A Level: Rearranging Equations into Y = mx + c

[Connor123]

Homework Statement

Hello, I recently started A level physics and have been given a HW to rearrange equations into Y = mx + c. I am struggling on what to do next once the independent variable as X is an inverse of the actual variable ( example below)
This may be a really easy question but id appreciate the help :)

Relevant Equations

c = fλ , rearrange into y = mx + c

c = fλ, where the dependent (y) variable is λ, and the independent variable (x) is f.

λ = c/f

c * 1/f = λ

Are you able to leave 1/f as equavalent to simply 'x', where the gradient would be c, or do you have to ensure 'x'/independent variable is actually f?

 

[PeroK]

Homework Statement: Hello, I recently started A level physics and have been given a HW to rearrange equations into Y = mx + c. I am struggling on what to do next once the independent variable as X is an inverse of the actual variable ( example below)
This may be a really easy question but id appreciate the help :)
Relevant Equations: c = fλ , rearrange into y = mx + c

c = fλ, where the dependent (y) variable is λ, and the independent variable (x) is f.

λ = c/f

c * 1/f = λ

Are you able to leave 1/f as equavalent to simply 'x', where the gradient would be c, or do you have to ensure 'x'/independent variable is actually f?

If I understand the question, you are simply asked to rename $\lambda \rightarrow y$ and $f \rightarrow x$.

 

[symbolipoint]

What you explain is completely obscure. No way to figure which variable is independent and which is independent in your equation c=fλ . Also not clear if you are or not, asking a simple algebra/arithmetic question.

 

[Connor123]

Sorry for lack of explanation, in this case dependent = λ while independent = f

so in the form y = mx + c, it would be 'λ = mf +c'

but what would the gradient be ?

 

[nasu]

The wavelength is inverse proportional to frequency. So, the x variable cannot be frequency.

 

[Connor123]

So simply put i've been given an equation that simply doesn't go into the form y = mx + c?

 

[PeroK]

Sorry for lack of explanation, in this case dependent = λ while independent = f

so in the form y = mx + c, it would be 'λ = mf +c'

but what would the gradient be ?

You have to decide what $c$ is in this case. It can't be both the constant in a linear equation and the speed of the wave. This is a weird question

If I was doing this question, I would look for $\lambda = af + b$, where we use $a, b$ instead of $m, c$.

Why use $m$ and $c$ in the first place? Why use $m$ for gradient? What's special about the letter $m$?

 

[nasu]

If you have a set of values for f and $\lambda$ you can plot $\lambda$ versus 1/f. So "y" will be $\lambda$ and "x" will be 1/f. The slope of the line of best fit will be the speed of the wave, c. There is no b here. Theoretically, the line should go through the origin.

 

[Connor123]

In this case we've been given no values, simply to change it, so as you say I assume the line would have no y - intercept, with the gradient of c. Thanks for all your help :)

 

[nasu]

I know you have no values. I just tried to explain what is the point of this exercise. The constant m (or c) is the slope of the line.

 

[PeroK]

So simply put i've been given an equation that simply doesn't go into the form y = mx + c?

Looks like it.

 

[haruspex]

Are you able to leave 1/f as equavalent to simply 'x', where the gradient would be c,

Yes, that is what you are expected to do.
In your experiment, or whatever, the independent variable may be f, but in graphing you are free to make the x coordinate 1/f.

 

[PeroK]

Yes, that is what you are expected to do.
In your experiment, or whatever, the independent variable may be f, but in graphing you are free to make the x coordinate 1/f.

I don't see that at all. $\frac 1 x$ is not the same as $x$.

 

[haruspex]

I don't see that at all. $\frac 1 x$ is not the same as $x$.

Sure, but "x" is being used in two different ways in post 1. In one usage it is the x coordinate in a graph of the form y=mx+c. In " and the independent variable (x) is f", the OP seems to be using it as equivalent to "the independent variable". I suspect it was not used that way in the actual problem statement.

 

[PeroK]

Sure, but "x" is being used in two different ways in post 1. In one usage it is the x coordinate in a graph of the form y=mx+c. In " and the independent variable (x) is f", the OP seems to be using it as equivalent to "the independent variable". I suspect it was not used that way in the actual problem statement.

I admit I got really confused by the question.