Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

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physics aqa a multiple choice question

Awave of frequency hHz travels at 8km-1 through a medium. What is the phase difference between 2 points 2km apart?

i know the answer is pi/2 rad

phase diff = 2pi/wavelength (x1-x2)

Use wavelength = speed / frequency

But i cant get the correct answer

Help much appreciated

boven78hx1

A wave of frequency hHz

Can you clarify, how many Hz?

There is a phase difference of

\displaystyle 2\pi

for every wavelength, this is true.

Thus

\displaystyle \phi = 2\pi \frac{x_2 - x_1}{\lambda}

.

Your question is written out sketchily, but if I assume that you are given

\displaystyle \frac{1}{\lambda} = 8 \mathrm{km}^{-1}

, then:

\displaystyle \phi = \frac{2\pi}{8 \mathrm{km}^{-1}} \times 2\mathrm{km} = \frac{\pi}{2}

If you sketch out the waveform you can see where all the math fits in.

OP

it is 5Hz

Hey again. Morbo killed the bird with a stone earlier; however, I'll kill it a second time with the aid of a diagram after lunch. That way you'll have a conceptual and physical method.

OP

where does 1/wavelength come from

boven78hx1

where does 1/wavelength come from

You gave a piece of data as

8\mathrm{km}^{-1}

, so I assumed it was

1 / \lambda

because that gave the right answer.

I did not realise you meant to write the following data:

\\ f = 5\mathrm{Hz}[br]\\ v = 8 \mathrm{km s}^{-1}

This means

\lambda = v / f = 1.6 \mathrm{km}

.

Now apply

\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{1.6 \mathrm{km}} \times 2 \mathrm{km} = 2\frac{1}{2}\pi

However, since phase difference only means anything for

0 \leq \phi < 2\pi

, we usually add/subtract multiples of

2\pi

until our expression of phase difference is in this range. In this case we subtract

2\pi

to get:

\phi = 2\frac{1}{2}\pi - 2\pi = \frac{\pi}{2}

Coincidentally, this gives the same answer as using the assumption I made earlier.

OP

cheers morbo
great help much appreciated

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