Year 10 Physics Wave Calculations

Year 10 Physics
Wave calculations

velocity = frequency x wavelength

Fill in the Blank

velocity = frequency × __________

Fill in the Blank

_______ _ = frequency × wavelength

Fill in the Blank

velocity = _________ × wavelength

Multiple Choice

What is the equation that links velocity, frequency and wavelength?

• velocity = frequency + wavelength

• velocity = frequency ÷ wavelength

• velocity = frequency × wavelength

• velocity = frequency × time

Dropdown

Wavelength is measured in ​ ​

Select answer

Dropdown

Frequency is measured in ​

Select answer

Dropdown

velocity is measured in ​ ​

Select answer

Calculate the velocity of a wave with a wavelength of 12m and a frequency of 3.0 Hz
v = fλ
v = 3 × 12
v= 36 m/s

Multiple Choice

Calculate the velocity of a wave with a wavelength of 12m and a frequency of 3.0 Hz

• 36m

• 36Hz

• 36m/s

• 36s

Calculate the velocity of a wave with a wavelength of 0.10 m and a frequency of 120 Hz
v = fλ
v = 120 × 0.1
v= 12 m/s

Multiple Choice

Calculate the velocity of a wave with a wavelength of 0.10m and a frequency of 120 Hz

• 12m/s

• 12Hz

• 12m

• 12s

Calculate the velocity of a wave with a wavelength of 1.0 x 10^-6 m and a frequency of 3.0 x 10¹⁴ Hz

In maths you learnt the when you multiply two powers, you add them

For example, 10³ × 10⁴
= 10⁷ and not 10¹²

So, 10^-6 × 10¹⁴
means -6 + 14 and not -6 × 14. So, 10⁸

Calculate the velocity of a wave with a wavelength of 1.0 x 10^-6 m and a frequency of 3.0 x 10¹⁴ Hz
(1 × 10^-6) × (3 × 10¹⁴) = 3 × 10⁸

Multiple Choice

Calculate the velocity of a wave with a wavelength of 1.0 x 10^-6 m and a frequency of 3.0 x 10¹⁴ Hz

• 3 × 10^-8 m/s

• 3 × 10^-8 m

• 3 × 10^-8 s

• 3 × 10⁸ m/s

Multiple Choice

Calculate the velocity of a wave with a wavelength of 1.0 x 10^-5 m and a frequency of 3.0 x 10¹⁰ Hz

• 3 × 10^-5 m/s

• 3 × 10^-5 m

• 3 × 10^-5 s

• 3 × 10⁵ m/s

Multiple Choice

Calculate the velocity of a wave with a wavelength of 1.0 x 10⁵ m and a frequency of 3.0 x 10¹⁰ Hz

• 3 × 10¹⁵ m/s

• 3 × 10^-15 m

• 3 × 10¹⁵ s

• 3 × 10^-15 m/s

A wave travelling at a speed of 3.0×10⁸ m/s typically refers to electromagnetic waves in vacuum. In vacuum, all electromagnetic waves, including visible light, radio waves, microwaves, X-rays, and gamma rays, travel at the speed of light, which is approximately 3.0×10⁸ m/s.

Multiple Choice

Rearrange the wave speed equation to make λ the subject

• λ = vf

• λ = v ÷ f

• λ = f ÷ v

Multiple Choice

Rearrange the wave speed equation to make f the subject

• f = λv

• f = v ÷ λ

• f = λ ÷ v

Calculate the answers to the following questions using the rearranged equations...

Multiple Choice

What is the wavelength of a wave with a velocity of 3.0 x 10⁸ and a frequency of 4.7 x 10¹²Hz . Choose the correct rearrangement

• v = fλ

• f = v ÷ λ

• λ = v ÷ f

Multiple Choice

What is the wavelength of a wave with a velocity of 3.0 x 10⁸ and a frequency of 4.7 x 10¹²Hz . Choose the correct substitution

• λ = (4.7 x 10¹²) × (3.0 x 10⁸)

• λ = (4.7 x 10¹²) ÷ (3.0 x 10⁸)

• λ = (3.0 x 10⁸) ÷ (4.7 x 10¹²)

Multiple Choice

λ = (3.0 x 10⁸) ÷ (4.7 x 10¹²) =

• 6.38 × 10⁻⁵ meters

• 6.38 × 10⁵ meters

Multiple Choice

What is the wavelength of a wave with a velocity of 330m/s and a frequency of 2.2 x 10⁴ Hz. Choose the correct substitution

• λ = (2.2 x 10⁴⁾ × 330

• λ = 2.2 x 10⁴ ÷ (2.2 x 10⁴)

• λ = 330 ÷ (2.2 x 10⁴)

Multiple Choice

λ = 330 ÷ (2.2 x 10⁴)

• 1.5 × 10⁻²

• 1.5 × 10²

Batman is developing some sonar technology for his echolocation goggles. He stands in a cave holding a speaker and a timer 99.0 m away

from the far end of the cave. Batman remembers that sound travels at about 330 m/s in air. What we need to know...

Dropdown

We have to remember the sound heard as an echo has travelled ​

Select answer

as far​.

Dropdown

The distance from the sound source to a wall is 10m. The sound wave travels

Select answer

when heard as an echo.

Multiple Choice

The distance a sound travels to a wall is 100 m. It bounces back as an echo. How far has the sound travelled when heard as the echo?

• 100m

• 50m

• 200m

• 400m

Remember:
speed = distance ÷ time

Match

Match the following result with the rest of the equation

• speed =

• time =

• distance =

distance ÷ time

distance ÷ speed

speed × time

• distance ÷ time

• distance ÷ speed

• speed × time

Multiple Choice

The distance a sound travels to a wall is 100m. It bounces back as an echo. Sound travels at 330m/s. How long did it take to get to the wall?

• 100 × 330

• 50 × 300

• 200 × 300

• 100 ÷ 330

Multiple Choice

The distance a sound travels to a wall is 100m. It bounces back as an echo. Sound travels at 330m/s. How long did it take to get to the wall?

• 0.15s

• 0.6s

• 0.3s

• 3s

The solution was time = distance ÷ speed
100* ÷ 330 = 0.3 (s)
*just to the wall

Multiple Choice

The distance a sound travels to a wall is 100m. It bounces back as an echo. Sound travels at 330m/s. How long did it take to be received as an echo?

• 3 300s

• 33 000s

• 0.3s

• 0.03s

The solution was distance ÷ time
200* ÷ 330 = 0.6 (s)
*to the wall was 100m and back again gives 200m

Batman is developing some sonar technology for his echolocation goggles. He stands in a cave holding a speaker and a timer 99.0 m away

from the far end of the cave. Batman remembers that sound travels at about 330 m/s in air.

Dropdown

The sound wave travels

Select answer

metres to the end of the cave

Dropdown

The sound wave travels ​

Select answer

metres to the end of the cave and back

Multiple Choice

The formula to work out how long the sound travels for is:

• time = distance ÷ speed

• time = speed ÷ distance

• time = speed × distance

Reorder

Reorder the following to create the formula needed to calculate the time it kook for the sound to travel to the end of the bat cave and back

• time

• =

• 198

• ÷

• 300

1

2

3

4

5

• 1
• 2
• 3
• 4
• 5

Multiple Choice

The time it takes the sound to travel 198 metres at 330m/s is...

• 1.2s

• 0.3s

• 0.6s

• 1.7s

The sound wave completes 1,200 wave cycles in the time it takes to travel to the far end of the bat cave and back.
Calculate the time period of one wave cycle.

We know that it took 0.6s. In that time, there were 1200 wave cycles...

Multiple Choice

The time for one wave cycle would be...

• 0.6 ÷ 1200

• 1200 ÷ 0.6

• 1200 × 0.6

This is because 1200 cycles were squeezed into just 0.6 seconds. The time for one of those waves would be very small... 0.6 ÷ 1200.
0.6 would be divided into 1200 pieces.

Math Response

0.6 ÷ 1200 = (seconds)

Type answer here

Correct Answer: 0.0005

Deg^°

Rad

Hence, calculate the frequency of the sound wave...

Frequency is the number of waves in 1 second. One wave takes 0.0005s
1s ÷ 0.0005

Math Response

1 ÷ 0.0005 =

Type answer here

Correct Answer: 2000

Deg^°

Rad

Fill in the Blank

The answer is 2000

Hence, calculate the wavelength of the sound wave to 2 sf

Dropdown

wavelength = ​

Select answer

÷ ​

Select answer

Dropdown

wavelength = ​ ​

Select answer

÷ ​ ​

Select answer

Dropdown

0.165 to 2 s.f is ​

Select answer

The Joker is developing some high-powered torches for his fight against Batman. He reads that the torch has a frequency of 5.30 x 10¹⁴ Hz and a wavelength of 5.66 x 10^-7 m.

Show that the speed of light is about 3.00 x 10⁸ m/s

Multiple Choice

Which formula do we need?

• speed = distance ÷ time

• velocity = fλ

Multiple Choice

velocity =

• 5.30 x 10¹⁴ Hz × 5.66 x 10^-7

• 5.30 x 10¹⁴ Hz ÷ 5.66 x 10^-7

5.30 x 10¹⁴ Hz × 5.66 x 10^-7
Let's deal with the powers first...
¹⁴ × ^-7
When we multiply powers we add them...

Multiple Choice

¹⁴ × ^{-7 =}

• 7

• 21

• 98

Now, lets multiply the integers... 5.3 × 5.66

Math Response

5.3 × 5.66 =

Type answer here

Correct Answer: 29.998

Deg^°

Rad

We have 29.998 x 10⁷
This is not showing that the speed of light is about 3.00 x 10⁸ m/s
Or is it???

Round 29.998 to get 30
So, 30 x 10⁷
Which is the same as 3 x 10⁸

v = fλ
If we increase the frequency will we increase the speed (velocity)?

Poll

v = fλ
If we increase the frequency will we increase the speed (velocity)?

• Yes

• No

v = fλ
If we increase the frequency will we increase the speed (velocity)?

Yes!

Imagine the frequency was originally 2 and the λ was 3. The v would be 6.

Increase the frequency to 3 and now the v is 9 (3 × 3)