 6 a ii:
I got 3/4 pi and 7/4 pi
However, the mark scheme says pi and 2pi. Surely for it to be pi and 2pi point X must be at the peak of the wave, right? Or am I misunderstanding it. Thanks!

Paper: http://filestore.aqa.org.uk/subjects/AQA-PHYA2-QP-JUN12.PDF
Mark Scheme: http://filestore.aqa.org.uk/subjects/AQA-PHYA2-W-MS-JUN12.PDF


If it was a moving wave you would be correct, however it is a stationary wave so the phase difference can only ever be nπn\pi where n is an integer.
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gigglej
OP
Original post by natninja
If it was a moving wave you would be correct, however it is a stationary wave so the phase difference can only ever be nπn\pi where n is an integer.


Ahh, that makes a lot of sense now! Thank you!!
Original post by gigglej
Ahh, that makes a lot of sense now! Thank you!!


It's pretty simplified but it works

A fuller version would be this:

The equation of a travelling wave is A(x,t)=A0cos(kx−ωt)A(x,t) = A_0cos(kx-\omega t)

where k is the wavevector and omega is the angular frequency. A stationary wave is made up of two identical travelling waves in opposite directions and therefore has an opposite k vector. The superposition of these two waves is given by:

A(x,t)=A0(cos(kx−ωt)+cos(−kx−ωt))A(x,t) = A_0(cos(kx-\omega t) + cos(-kx - \omega t))

since cos(x)=eix+e−ix2cos(x) = \frac{e^{ix}+e^{-ix}}{2}

we can write this as:

A(x,t)=A02(eikx−iωt+e−ikx+iωt+e−ikx−iωt+eikx+iωt)A(x,t) = \frac{A_0}{2}(e^{ikx-i\omega t} + e^{-ikx + i\omega t} + e^{-ikx - i\omega t} + e^{ikx + i\omega t})

factorising out gives us:

A(x,t)=A02[(eikx+e−ikx)(eiωt+e−iωt)]A(x,t) = \frac{A_0}{2}[(e^{ikx} + e^{-ikx})(e^{i\omega t} + e^{-i\omega t})]

converting back to trig functions:

A(x,t)=A0cos(kx)cos(ωt)A(x,t) = A_0cos(kx)cos(\omega t)

This means that particles will either have an amplitude of zero or they will be exactly in phase or exactly out of phase depending on whether cos(kx)cos(kx) is positive or negative.

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