explain-how-to-determine-the-frequency-f-of-a-wave-what-would-you-need-to-know

Name: explain how to determine the frequency f of a wave what would you need to know
Uploaded: 2020-10-18
Duration: 49 s
Channel: Numerade
Description: Explain how to determine the frequency, $f$, of a wave. What would you need to know?

Question

Explain how to determine the frequency, $f$, of a wave. What would you need to know?

Madi Sousa · Accepted Answer

this problem is very similar to the first one in this chapter of your textbook where we wanted to know an equation to find wavelength. And now we want to know an equation to find frequency. So all we can do is rearrange the equation we had before. We first said that Lambda is equivalent to C over f. So we want F, which is our frequency so f would be equivalent to see over Lambda. Remember that lambda is our wavelength. C is the speed of light and f is our frequency. So using this equation, we have found frequency. And we can use this equation to find the frequency of our sample or during our experiment. So I hope that this made sense and how I hope this. It&#x27;s clear how we got to one equation to the one for frequency.

Rachel Hochberg · Answer

a wave can be characterized by its amplitude and its wavelength. So I&#x27;m going to draw a wave. This is our origin or central line. Our wave goes above and below that origin or central line. The highest point on the wave is called the crust. The lowest point on the wave is called the trowel, or truck. The vertical heights from the origin or the central line to the top of the crest or from the origin or central line down to the trowel or truck is called the amplitude. The wavelength is the distance between two adjacent crests, or between two analogous points. So right here we have a second crust. This distance is our wave length wavelength has a Greek letter symbol that is associated with it, and that is the Lambda. It can also be between two analogous points, meaning two of the same type of points on the waves. So let&#x27;s pick this point right here, so it&#x27;s after it&#x27;s gone through the trowel or trawl. It&#x27;s hitting the central line for the second time, so we go up, we go all the way back down, and then here it is going to cross over after hitting the truck again the central or origin line so this distance would also be classified as a wave length wavelength can be measured in meters Nana meters, micro meters and and sometimes kilometers or kilometers.

Jheremiah Simon · Answer

okay for this problem, we are given the wavelength of a wave to be 650 nano meters, which is about 6.5 times 10 to the negative seven meters. And we need to figure out the frequency of this wave and classify what type of wave it is. So to get the frequency, we will just divide the speed of light by the wavelength. So three times tend to the eight divided by 6.5 times 10 to the negative seven. And you should get something around 4.62 times 10 to the 14 six to tubs tend to be 14 hurt. And if we look up where this wave length and frequency is located, we will see that it is just under the value for red in the visible spectrum. So this is what we would call infrared light. That&#x27;s it. So here is your frequency that you were looking for and it is infrared light

Susanna Weber · Answer

in this question, we can see that See does depend, of course, on the wavelength of the wave and the frequency of the wave. Um, but we want to figure out here is that so? This means that for any Lichtman and magnetic wave, it can have any wavelength. And it could have any frequency as long as they together multiply to be see the speed of light. This is a universal constant. So that means that frequency and wavelength are in versus of each other. And that is what the speed of light depends on.

Kyle Godbey · Answer

question ten is a pretty short one. We want to find the frequency of light with given away flanks here. And uh, maybe she do. You read. It&#x27;s fitting in this problem, I think, and we can just simple use that the frequency is speed of light divided by that wavelength. And if you plug it in, you get that. The frequency is four point two times ten to the fourteen hurts. And that part is simple. And so is the next one. Where the question also asked what? How would we classify this light? Since it&#x27;s wavelength is seven hundred twenty nano meters, that is red visible light. It is near the end of the spectrum, so nearing the infrared, but we can still see seven hundred twenty nine meters.

Kim Trang Nguyen · Answer

question number one. What is great? It&#x27;s Langway length frequency and Emily, too, you see in the screen, and I brought the question. Answer already. Weight is five waiting. This stubborn by wish energy is transmitted way. Lend frequency and enmity, or these are characteristic off the wave. So what is wailing way less the distance between two identical boys on successive ways? All you can say, that&#x27;s it. It&#x27;s done between two. A trace and Crestar to address and stuff Andrea. Frequency frequency doesn&#x27;t mean the number off ways passing through a particular point in one second M V two MMBtu the height or CREss or death off drop offish way or you, can you? The definition. The book. The definition, the book say, is a vertical distance from the midline of wayto that bit off draft that the same meaning. And if you had, If the way has higher M the Greater MD deal, that light is brighter. If the way has a shock. The MD do small M V two That&#x27;s it. The lie. So the light is dim. A fire Zambo I have this way and the waves on the right side. I just I could do it. I did draw me this way. You got the Emnid deal. It&#x27;s greater than the ability of the way on the left side. So this weighs. Had the light is brighter than the way on the left side because the employees to is greater.

Problem

Draw sketches of two waves to show how green ligh…

Problem 2 Easy Difficulty

Explain how to determine the frequency, $f$, of a wave. What would you need to know?

Answer

$\lambda=\frac{c}{f}$
or $\mathrm{f}=\frac{c}{\lambda}$

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$\lambda=\frac{c}{f}$or $\mathrm{f}=\frac{c}{\lambda}$

Living by Chemistry

Stephen P.

Allea C.

Morgan S.

Karli S.

Liquids - Intro

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Angelica M. StacyLiving by Chemistry

Stephen P.

Allea C.

Morgan S.

Karli S.

$\lambda=\frac{c}{f}$
or $\mathrm{f}=\frac{c}{\lambda}$

Angelica M. Stacy

Living by Chemistry