Back
to top
Waves and harmonic frequncy
Watch
Announcements
Page 1 of 1
Go to first unread
Skip to page:
(b) A small speaker emitting a note of 420 Hertz is placed over the open upper end of avertical tube which is full of water. When the water is gradually run out from thebottom of the tube the air column can be heard to resonate at two distinct points. Thefirst is when the water surface is 0.176 metres below the top of the tube, and thesecond is when it is 0.583 metres below the top. Showing all steps in yourcalculations, find
(i) the speed of sound in air and
(ii) the end correction.
I dont know where to start really need help with htis one
(i) the speed of sound in air and
(ii) the end correction.
I dont know where to start really need help with htis one
0
reply
Report
#2
Firstly, let's assume that these are the first two modes of vibration.
For a tube like this (one closed and one open end) that means they look like the first two patterns on this diagram
![Image]()
Now we need to use the relationship that connects Lamda to L, the length of the tube, for these two modes.
These are, for the first one Lamda = 4L and for the second, Lamda = 4L/3 (if you can't remember where that comes from, let me know)
So, we can find the wavelength if we know the length of the tube and the end correction* so for now let's say the end correction is x.
*In practice the antinode of the wave is NOT exactly at the end of the tube, it's usually a couple of cm outside the tube - this is what's mean by end correction, it's the distance of the antinode from the end of the tube
That means that for the first pattern, the length we need to use is not 0.176 but (0.176 + x) and so wavelengh is then 4(0.176 + x)
Now, do the same for the 2nd mode and remember that these two wavelengths must be the same. So, now you have two expressions for wavelength, set them equal to each other, rearrange and solve for x.
Once you've got x, you can subs it back in to find the wavelength, and once you've got that, you can use it with the frequency to find the wavespeed.
If you need more step by step help, let me know.
FWIW I got the end correction to be 0.0275 which gave me a wavelength of 0.814 and a speed of sound of 342 to 3sf.
For a tube like this (one closed and one open end) that means they look like the first two patterns on this diagram
Now we need to use the relationship that connects Lamda to L, the length of the tube, for these two modes.
These are, for the first one Lamda = 4L and for the second, Lamda = 4L/3 (if you can't remember where that comes from, let me know)
So, we can find the wavelength if we know the length of the tube and the end correction* so for now let's say the end correction is x.
*In practice the antinode of the wave is NOT exactly at the end of the tube, it's usually a couple of cm outside the tube - this is what's mean by end correction, it's the distance of the antinode from the end of the tube
That means that for the first pattern, the length we need to use is not 0.176 but (0.176 + x) and so wavelengh is then 4(0.176 + x)
Now, do the same for the 2nd mode and remember that these two wavelengths must be the same. So, now you have two expressions for wavelength, set them equal to each other, rearrange and solve for x.
Once you've got x, you can subs it back in to find the wavelength, and once you've got that, you can use it with the frequency to find the wavespeed.
If you need more step by step help, let me know.
FWIW I got the end correction to be 0.0275 which gave me a wavelength of 0.814 and a speed of sound of 342 to 3sf.
Last edited by phys981; 1 month ago
1
reply
(Original post by phys981)
Firstly, let's assume that these are the first two modes of vibration.
For a tube like this (one closed and one open end) that means they look like the first two patterns on this diagram
![Image]()
Now we need to use the relationship that connects Lamda to L, the length of the tube, for these two modes.
These are, for the first one Lamda = 4L and for the second, Lamda = 4L/3 (if you can't remember where that comes from, let me know)
So, we can find the wavelength if we know the length of the tube and the end correction* so for now let's say the end correction is x.
*In practice the antinode of the wave is NOT exactly at the end of the tube, it's usually a couple of cm outside the tube - this is what's mean by end correction, it's the distance of the antinode from the end of the tube
That means that for the first pattern, the length we need to use is not 0.176 but (0.176 + x) and so wavelengh is then 4(0.176 + x)
Now, do the same for the 2nd mode and remember that these two wavelengths must be the same. So, now you have two expressions for wavelength, set them equal to each other, rearrange and solve for x.
Once you've got x, you can subs it back in to find the wavelength, and once you've got that, you can use it with the frequency to find the wavespeed.
If you need more step by step help, let me know.
FWIW I got the end correction to be 0.0275 which gave me a wavelength of 0.814 and a speed of sound of 342 to 3sf.
Firstly, let's assume that these are the first two modes of vibration.
For a tube like this (one closed and one open end) that means they look like the first two patterns on this diagram
Now we need to use the relationship that connects Lamda to L, the length of the tube, for these two modes.
These are, for the first one Lamda = 4L and for the second, Lamda = 4L/3 (if you can't remember where that comes from, let me know)
So, we can find the wavelength if we know the length of the tube and the end correction* so for now let's say the end correction is x.
*In practice the antinode of the wave is NOT exactly at the end of the tube, it's usually a couple of cm outside the tube - this is what's mean by end correction, it's the distance of the antinode from the end of the tube
That means that for the first pattern, the length we need to use is not 0.176 but (0.176 + x) and so wavelengh is then 4(0.176 + x)
Now, do the same for the 2nd mode and remember that these two wavelengths must be the same. So, now you have two expressions for wavelength, set them equal to each other, rearrange and solve for x.
Once you've got x, you can subs it back in to find the wavelength, and once you've got that, you can use it with the frequency to find the wavespeed.
If you need more step by step help, let me know.
FWIW I got the end correction to be 0.0275 which gave me a wavelength of 0.814 and a speed of sound of 342 to 3sf.
Thansk so much for your reply its really really helped, i couldnt at all figrure it out
0
reply
Report
#4
I think the 2L/n thing you are thinking of is for a wave with either two fixed/closed ends or two open ends.
The patterns are a bit messier when there's one fixed end and one open.
The 420Hz you only need once you get the wavelenth, and that's to work out the speed using the wave equation.
The patterns are a bit messier when there's one fixed end and one open.
The 420Hz you only need once you get the wavelenth, and that's to work out the speed using the wave equation.
0
reply
(Original post by phys981)
I think the 2L/n thing you are thinking of is for a wave with either two fixed/closed ends or two open ends.
The patterns are a bit messier when there's one fixed end and one open.
The 420Hz you only need once you get the wavelenth, and that's to work out the speed using the wave equation.
I think the 2L/n thing you are thinking of is for a wave with either two fixed/closed ends or two open ends.
The patterns are a bit messier when there's one fixed end and one open.
The 420Hz you only need once you get the wavelenth, and that's to work out the speed using the wave equation.
youre a life saver
0
reply
Report
#6
This isn't the neatest diagram but is meant to help you see how the lamda and L are related
![Image]()
0
reply
(Original post by phys981)
This isn't the neatest diagram but is meant to help you see how the lamda and L are related
![Image]()
This isn't the neatest diagram but is meant to help you see how the lamda and L are related
Did you get x = -0.027
what is the units for this?
0
reply
Report
#8
That's a distance, so it's in metres, (like the 0.176 and 0.583 you used to calculate it).
0
reply
(Original post by phys981)
That's a distance, so it's in metres, (like the 0.176 and 0.583 you used to calculate it).
That's a distance, so it's in metres, (like the 0.176 and 0.583 you used to calculate it).
0
reply
Report
#10
(Original post by josephine130801)
but is the value right? does this mean that the antinode is just before the tube ends rather than adfter?
but is the value right? does this mean that the antinode is just before the tube ends rather than adfter?
The antinode is just outside the tube, not inside it. I have difficulty seeing why this happens but know it's true!
Very many questions on this ignore the end effects I think.
0
reply
(Original post by phys981)
Yes, it's right. I kept it as 0.0275 (I got 2x = 0.055)
The antinode is just outside the tube, not inside it. I have difficulty seeing why this happens but know it's true!
Very many questions on this ignore the end effects I think.
Yes, it's right. I kept it as 0.0275 (I got 2x = 0.055)
The antinode is just outside the tube, not inside it. I have difficulty seeing why this happens but know it's true!
Very many questions on this ignore the end effects I think.
0
reply
Report
#12
(Original post by josephine130801)
But my value is a negative, does it mattwer?
But my value is a negative, does it mattwer?
0
reply
(Original post by phys981)
No, it doesn't (it would, but it shouldn't be negative here and the antinode is always outside, not inside). I don't think mine was negative - it might be just a slip when rearranging the equations?
No, it doesn't (it would, but it shouldn't be negative here and the antinode is always outside, not inside). I don't think mine was negative - it might be just a slip when rearranging the equations?
i get:
4(0.176 + x) = ((4(0.583+x))/3) => 0.704 = (2.332 + 4x)/3 - 4x
=>0.704 - 0.777 = 4/3x - 4x => -0.073 = 2.667x
= -0.027 = x
0
reply
Report
#14
4(0.176 + x) = ((4(0.583+x))/3) - this is your first line
Here I cancelled the 4s and multiplied by 3 to get rid of the fraction bit
So 3 ( 0.176 + x ) = 0.583 + x
0.528 + 3x = 0.583 + x
3x - x = 0.583 - 0.528
2x = 0.055
Here I cancelled the 4s and multiplied by 3 to get rid of the fraction bit
So 3 ( 0.176 + x ) = 0.583 + x
0.528 + 3x = 0.583 + x
3x - x = 0.583 - 0.528
2x = 0.055
0
reply
(Original post by phys981)
4(0.176 + x) = ((4(0.583+x))/3) - this is your first line
Here I cancelled the 4s and multiplied by 3 to get rid of the fraction bit
So 3 ( 0.176 + x ) = 0.583 + x
0.528 + 3x = 0.583 + x
3x - x = 0.583 - 0.528
2x = 0.055
4(0.176 + x) = ((4(0.583+x))/3) - this is your first line
Here I cancelled the 4s and multiplied by 3 to get rid of the fraction bit
So 3 ( 0.176 + x ) = 0.583 + x
0.528 + 3x = 0.583 + x
3x - x = 0.583 - 0.528
2x = 0.055
, i got the same final answers as you did, really appreciate your help!!!!!
0
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
