waves help
its question 26aii on this
http://pmt.physicsandmathstutor.com/download/Physics/A-level/Past-Papers/OCR-A/AS-Paper-1/June%202016%20QP%20-%20Paper%201%20OCR%20Physics%20(A)%20AS-Level.pdf
the answers are on here
http://pmt.physicsandmathstutor.com/download/Physics/A-level/Past-Papers/OCR-A/AS-Paper-1/June%202016%20MS%20-%20Paper%201%20OCR%20Physics%20(A)%20AS-Level.pdf
i dont understand why theyve divided 2/40 and then multiplied it by 60 like why does the 60 come into it and why is 2/40?
http://pmt.physicsandmathstutor.com/download/Physics/A-level/Past-Papers/OCR-A/AS-Paper-1/June%202016%20QP%20-%20Paper%201%20OCR%20Physics%20(A)%20AS-Level.pdf
the answers are on here
http://pmt.physicsandmathstutor.com/download/Physics/A-level/Past-Papers/OCR-A/AS-Paper-1/June%202016%20MS%20-%20Paper%201%20OCR%20Physics%20(A)%20AS-Level.pdf
i dont understand why theyve divided 2/40 and then multiplied it by 60 like why does the 60 come into it and why is 2/40?
You have calculated the speed, with the equation:
v = f * lambda
We want the absolute uncertainty.
Since this is an A-Level Q and they're probably working on the assumption you've done practical physics, it's safe to assume you know about the propagation of errors. You should at least, I hope... in any case, I'll just run you through it with an arbitrary example, rather than using this one, just so that it doesn't just give you the answer.
Consider the function f(x) = 2x^2 + 4x.
It's an arbitrary function describing some variable f that depends on x. It might be tension and angle, torsion and density, whatever it might be. We want to know the uncertainty in f for some value of x, assuming we know the uncertainty in x.
Let's throw the derivative into the mix.
df(x)/dx = 4x + 4.
Can you rearrange that? It looks like you can.
df(x) = dx * (4x + 4)
That looks pretty satisfying, wouldn't you say?
If we make an approximation to this, i.e. for some small Df (D implies the triangle. I can't draw triangles in text.) there exists a Dx...
Df = Dx * (4x + 4)
This Df exists for any value x that we find in our domain. Now we can suddenly calculate the uncertainty in f for our value x.
Multivariate Functions/etc
Not really "multivariate," I just like slapping fancy words around. For the speed equation you have,
v = f * lambda
We can treat f as a constant. It has no uncertainty, it's as good as the number 4 in my example above.
Why is lambda not constant?
It has an uncertainty, and it can vary over some range. That's the simple way of saying it.
The factor of 2
Write out your equation for the speed, v.
The uncertainty in lambda is not 2 cm. Lambda is equal to double the value of whatever you've been given. You have been given the half-wavelength, which I will call x.
Lambda = 2x
D(Lambda) = 2 * D(x)
I'll leave the rest to you. God speed.
v = f * lambda
We want the absolute uncertainty.
Since this is an A-Level Q and they're probably working on the assumption you've done practical physics, it's safe to assume you know about the propagation of errors. You should at least, I hope... in any case, I'll just run you through it with an arbitrary example, rather than using this one, just so that it doesn't just give you the answer.
Consider the function f(x) = 2x^2 + 4x.
It's an arbitrary function describing some variable f that depends on x. It might be tension and angle, torsion and density, whatever it might be. We want to know the uncertainty in f for some value of x, assuming we know the uncertainty in x.
Let's throw the derivative into the mix.
df(x)/dx = 4x + 4.
Can you rearrange that? It looks like you can.
df(x) = dx * (4x + 4)
That looks pretty satisfying, wouldn't you say?
If we make an approximation to this, i.e. for some small Df (D implies the triangle. I can't draw triangles in text.) there exists a Dx...
Df = Dx * (4x + 4)
This Df exists for any value x that we find in our domain. Now we can suddenly calculate the uncertainty in f for our value x.
Multivariate Functions/etc
Not really "multivariate," I just like slapping fancy words around. For the speed equation you have,
v = f * lambda
We can treat f as a constant. It has no uncertainty, it's as good as the number 4 in my example above.
Why is lambda not constant?
It has an uncertainty, and it can vary over some range. That's the simple way of saying it.
The factor of 2
Write out your equation for the speed, v.
The uncertainty in lambda is not 2 cm. Lambda is equal to double the value of whatever you've been given. You have been given the half-wavelength, which I will call x.
Lambda = 2x
D(Lambda) = 2 * D(x)
I'll leave the rest to you. God speed.
im really sorry but i still dont understand lol why did they divide 2/40 and then multiply is by 60
Original post by Callicious
You have calculated the speed, with the equation:
v = f * lambda
We want the absolute uncertainty.
Since this is an A-Level Q and they're probably working on the assumption you've done practical physics, it's safe to assume you know about the propagation of errors. You should at least, I hope... in any case, I'll just run you through it with an arbitrary example, rather than using this one, just so that it doesn't just give you the answer.
Consider the function f(x) = 2x^2 + 4x.
It's an arbitrary function describing some variable f that depends on x. It might be tension and angle, torsion and density, whatever it might be. We want to know the uncertainty in f for some value of x, assuming we know the uncertainty in x.
Let's throw the derivative into the mix.
df(x)/dx = 4x + 4.
Can you rearrange that? It looks like you can.
df(x) = dx * (4x + 4)
That looks pretty satisfying, wouldn't you say?
If we make an approximation to this, i.e. for some small Df (D implies the triangle. I can't draw triangles in text.) there exists a Dx...
Df = Dx * (4x + 4)
This Df exists for any value x that we find in our domain. Now we can suddenly calculate the uncertainty in f for our value x.
Multivariate Functions/etc
Not really "multivariate," I just like slapping fancy words around. For the speed equation you have,
v = f * lambda
We can treat f as a constant. It has no uncertainty, it's as good as the number 4 in my example above.
Why is lambda not constant?
It has an uncertainty, and it can vary over some range. That's the simple way of saying it.
The factor of 2
Write out your equation for the speed, v.
The uncertainty in lambda is not 2 cm. Lambda is equal to double the value of whatever you've been given. You have been given the half-wavelength, which I will call x.
Lambda = 2x
D(Lambda) = 2 * D(x)
I'll leave the rest to you. God speed.
v = f * lambda
We want the absolute uncertainty.
Since this is an A-Level Q and they're probably working on the assumption you've done practical physics, it's safe to assume you know about the propagation of errors. You should at least, I hope... in any case, I'll just run you through it with an arbitrary example, rather than using this one, just so that it doesn't just give you the answer.
Consider the function f(x) = 2x^2 + 4x.
It's an arbitrary function describing some variable f that depends on x. It might be tension and angle, torsion and density, whatever it might be. We want to know the uncertainty in f for some value of x, assuming we know the uncertainty in x.
Let's throw the derivative into the mix.
df(x)/dx = 4x + 4.
Can you rearrange that? It looks like you can.
df(x) = dx * (4x + 4)
That looks pretty satisfying, wouldn't you say?
If we make an approximation to this, i.e. for some small Df (D implies the triangle. I can't draw triangles in text.) there exists a Dx...
Df = Dx * (4x + 4)
This Df exists for any value x that we find in our domain. Now we can suddenly calculate the uncertainty in f for our value x.
Multivariate Functions/etc
Not really "multivariate," I just like slapping fancy words around. For the speed equation you have,
v = f * lambda
We can treat f as a constant. It has no uncertainty, it's as good as the number 4 in my example above.
Why is lambda not constant?
It has an uncertainty, and it can vary over some range. That's the simple way of saying it.
The factor of 2
Write out your equation for the speed, v.
The uncertainty in lambda is not 2 cm. Lambda is equal to double the value of whatever you've been given. You have been given the half-wavelength, which I will call x.
Lambda = 2x
D(Lambda) = 2 * D(x)
I'll leave the rest to you. God speed.
Original post by Zebragirl
im really sorry but i still dont understand lol why did they divide 2/40 and then multiply is by 60
The fractional uncertainty in the speed, Dv / v, equals the fractional uncertainty in the wavelength, D(lambda) / lambda.
You can derive that relationship if you use the propagation of errors and solve for the ratio.
They just skip the propagation step and use the ratio they teach you in the practical segment of your syllabus, presumably.
omg thank you
Original post by Callicious
The fractional uncertainty in the speed, Dv / v, equals the fractional uncertainty in the wavelength, D(lambda) / lambda.
You can derive that relationship if you use the propagation of errors and solve for the ratio.
They just skip the propagation step and use the ratio they teach you in the practical segment of your syllabus, presumably.
You can derive that relationship if you use the propagation of errors and solve for the ratio.
They just skip the propagation step and use the ratio they teach you in the practical segment of your syllabus, presumably.
Original post by Zebragirl
omg thank you
Just a warning, in case that was taken as a fact, it's just specific for this case. If the frequency has an uncertainty, that's not true, and then error propagation really does come in handy.
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