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I can calculate the speed, frequency or wavelength of a wave using the wave equation.

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This content is © Oak National Academy Limited (2023), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

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Key learning points

The speed of wave can be caluated using wave speed = frequency × wavelength.
The speed of a wave can be caluated using wave speed = distance ÷ time.
The choice of equation to use to caluate wave speed depends on the information provided in the question.
Prefixes can be used with units to show very large or very small values.
The frequency of a wave does not change as it crosses a boundary from one wave medium into another.

Common misconception

In, v = f × λ, speed is dependent on frequency and wavelength, rather then the medium.

Focus on speed being property of a medium and equation not being as an equation with three interdependent variables.

Keywords

Wave equation - wave speed = frequency × wavelength (v = f × λ)
Wave speed - The speed of a wave through a medium. Given by the wave equation: wave speed = frequency × wavelength (v = f × λ)
Frequency - The number of waves produced each second. Frequency is measured in Hertz (Hz)
Wavelength - The distance between one wave crest (or trough) and the next is the wavelength of the wave.
Prefix - A prefix is added to units to represent large or small values , for example 1 km = 1000 m and 1 mm = 0.001 m

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Hello there, I'm Mr. Forbes, and I'll be leading you through this lesson from the Measuring Waves unit.

The lesson's all about wave speed equations.

There are a pair of equations that we can use to calculate the speed of a wave based on properties like frequency, wavelength, distance, or time.

And this lesson will allow us to see which equation to use depending on which information we're given.

By the end of this lesson you're gonna be able to calculate the speed of a wave based on wavelength and frequency or distance and time.

To be able to do that successfully, you'll need a calculator, so make sure you have one before you move on in the lesson.

Once you're ready, we'll start.

So let's go.

You're going to need to understand a few keywords to be able to successfully complete this lesson, and they're listed here, wave speed, frequency, wavelength, prefix and the wave equation.

I'll show you some definitions of those keywords and you can refer back to them at points during the video if you'd like to.

And if you're not clear about exactly what they mean.

Here they are.

The lesson's in three parts.

The first part is going to be about selecting the correct equation based upon what information we've been provided with.

So, we might be provided with frequency and wavelength, in which case we'll choose one equation.

Or, we might be provided with distance and time, and we'll select the other equation to use.

The second part goes a little bit further and looks at waves that have got values with prefixes, so distances, kilometres or frequencies in megahertz.

And the final part, we'll look at how to calculate wavelength and frequency from the wave equation so that we can rearrange it and find those values.

We're gonna move straight onto the first part, selecting and using wave equations Before starting this lesson, you should have already seen the wave equation.

If you haven't, I'll just go through it now to check that you've got a basic understanding of it.

The wave speed equation is this.

Wave speed equals frequency times wave length.

So basically, if you know the frequency and the wave length from a wave, you'd be able to calculate wave speed just by multiplying those two values together.

We often write this as a set of symbols to save a little bit of effort.

And the symbols we use are V, F and that unusual symbol that are Lambda.

V is the wave speed measured in metres per second.

F is the frequency measured in hertz or how many waves per second.

And lambda is the wavelength measured in metres.

I'm gonna show you an example of the wave speed equation and how it's used.

I have a basic question here about waves.

What is the speed of a wave with a wavelength 1.

5 metres and a frequency of 6.

0 hertz? The process I like to use to solve these is to always follow the same pattern.

I'd like to write down the equation first, so I'll do that.

Wave speed equals frequency times wavelength.

And then, I look at the question and try and identify those variables clearly.

I've got a frequency of 6.

0 hertz in the question and a wavelength of 1.

5 metres.

So, I write those down instead of the words just like this.

Wave speed equals 6.

0 times 1.

5, and a final stage is simply to multiply those two together to get a wave speed.

Wave speed is 9.

0 metres per second.

I've got a second question, and I'd like you to have a go at this, have you to read through with me.

What is the speed of a wave with a wavelength of 0.

25 metres and a frequency of 80 hertz? So, I'd like you to follow the same procedure as I did and come up with a wave speed.

So, pause the video, solve there question and then restart.

Welcome back.

Hopefully you followed this procedure.

You wrote down the wave speed equation.

Wave speed is frequency times wavelength.

You then substituted in the two values, the frequency there is 80 hertz, and the wavelength is 0.

25 metres.

Then you solve it by multiplying those two together, making sure that you use the unit at end there.

So 80 times not 0.

25 is 20 metres per second.

You should also know a different way of calculating the speed of anything, and that is from the distance and time information.

So, you should have seen this equation before in the past.

Distance travelled equals speed times time.

In symbols, we write that as S equals V times T where S is the distance travelled in metres.

V is the wave speed in metres per second, and T is the time in seconds.

I'm gonna lead you through an example of using that equation very similar to the earlier one.

What is the speed of a wave if it travels 40 metres in 5.

0 seconds? So the stage is very similar.

We write down the equation distance travelled equals speed times time.

And we look for that information and substitute it in.

So, the distance travelled is 40 metres, and the time is five seconds.

So write those two values in, and I come up with a relationship like this.

I'm gonna divide both sides by 5.

0 in order to get speed on its own on one side.

So, it becomes that, 40 divided by 5.

0 equals the speed.

And then, I'm gonna solve it and get speed equals 8.

0 metres per second.

As before, I've got an example for you to do.

So, I'd like you to answer this question, what's the speed of a wave if it travels 270 metres in 45 seconds? Pause the video and go through the same process I did, and then restart once you've got an answer.

Excellent.

Okay, let's have a look.

You should have written down the distance travelled equals speed times time equation, and then looked for the values in the question.

We have a distance of 270 metres and a time of 45 seconds.

So, we write those two values in.

We rearrange it slightly, moving the 45 over the other side by dividing both sides by 45.

And we get this, 270 divided by 45 as speed.

And the final stage is to do that calculation.

Speed equals 6.

0 metres per second.

So, well done if you got that.

Right.

So, you've seen we've got two different equations we can use to calculate the wave speed.

We've got a version we can use if we have frequency and wavelength, and that is wave speed equals frequency times wavelength.

And we have a second equation we can use if we've got distance travelled and time, and that is distance travelled equals speed times time.

The next couple of slides I'm gonna show you how to select which of those equations to use depending on what information you've seen.

Okay, now as we know how to select the wave equations, we're gonna solve a few questions with them, identifying the data, selecting the equation, and doing the calculation.

So, here's a first example.

An earthquake produces waves with a frequency of 0.

5 hertz.

The waves are detected by a seismograph 2,800 metres away from the source.

And it's eight seconds later that those waves are detected.

We've been asked to calculate the wave speed.

So, the process I go through is identifying the data and writing it down separately and clearly, so I can use it in the calculations.

So, first of all we'll look through, and I can identify the frequency in that question.

So, I'll write it separately over there on the left.

Frequency is not 0.

5 hertz.

I'll look through again and try and identify more data and I can see a time.

I've got time of eight seconds, so I'll write that down as well.

And a third piece of information I think.

There we go.

We've got 2,800 metres, so the distance it's travelled is 2,800 metres.

So, I can then look quite easily at that and say, \"Well I've got distance and time, so I'm gonna select the equation that uses those.

Distance travelled equals speed times time.\" Putting the values in is fairly simple then.

I just write them in as a replacements for the words.

Got distance travelled 2,800, still have the speed there, and then I've got a time of 8.

0 seconds.

I'll adjust that equation to get a value for the speed.

The speed is 2,800 divided by 8.

I divided both sides of the equation by eight there.

And finally I'll solve that and get a speed of 350.

I'm not quite finished though.

I haven't got a unit for it.

So, look back and try and decide what the unit for speed is.

Well, it's got metres and seconds, so I've got a unit of metres per second here.

Here's the second example.

We're gonna do the same process again, identifying the data in the question clearly and then solving the the question.

A ripple tank is used to make waves with a frequency of 2.

5 hertz.

The waves have a wave length of 3.

0 centimetres.

and reach the end of the tray 4.

0 seconds.

Calculate the wave speed.

So as before, I'll write down the values I can find in the question clearly, so it can use them to select the equation.

I've got a frequency of 2.

0 hertz, so I'll write that down.

I have a time of 4.

0 seconds.

And I've got a wavelength of 3.

0 centimetres, which is 0.

03 metres, and I'll use that bit later.

So, looking at that data, frequency and wavelength are the two things I think I can use to calculate the wave speed.

So, write down the appropriate equation.

Wave speed is frequency times wavelength.

I put the two values in.

5 hertz was the frequency 0.

03 metres was the wavelength.

So, I got a wave speed for that.

Do the calculation.

Wave speed equals 0.

075, and again I've got to write down the units.

It was metres and seconds I used in the calculation, so the wave speed is metres per second.

Now, it's time for a check.

I'd like you to try and solve this one, selecting the equation and coming up with a solution.

A wave in a spring has a wavelength of 0.

25 metres and a frequency of 1.

4 hertz.

It takes 5.

0 seconds to travel the full length of the spring.

What's the wave speed? So, I'd like you to pause the video, and then solve it, and then restart, and we'll go through the solution.

Welcome back.

Hopefully you selected the correct answer which was not 0.

35 metres per second.

I'm gonna lead you through how we did that.

Looking at the question, you've got three pieces of information, but only two of them are needed to be able to solve it.

We can use wave speed equals frequency times wavelength, because the wavelength has been given and the frequency's been given.

So, substituting these two values, we have a frequency of 1.

4 hertz and a wavelength of 0.

25 metres.

Multiplying those two together gives us the wave speed of 0.

35 metres per second.

So well done if you've got that.

Here's the second example.

A wave with a frequency of 1.

5 hertz travels along the rope of length 5.

4 metres taking 3.

0 seconds to travel from one end to the other.

What is the wave speed? Again, I'd like you to pause the video and try and solve it by selecting the correct equation and using the values to get an answer.

Restart as soon as you've got them.

Welcome back.

Hopefully you've got the right answer which is 1.

8 metres per second.

You should have been able to calculate that based upon the data in the question and selecting right equation.

The equation we needed to use here was distance travelled equals speed times time, because even though you've got frequency provided in the question, you don't have a wavelength, so you can't use the other equation.

Looking at the values in the question, had a distance travelled of 5.

4 metres, and we had a time of 3.

0 seconds.

So, we could put the values in like that, adjust the equation to get speed.

So, we've got 5.

4 divided by 3.

0 is equal to the speed, giving us a speed of 1.

8 metres per second.

So, well done if you got that.

Okay, it's time for you to do the first task.

And you're gonna calculate the wave speed for three different waves, selecting the most appropriate equation to come to another solution.

So, here are three questions.

A stone thrown into a pond produces a wave with a wavelength of 0.

2 metres, which takes 12 seconds to travel a distance of 3.

0 metres.

And you're gonna calculate the wave speed for that.

The second question, during an experiment, four waves each second pass by a point on a spring.

These have a wavelength of 0.

30 metres.

And the third one, a sound wave travels through the metal bar of length 1.

5 metres.

The wave has a frequency of 650 hertz and takes 0.

0003 seconds to pass along the bar.

So, some fairly difficult numbers to use there, but I'd like you to try and solve all three of those.

Pause the video, work through the solutions, and then restart, and I'll try and lead you through the examples again.

Welcome back.

Let's have a look at the solution to the first of those questions.

And that was to calculate the wave speed for this.

A stone thrown in a pond produces waves with a wavelength of 0.

2 metres taking 12 seconds and travelled a distance of 3.

9 metres.

So, I identify the data in that question clearly over here on the left.

I've got an unknown frequency.

There's there's no information about frequency in that question at all.

So I'll note that.

So I know I'm not gonna be able to use the equation that involves frequency.

I've got a wavelength of 0.

2 metres identified there.

I've got a time of 12 seconds, and I've got a distance of 3.

0 metres.

So, selecting the correct equation would be this, distance travelled equals speed times time.

I'll put the values in from the list.

I've got a distance travelled of 3.

0 and a time of 12.

A rearrange slightly, 3.

0 divided by 12 is the speed, and then I solve that to get a speed of not 0.

25 metres per second.

So, congratulations if you got that first one.

Here's the second of the examples.

During an experiment, four waves each second pass by a point on a spring.

These have a wavelength of 0.

30 metres.

So again, I'm gonna identify the data.

I've got a frequency of four hertz there, four waves each second, four hertz.

I've got a wavelength of 0.

3 metres, and so I can select the most appropriate equation based upon just those two pieces of information, frequency in wavelength.

I've also got a time here of one second, but I don't think I'm gonna be needing it.

I've got an unknown distance, so I'm not gonna be able to use the distance equals speed times time equation.

So, writing the correct equation, wave speed equals frequency times wavelength.

Selecting the two values I've identified.

Got frequency of four, a wavelength of 0.

30.

And I can solve that quite easily, multiplying them to give a wave speed of 1.

2 metres per second.

This is the third example.

Sound wave travels through a metal bar of length, 1.

5 metres.

It's got frequency of 650 hertz, and it takes 0.

003 seconds to travel along the bar.

So again, identify the data.

Frequency of 650 hertz.

Unknown wavelengths, so I'm probably not gonna be able to use that wavelength equation.

Got a time of 0.

0003 seconds, and I've got a distance of 1.

5 metres.

So, it's fairly clear I'm gonna be using a distance and time equation.

Distance travelled equals speed times time.

Putting the values in 1.

5 equals speed times 0.

003, and then rearranging 1.

5 divided by 0.

0003 equals speed.

Give me a final answer.

Speed of 5,000 metres per second.

So, the speed of sound through that metal bar, 5,000 metres per second.

Well done if you've got that one.

They were fairly tricky values.

Okay, we're moving on to the second part of the lesson, which is about prefixes for wave quantities.

Prefixes are just things we use in physics to show large or small numbers more easily.

And you'll have seen quite a lot of examples of prefixes before.

So, things like instead of writing 4,000 metres, you might just write down four kilometres.

And the K in that, in the four kilometres is standing for 1,000, and that's a prefix, a standard prefix we use in physics.

We also use it for kilogramme, where it's 1,000 grammes, kilonewton, which would be 1,000 Newtons.

We also use prefixes for small numbers.

One of the most common ones you'll have seen will be milli.

So for millimetres we can write that, and that's 0.

002 metres, just written as two millimetres.

Makes it much, much easier to write down.

Now waves can have a very wide range of wavelengths and a very wide range of frequencies.

And so, we use prefixes quite a lot to describe them.

I've got a set of waves here, radio waves, waves in spring, ripple in water, some infrared radiation and visible light, the most common electromagnetic radiation we discuss.

And there are prefixes we use for those quite commonly, and they're here.

You can see there's quite a lot of prefixes there.

Some of them you'll be familiar with already.

We use K kilo for 1,000, C, centi for hundredth, m, milli for a thousandth, and then we've got two newish ones.

They're micro and nano, and I'll give you some examples of those in use.

Radio waves have wavelengths of several hundred or a few thousand metres.

So a radio wave example, there's got a wave of 2000 metres or two kilometres.

When you've done experiments with waves in springs, you'll see they're much shorter than a metre.

You might have quite short wavelengths.

So a wavelength of seven centimetres or 0.

07 metres.

The ripples in the ripple tank experiments you might have done might only be a few millimetres from peak to peak.

So a wavelength of three millimetres, which is a simpler way of writing 0.

03 metres.

And then, the more complex examples, infrared radiation has a very short wavelength.

It's measured in micrometres, and a micrometre is a millionth of a metre.

And so, 16 micrometres is 16 millionths of a metre or 0.

000016 metres.

So, you can see quite easily that 16 micrometres is much easier to write.

And that's a bit clearer again for nanometers, which is a billionth of a metre.

We have a nanometers used to measure the wavelength of visible light.

635 nanometers is 0.

000000635 metres, very small distance indeed.

And it's far easier to write down nanometers than to write down all those zeros.

I quite commonly mess that up and miss out a zero here and there, and that gives me the wrong answer.

So, quick check of that.

I'd like you to decide which is the best way of writing out, 0.

000005 metres using a prefix.

I'd like you to pause, select, and then restart the video.

Okay, the correct answer was five micrometres.

That is five millionths of a metre.

Five micrometres is the best way of doing it.

Five millimetres and five nanometers of both incorrect.

So, well done if you've got that.

Another quick check for you here.

I've got a set of wavelengths and a set of wavelengths using prefixes, which is a an easier way of writing them down.

And I'd like you to draw lines that connect the wavelengths to the same value using the prefix.

I'd like you to pause the video, draw the four lines, and then restart and we'll check your answers.

Welcome back.

Here's the examples.

Let's have a look.

We've got not 0.

05 metres, and that is five millimetres.

It's 5000th of a metre.

And the second example, 0.

05 metres, that's probably the easiest of them there maybe.

That's five centimetres, and then I've got 5,000 metres, that's five kilometres.

That only leaves one, and that 0.

000050 metres is 50 micrometres, that top one there.

Brilliant, if you've got all of those problems. Waste can also have a wide range of frequencies, so we use prefixes to record those as well.

Here's three example waves, sound waves, radio waves and infrared radiation and the prefixes we commonly use for them.

Sound waves often have frequencies measured in thousand hertz, so kilos the most common prefix.

Two kilohertz there, 2000 hertz.

Radio waves have frequencies in a millions of hertz, so mega is the most commonly used prefix there.

And three megahertz, 3 million hertz.

And infrared radiation that's got much, much higher frequency.

So, we measure it in gigahertz with giga is billion.

So, we've got 8.

4 gigahertz, it's 8.

4 billion hertz.

Now you'll have to use those prefixes in calculations.

This is the set of stages I like to use to make sure I go through the process and get the right answer each time.

First thing I do is just as we did before, identify the variables in the question.

Then, if any of those variables have got prefixes, I like to change those to the base units.

So, I convert kilohertz to hertz.

I convert millimetres to metres.

So, 40 kilohertz I'd convert to 40,000 hertz.

Then, I select the equation based upon the data I've seen.

You've done that earlier in the lesson.

And finally I calculate the answer.

And, as usual, I double check that I've got units at the end of any answer so I get all the marks.

Here's an example you use in prefixes, and we're gonna try and solve this together.

We've got a sound wave in water with a frequency of 45 kilohertz and a wavelength of 3.

3 centimetres.

Both of those I've got prefixes.

We've got 45 kilohertz and 3.

3 centimetres.

We're gonna calculate the speed of the wave.

So, just as before I'm gonna identify the variables.

I've got a frequency of 45 kilohertz, and as I said I'd like to convert that into the base unit of 45,000 hertz.

So, I'm gonna write that down there.

And then, I look at the wavelength.

I've got a wavelength of 3.

3 centimetres, and I'm gonna write that down in metres as well, 0.

033 metres.

And now, all I have to do is write the equation, put the values in and do the calculation itself.

And that gives me a wave speed of 1485, 1,485.

I haven't got a unit yet, so I'll look back, and I see I've got a frequency of hertz and I've got a wavelength of metres.

That's gonna gimme a speed in metres per second.

Now, the value's given in the question though, I've only got two significant figures and my answer's got four, which I think is a bit too many.

So, I'm gonna round off to two figures just to give a final answer.

And that rounds up to 1,500 metres per second.

Okay, here's the next task, task B.

I'm gonna ask you to calculate the wave speed for two waves.

First, an earthquake producing a wave with a wave of the 2.

5 kilometres travelling a distance of 300 kilometres in one minute.

And the second a microwave oven producing microwaves with a frequency of 2.

4 gigahertz and a wavelength of 12.

5 centimetres.

You'll have to select the appropriate equation, and you'll have to convert those values into base units to be able to get the right answer.

So, pause the video, try and solve those two, and I'll lead you through them in a moment.

Welcome back.

Let's have a look at the first of those two.

We've got an earthquake producing a wave with a wave length of 2.

5 kilometres travelling a distance of 300 kilometres in one minute.

I'm going to identify all the data from the question and write it down nice and clearly.

I've got a frequency, well no I haven't.

There's no frequency mentioned there, so it's unknown.

I've got a wavelength of 2.

5 kilometres, and I'm gonna write that down as 2,500 metres, converting it from the prefix value there to the base units.

The next bit of information, I've got a time and although it says one, I need to be using times in seconds, so I'm gonna write that down as 60 seconds.

And finally I've got a distance of 300 kilometres, that's 300,000 metres.

Looking at all that data you should be able to select the correct equation.

And, that's distance travelled equals speed times time.

Putting the values in, I've got a distance travelled of 300,000 metres and the time of 60 seconds.

Rearranging that, I should be able to get the expression for the speed.

300,000 divided by 60 is a speed, so the speed is 5,000 metres per second.

So, earthquakes waves travel at 5,000 metres per second or five kilometres per second.

Here's the second of the questions.

A microwave oven producing microwaves with a frequency of 2.

4 gigahertz and a wavelength of 12.

5 centimetres.

So, as before, I identified the data and changed it to base units.

I've got a frequency of 2.

4 gigahertz.

That's written down as that.

I've got a wavelength of 12.

5 centimetres 0.

125 metres.

So, what I do is just use the correct equation.

I've got frequency of wavelength, so I can use this.

Write in the two values, and that's multiplied together on a calculator will give me a final answer of 300 million metres per second.

So, well done if you've got that answer.

Again, tricky units to use, but you need practise with them.

Okay, now it's time to move on to the final cycle of the lesson finding frequency in wavelength.

And in this we're gonna use the wave equation, but this time we are gonna find a frequency based upon the wave speed and the wavelength or the wavelength based upon the frequency and the wave speed.

When a wave moves from one medium to another across a boundary, it'll change its speed.

And that change in speed will cause a change in one of the other properties of the wave.

The frequency of the wave doesn't change, it's the wavelength that actually changes.

If a wave moves between mediums, we get effects like this.

If the wave speed increases then the wavelength will increase as well.

And if the wave speed decreases, then the wavelength will decrease.

The frequency will stay the same during those boundary changes.

Now, quick check of that.

Is this true or false? When a wave slows down that boundary, the wavelength stays the same.

I'd like you to pause and make selection and then restart.

Okay, that was false.

Excellent if you got that, but I'd like you to justify your answer.

I'd like you to decide why that was false.

Which of these two is it? Is it because the wavelength decreases, because the frequency stays the same, or the wavelength decreases because the frequency increases? Again, pause, make a selection and then restart.

Okay, that was the wavelength decreases, because the frequency stays the same.

If you have a change in the wave speed and the frequency is not changing, then you must have a change in the wavelength.

Now, you've seen this equation many times before in the lesson.

We're gonna be able to use it to calculate the frequency if we know the wave speed and the wavelength.

During this next section, I'm probably gonna be using the symbols a bit more, because writing out that equation each time is a bit long-winded.

Okay, we're gonna try and find the frequency of a wave based upon the wave speed and the wavelength.

So, I'm gonna try and answer this question.

Calculate the frequency of a wave if it's wave speed is 5.

0 metres per second, and it's wavelength is not 0.

25 metres.

First thing I'm gonna do is see what data I have and make sure I select the right equation.

I've got here a wave speed, five metres per second and a wavelength, and I've also got frequency.

So, I'm gonna use the wave speed equation, and that's this one V equals F lambda.

And as I said, I've decided to use symbols for this section.

And then, I'm going to substitute the values in.

So, I'm gonna substitute the wavelength, and I'm gonna substitute in the wave speed, and get this equation, 5.

0 equals F times 0.

25.

I'm going to adjust that and get 5.

0 divided by not 0.

25 is equal to the frequency F.

And finally, I'm gonna solve that and get a frequency of 20 hertz.

Okay, I've got a second example, and I'd like you to solve this one.

I'd like you to follow exactly the same procedure writing down equation, substitute in the values and coming up with a solution.

Pause the video, go through that process and restart, and I'll show you how you should have solved it.

Welcome back.

Here's the solution.

Again, we do the same as before.

We identify the two pieces of data.

They're here.

We write down the equation, V equals F lambda.

Substitute in the two values we identified.

The V is 333, and the lambda is 0.

09 metres.

Put those in.

Adjust the equation to get F on its own on one side.

So, that's 333 divided by 0.

09 equals F.

And solve that to give a final frequency of 3,700 hertz.

Okay, a quick check for you to try that again.

What's the frequency of a wave if its wave speed is 60 metres per second, a wavelength 2.

5 metres.

I'd like you to pause and solve it, and then I'll go through that solution when you restart.

Welcome back.

Let's have a look at the solution for that.

24 hertz is the answer you should have come up with.

So congratulations if you've got that.

The process again, write down the equation, substitute the two values in, there's 60 for the wave speed, 2.

5 for the wavelength and that gives a solution.

60 divided by 2.

5 is equal to the frequency.

The frequency is 24 hertz.

So, congratulations again.

So far, we've used this equation to find the wave speed and we've used it to find a frequency.

Now obviously we can use it to find the wavelength again if we know those two variables.

So, we're gonna go through a few examples of trying that.

Here's the first of those examples.

Calculate the wavelength of a wave if its wave speed is 5.

0 metres per second and its frequency is 2.

5 hertz.

I'm gonna go through my normal process, identifying the data.

I've got a wave speed of 5.

0 millimetres per second.

I've got a frequency of 2.

5 hertz.

Writing down the equation.

Putting those two values I've circled into that equation in the correct places.

So, I've got a speed of 5.

0, and I've got a frequency of 2.

So, I've got 5.

0 equals 2.

5 times lambda.

And then, adjusting that to get Lambda on its own, dividing both sides of the equation by 2.

5 there.

So, 5.

0 divided by 2.

5 is equal to lambda the wavelength.

And finally I can do that calculation and get my wavelength is equal to 2.

9 metres.

As before, there's an example for you to do.

Here it is.

I'd like you to read through that, try and solve it in exactly the same process as I used before.

So, pause the video, solve the question, and then restart.

Hello again.

Let's have a look.

We can identify the two pieces of information.

We can write down the appropriate equation and substitute those two circled values in the correct place.

Frequency of 18 hertz, wave speed of 90 metres per second.

So, those it's go in those positions.

Adjust the equation to get Lambda on its own.

The wavelength on its own there.

19 divided by 18 is equal to the wavelength.

And finally solve it.

So, the wavelength you should have got is 5.

9 metres.

Well done if you've got that.

Here's another check on that.

What's the wavelength of a wave if it's wave speed is 4,000 metres per second and a frequency is 25 kilohertz.

Carefully, that one.

I'd like you to try and solve that one.

Pause the video, come up with a solution and restart.

Okay, welcome back.

The solution you should have come up with is 0.

16 metres.

Let's go through that carefully to see why that was the right answer.

Writing down the equation, we've got V equal to f lambda.

And then substituting the values.

Now the values here, 4,000 metres per second, fairly straightforward, but the frequency was 25 kilohertz, 25,000 hertz.

So, I've written that that in in full.

I rearranged that equation to get Lambda on its own.

4,000 divided by 25,000 equals lambda.

And finally, do that division and get Lambda is 0.

16 metres.

So, well done if you've got that was the tricky bit of identifying the prefix.

Right, so we're at the final task here, and that's going to allow you to show your knowledge of all of the calculations you've done in this last section.

I'd like you to use the wave equation to complete this table.

It shows wave speed, wavelength and frequency for variety of waves.

And there's some gaps in that table, and you'll have to use the equation to calculate all three different things, wave speed, wavelength and frequency in the different waves.

So, pause the video, use the equation and the same process as you've used before to try and solve that.

And then, you can restart, and we'll go through the solution.

Hi, again.

And here's the solution to that table.

If you followed the same procedure, you should have come up with these answers.

Water wave, the wave speed was 1.

8 metres per second.

For the wave in the spring, the frequency of 5.

0 hertz.

The sound in air, the frequency was 5,500 hertz.

The ultrasound in water had a wavelength of 0.

02 metres.

And the ultrasound in steel had a wavelength of 0.

085 metres.

Congratulations if you've got all of those.

Well done, you've completed the final task.

And we've reached the end of the lesson.

And here's a summary of the information that you should have got.

We spent a lot of time using the wave speed equation, and that is wave speed equals frequency times wavelength or V equals F lambda.

And the other key piece of information you should got from this lesson is about prefixes and how we use them.

And I've got a range of example prefixes there, and you should get familiar with each of those.

You should try to memorise each of these.

Well done for reaching the end of the lesson.

And I'll see you again in a future one.

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