update relation

Author: eahussein

resources/links/.ipynb_checkpoints/relation between freq and the others-checkpoint.md

@@ -12,12 +12,9 @@ Reference:
     <img width="33%" src="stringWave.png"> 
 </p>
 
-- **Frequency** $f$ describes how often a wave arrives at a certain point, and it gets measured by Hertz (Hz) = $\frac{num of cycles}{t (sec/min/day)}$ , which describes as  the number of complete oscillations per unit time.
+- **frequency** $f$ describes how often a wave arrives at a certain point, and it gets measured by Hertz (Hz) = $\frac{num of cycles}{t (sec/min/day)}$, which describes the number of complete oscillations per unit time.
 	- $f = 1/t$, $t$ is the period a wave takes to complete one Oscillation.
-	- So as the number of cycles increases, the time it takes to complete one cycle decreases, so they are _Inversely proportional_
-- **Wavelength** $\lambda$ is the measure of two similar nearest points in the wave, and it gets measured by the distance.
+	- So, as the number of cycles increases, the time it takes to complete one cycle decreases, so they are _Inversely proportional_
+- **wavelength** $\lambda$ is the measure of two similar nearest points in a wave, measured by the distance.
 - **Wave speed** $v$ = $f \times \lambda$
-	- Since the speed of light is constant, then the longer the wavelength $\lambda$ the smaller is the frequency $f$. However it does change through different mediums. (not sure how will this work in space, will need to ask Prof). I would assume that in space the speed of light does not get effected.
--  Frequencies]] does not change through a ***medium***, but the wavelength does, but why ?
-	-  Giving this equation $v = f \times \lambda$, it shows that the $v \propto \lambda$.
-	- Therefore, going through a thicker medium, the speed will be slower and the wavelength will be shorter leaving the frequency unchanged. meaning Both $v$ and $\lambda$ will decrease leaving the frequency **_unchanged_**.
+	- Giving that speed of light is constant, the longer the wavelength $\lambda$, the smaller the frequency $f$. 

resources/links/relation between freq and the others.md

@@ -12,12 +12,9 @@ Reference:
     <img width="33%" src="stringWave.png"> 
 </p>
 
-- **Frequency** $f$ describes how often a wave arrives at a certain point, and it gets measured by Hertz (Hz) = $\frac{num of cycles}{t (sec/min/day)}$ , which describes as  the number of complete oscillations per unit time.
+- **frequency** $f$ describes how often a wave arrives at a certain point, and it gets measured by Hertz (Hz) = $\frac{num of cycles}{t (sec/min/day)}$, which describes the number of complete oscillations per unit time.
 	- $f = 1/t$, $t$ is the period a wave takes to complete one Oscillation.
-	- So as the number of cycles increases, the time it takes to complete one cycle decreases, so they are _Inversely proportional_
-- **Wavelength** $\lambda$ is the measure of two similar nearest points in the wave, and it gets measured by the distance.
+	- So, as the number of cycles increases, the time it takes to complete one cycle decreases, so they are _Inversely proportional_
+- **wavelength** $\lambda$ is the measure of two similar nearest points in a wave, measured by the distance.
 - **Wave speed** $v$ = $f \times \lambda$
-	- Since the speed of light is constant, then the longer the wavelength $\lambda$ the smaller is the frequency $f$. However it does change through different mediums. (not sure how will this work in space, will need to ask Prof). I would assume that in space the speed of light does not get effected.
--  Frequencies]] does not change through a ***medium***, but the wavelength does, but why ?
-	-  Giving this equation $v = f \times \lambda$, it shows that the $v \propto \lambda$.
-	- Therefore, going through a thicker medium, the speed will be slower and the wavelength will be shorter leaving the frequency unchanged. meaning Both $v$ and $\lambda$ will decrease leaving the frequency **_unchanged_**.
+	- Giving that speed of light is constant, the longer the wavelength $\lambda$, the smaller the frequency $f$.