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# Frequency & wavelength question? watch

1. So I dont know what it is about this question but the use of Voltage throws me off, here it is:
http://postimg.org/image/vlaxj7519/
Could someone who knows what they're doing walk me through this please? I'm sure its easy but dont seem to know what to do!
2. From what I can see the voltage stuff doesn't really matter until part c), and even then you don't need to do any calculations.

A) Read the graph and think about the relationship between frequency and time.

B) Think about the relationship between frequency and wavelength.

C) Think about what it means for the frequency to halve and the sound to be larger - this is where your voltage comes in.

Edited out the wrong hints >.<
3. The thing about voltage is just to do with the fact that the oscilloscope receives the information about the wave being measured as variations in voltage. You don't need to worry about it! It's just how the oscilloscope works.

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4. (Original post by SeanFM)
From what I can see the voltage stuff doesn't really matter until part c), and even then you don't need to do any calculations.

A) Read the graph and think about the relationship between frequency and time.

B) Think about the relationship between frequency and wavelength.

C) Think about what it means for the frequency to halve and the sound to be larger - this is where your voltage comes in.

Edited out the wrong hints >.<
But we're not given any time values?
Please just walk me through, my best guess is that ms is it's speed and thus speed = wavelength x frequency but we dont know either, so how?
5. (Original post by Danny.L)
But we're not given any time values?
Please just walk me through, my best guess is that ms is it's speed and thus speed = wavelength x frequency but we dont know either, so how?
You're kind of along the right lines - what you've suggested is important for part b, though for that question you should use the speed of light in that equation rather than the speed of the wave.

You're given a scale for the graph - each 'block' represents 0.5ms.

One complete wave is the time it takes to get from a fixed displacement to the same fixed displacement. The best thing to look at on the graph would be the two peaks - the two high points at (2,4) and (5,4).

So how long does it take to get from one peak to the next? Use the fact that each line is 0.5ms. Hopefully you'll know that frequency is the inverse of time, so use the time you've found and invert it to get the frequency.

6. (Original post by SeanFM)
You're kind of along the right lines - what you've suggested is important for part b, though for that question you should use the speed of light in that equation rather than the speed of the wave.

You're given a scale for the graph - each 'block' represents 0.5ms.

One complete wave is the time it takes to get from a fixed displacement to the same fixed displacement. The best thing to look at on the graph would be the two peaks - the two high points at (2,4) and (5,4).

So how long does it take to get from one peak to the next? Use the fact that each line is 0.5ms. Hopefully you'll know that frequency is the inverse of time, so use the time you've found and invert it to get the frequency.

Okay so it's 1.5ms between each wave. So I guess that means its 3 seconds between each wave? as if each block represents 1 meter per second then 3 blocks = 3 seconds?
Frequency = 1/period
speed = 3*10^8
But this is where I am getting confused. You surely cant just add up the blocks and get time, as it isnt time, I am just incredibly confused by what "0.5ms" actually represents? Like it isn't time as that would just be 1second, it isn't just distance or it'd be 0.5m it is its speed, so each block represents its speed? but then if we tally it up to 3 blocks the wave is at 1.5ms?
My best shot in the dark is that each block is 0.5m as well as representing 1 second. So the displacement = 1.5m.
Really confused, like I have never seen a graph like this before and it has thrown me off, and tbh I am geting incredibly frustrated here, so please put me out of my misery
7. (Original post by Danny.L)
Okay so it's 1.5ms between each wave. So I guess that means its 3 seconds between each wave? as if each block represents 1 meter per second then 3 blocks = 3 seconds?
Frequency = 1/period
speed = 3*10^8
But this is where I am getting confused. You surely cant just add up the blocks and get time, as it isnt time, I am just incredibly confused by what "0.5ms" actually represents? Like it isn't time as that would just be 1second, it isn't just distance or it'd be 0.5m it is its speed, so each block represents its speed? but then if we tally it up to 3 blocks the wave is at 1.5ms?
My best shot in the dark is that each block is 0.5m as well as representing 1 second. So the displacement = 1.5m.
Really confused, like I have never seen a graph like this before and it has thrown me off, and tbh I am geting incredibly frustrated here, so please put me out of my misery
You're right that each wave takes 1.5ms (You don't need to double it or anything)- so in one second you'd get 1/0.0015 of them passing through a fixed point, which gives you the frequency. (The 0.0015 comes from it being 1.5 milliseconds - divide the number by 1000 to give you the number of seconds)

Here, ms stands for milliseconds, and so it is time. m/s is speed.

You've got the frequency and the speed of light - you should be able to find the wavelength.
I think I have made an error: that it is supposed to be the speed of sound.

You're almost there and you're more likely to remember how it's done if/when you see a graph question come up again.
8. (Original post by SeanFM)
You're right that each wave takes 1.5ms (You don't need to double it or anything)- so in one second you'd get 1/0.0015 of them passing through a fixed point, which gives you the frequency,

Here, ms stands for milliseconds, and so it is time. m/s is speed.

You've got the frequency and the speed of light - you should be able to find the wavelength.

You're almost there and you're more likely to remember how it's done if/when you see a graph question come up again.
oh my god!
if it's milliseconds I think I'll be okay!
That's embarassing
9. (Original post by SeanFM)
You're right that each wave takes 1.5ms (You don't need to double it or anything)- so in one second you'd get 1/0.0015 of them passing through a fixed point, which gives you the frequency. (The 0.0015 comes from it being 1.5 milliseconds - divide the number by 1000 to give you the number of seconds)

Here, ms stands for milliseconds, and so it is time. m/s is speed.

You've got the frequency and the speed of light - you should be able to find the wavelength.

You're almost there and you're more likely to remember how it's done if/when you see a graph question come up again.
so 1/0.0015=666.6 (rec)
now for the wavelength we do wavelength = v/f
so (3*10^8)/666.6 (rec) = 450000 m That sounds massive to me, so hesitant to put it
10. (Original post by SeanFM)
You're right that each wave takes 1.5ms (You don't need to double it or anything)- so in one second you'd get 1/0.0015 of them passing through a fixed point, which gives you the frequency. (The 0.0015 comes from it being 1.5 milliseconds - divide the number by 1000 to give you the number of seconds)

Here, ms stands for milliseconds, and so it is time. m/s is speed.

You've got the frequency and the speed of light - you should be able to find the wavelength.

You're almost there and you're more likely to remember how it's done if/when you see a graph question come up again.
Is that right?
Also, could use your help on another thread. This one is much more tricky as I havent learnt it yet!
http://www.thestudentroom.co.uk/show....php?t=3112411
11. (Original post by Danny.L)
so 1/0.0015=666.6 (rec)
now for the wavelength we do wavelength = v/f
so (3*10^8)/666.6 (rec) = 450000 m That sounds massive to me, so hesitant to put it
Oh dear, I'm terribly sorry. I think you're supposed to use the speed of sound. No wonder the wavelength seems so big. Doh!
12. (Original post by SeanFM)
Oh dear, I'm terribly sorry. I think you're supposed to use the speed of sound. No wonder the wavelength seems so big. Doh!
ahah thats 340m/s right?
13. (Original post by Danny.L)
ahah thats 340m/s right?
343.2, so yeah. 2 s.f should be good enough
14. (Original post by SeanFM)
343.2, so yeah. 2 s.f should be good enough
Thanks, btw help would be appreciated on here.
Got myself stuck on another question! D:<
http://www.thestudentroom.co.uk/show....php?t=3112411
15. (Original post by Danny.L)
Thanks, btw help would be appreciated on here.
Got myself stuck on another question! D:<
http://www.thestudentroom.co.uk/show....php?t=3112411
I've had a look but it's not something I've studied before so I don't want to lead you astray. Good luck though

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