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Path Difference

I've got as far as calculating the Path Difference (28 cm) and the wavelengths at the particular frequencies stated.

However, I have no idea how to proceed further.

I'm referring the the (b) part.

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Reply 1

Ari Ben Canaan

Additionally, when sound diffracts through a barrier what would be observed ?

Reply 2

int_applicant

Original post by Ari Ben Canaan

I've got as far as calculating the Path Difference (28 cm) and the wavelengths at the particular frequencies stated.

However, I have no idea how to proceed further.

I'm referring the the (b) part.

How is the path diffrerence 28cn....?..you mean 128cm? I got 128cm, using pythagoras( spelling is not right, but you get the gist) theorem........

Reply 3

int_applicant

Original post by Ari Ben Canaan

Additionally, when sound diffracts through a barrier what would be observed ?

I suppose when a wave diffracts you will have regions of constructive interference and regions of destructive interference along the wavefronts....so hence we can predicts that you will here difference sounds of varying pitch or loudness...basically high and low pitch sounds... that will fade out with increasing distance

Reply 4

int_applicant

Original post by Ari Ben Canaan

I've got as far as calculating the Path Difference (28 cm) and the wavelengths at the particular frequencies stated.

However, I have no idea how to proceed further.

I'm referring the the (b) part.

sorry, you are right...then you subtracting it from 100 to get 28cm......I am kinda stuck just like you..but hold on, I amstill looking at it....

Reply 5

int_applicant

Original post by Ari Ben Canaan

I've got as far as calculating the Path Difference (28 cm) and the wavelengths at the particular frequencies stated.

However, I have no idea how to proceed further.

I'm referring the the (b) part.

Hey mate...the answer is 2 minima....what you do is.. you get the wavelength of each frequency at 1 and at 4.....then you know the distance between two minima points is lambda/2.......that should do it......find the distance between minima points up until the wavelength of the freq of 4.....

Hope this helps...let me know if you have further questions...

Reply 6

Stonebridge

Original post by Ari Ben Canaan

I've got as far as calculating the Path Difference (28 cm) and the wavelengths at the particular frequencies stated.

However, I have no idea how to proceed further.

I'm referring the the (b) part.

Yes, the path difference is 28cm (128-100)

One way to think about this is to calculate the wavelength of the sound at 1000Hz. This comes out as 33cm.
When increased to 4000Hz the wavelength will be 8.25cm.
You know you will get a maximum when the 28cm is a whole number of waves.
So we start with a wavelength of 33cm and gradually decrease it to 8.25cm
So as you increase the frequency, and decrease the wavelength,
the first maximum will come at wavelength = 28cm. (One whole wavelength difference)
the next maximum will be at 14cm (2 wavelengths)
the next at 7cm (three wavelengths and a little over the 4000Hz for this)
In between these you will get the minima when there is 1½ and 2½ waves.
That would be at wavelength 21cm (half way between 28 and 14) and 10.5cm (half way between 14 and 7)

Reply 7

int_applicant

Original post by Stonebridge

Yes, the path difference is 28cm (128-100)

One way to think about this is to calculate the wavelength of the sound at 1000Hz. This comes out as 33cm.
When increased to 4000Hz the wavelength will be 8.25cm.
You know you will get a maximum when the 28cm is a whole number of waves.So we start with a wavelength of 33cm and gradually decrease it to 8.25cm
So as you increase the frequency, and decrease the wavelength,
the first maximum will come at wavelength = 28cm. (One whole wavelength difference)
the next maximum will be at 14cm (2 wavelengths)
the next at 7cm (three wavelengths and a little over the 4000Hz for this)
In between these you will get the minima when there is 1½ and 2½ waves.
That would be at wavelength 21cm (half way between 28 and 14) and 10.5cm (half way between 14 and 7)

what exactly does this statement mean?...... 28cm is just the difference in the distance travelled by both waves to reach point M

This is not accurate. The first maxima would be at wavelenth 33/4cm

I am not sure I understand your reasoning of this question

Reply 8

Stonebridge

Original post by int_applicant

I am not sure I understand your reasoning of this question

Does this help?
I've done an exact calculation of the wavelengths (and frequencies) of the two minima, those in my first post were (very) rough guesses.