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I know that:

$\frac{1}{T} = f$

For some oscillation or sinusoidal wave. For instance, a period of 2 has a frequency of 1/2

And, in the physics I have taken the formula would be like the one given above for period and frequency. So, I am confused as to the difference between the relation of frequency to wavelength and frequency to period.

For instance, the relation for EM waves is

$f = \frac{c}{\lambda}$ where c is the speed of light.

I get that T and $\lambda$ are different things but I don't get why this leads to the different relations above.

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  • $\begingroup$ As the frequency in time $\:\nu=1/T \:$ gives how many cycles are executed by $E(x_{0},t)$ per unit time $t$ at a specific space point $x_{0}$, so the frequency in space $\:1/\lambda \:$ gives how many cycles are executed by $E(x,t_{0})$ per unit length in space $x$ at a specific time moment $t_{0}$ : Significance of wave number?. $\endgroup$
    – Voulkos
    Commented Feb 21, 2022 at 15:07

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The wavelength is how far the wave propagates in one period.

So it is the period ($T$ or $1/f$) times the velocity of propagation ($c$ for EM waves in a vacuum, but possibly some other $v$ for other waves or other media).

$$\lambda = \left(\frac{1}{f}\right)\left(v\right)$$

This is just the same as the formula you learned for massive objects travelling at constant velocity $$s=vt$$ where $s$ is the distance travelled, $v$ is the velocity, and $t$ is the time interval you're considering.

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The speed of light in a vacuum is constant. However, the speed of light in various media may change depending upon frequency (or equivalently upon wavelength). This phenomenon is known as dispersion. The relationship between frequency and wavelength is known as the dispersion relationship for a given medium or system.

An example of dispersion can be found in an ordinary glass prism. The velocity of light through a medium is related to the index of refraction of that medium. When the speed of light depends upon frequency or wavelength, so does the index of refraction. When light enters glass at an oblique angle, the amount the light is "bent" depends upon the index of refraction of the glass (and also of the medium through which it was traveling before striking the glass). So, the amount that light "bends" when entering glass at an oblique angle depends upon the wavelength or frequency of that light. Thus, when white light strikes a prism, the components of that light spread out, each component being bent by a different amount depending upon wavelength. That is a very visible example of dispersion.

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I know that:
$\frac{1}{T}=f$
For some oscillation or sinusoidal wave. For instance, a period of $2$ has a frequency of $1/2$

I'm not sure I got your problem here. May be part of the problem arises from omitting the units. So it is not $T=2$ and $f=\frac{1}{2}$, but rather it is $T=2 \text{ s}$ and $f=\frac{1}{2}\text{ Hz}$.

I get that $T$ and $\lambda$ are different things but I don't get why this leads to the different relations above.

Yes, $T$ and $\lambda$ are indeed very different things. Most importantly they have different units. $T$ is a time (measurable in seconds), and $\lambda$ is a length (measurable in meters). Therefore it should be no surprise, that their relations to frequency $f$ must be different. $$f=\frac{1}{T}$$ $$f=\frac{c}{\lambda}$$

Examples:
Let the period be $T=2\text{ s}$. Then you get the frequency $$f=\frac{1}{T}=\frac{1}{2\text{ s}}=0.5\text{ s}^{-1}=0.5\text{ Hz}$$ Let the wavelength be $\lambda=2\text{ m}$. Then you get the frequency $$f=\frac{c}{\lambda}=\frac{3\cdot 10^8\text{ m/s}}{2\text{ m}} =1.5\cdot 10^8\text{ s}^{-1}=150\text{ MHz}$$ Notice how the units are part of the calculations.

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  • $\begingroup$ So on something like a sin(x) graph where the period is 2π is the 'wavelength' also 2π because there are not necessarily units? In other words, the x-axis is just units of x, not meters or seconds or otherwise. Or am I thinking about it wrong. $\endgroup$
    – Hydrolox
    Commented Feb 21, 2022 at 15:38
  • $\begingroup$ @Hydrolox The function argument in $\sin(...)$ must always be dimension-less, but not seconds or meters. Hence $\sin(x)$ would be wrong. Correct would be $\sin(\frac{2\pi t}{T})$ or $\sin(\frac{2\pi x}{\lambda})$. $\endgroup$
    – Thomas Fritsch
    Commented Feb 21, 2022 at 15:50
  • $\begingroup$ So calling it period or wavelength then depends on the units you use in the sin function? e.g t and x above $\endgroup$
    – Hydrolox
    Commented Feb 21, 2022 at 16:20
  • $\begingroup$ @Hydrolox yes, correct $\endgroup$
    – Thomas Fritsch
    Commented Feb 21, 2022 at 16:23
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This is the wave equation

$${\frac {\partial ^{2}}{\partial {t}^{2}}}A \left( x,t \right) -{\nu}^{ 2}{\frac {\partial ^{2}}{\partial {x}^{2}}}A \left( x,t \right)=0 $$

you can check that the solution is (with $~\nu=\frac{\omega}{k}~$)

$$A \left( x,t \right) =A_{{0}}\sin \left( \omega\,t+k\,x \right)$$

thus for a fix x value you obtain periodic time function where the period $~T=\frac{2\pi}{\omega}=\frac{1}{f}~[s]$ and for a fic t you obtain periodic function of x with the period $~\lambda=\frac{2\pi}{k}~ [m]~$

with

$$\nu=\frac{\omega}{k}=\frac{2\pi/T}{2\pi/\lambda}=\frac{\lambda}{T}=\lambda\,f$$

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