Waves: transverse, longitudinal and wave speed
By the end of this chapter you will be able to
- Classify waves as transverse or longitudinal, and explain the direction of vibration
- Label a wave diagram with amplitude and wavelength, and find the period from the frequency
- Calculate wave speed, frequency or wavelength using v = f lambda
- Calculate the distance from an echo time, and explain ultrasound uses
You should already know
- Energy transfers by waves (Chapter 1)
- Speed, distance and time (Chapter 16)
Where this leads
- Reflection, refraction and lenses (Chapter 18)
- The electromagnetic spectrum (Chapter 19)
GCSE Physics. Content follows the GCSE (9-1) subject content for England (Edexcel and AQA). Content references: Distinguish transverse and longitudinal waves, with examples, and the direction of vibration, Describe amplitude, wavelength, frequency (in hertz) and period, and use T = 1 / f, Use the wave equation v = f lambda, Describe reflection and refraction of waves at a boundary (idea only), Describe sound as a longitudinal wave, its speed in air, ultrasound uses, and measuring the speed of sound by echo.
Exercise P Do you remember?
Method. Multiply.
- Multiply. 5 x 2 = 10.10
- Write the answer. 10
Check. ✓
Method. Divide.
- Divide. 1 / 50 = 0.02.0.02
- Write the answer. 0.02
Check. ✓
Method. Divide.
- Divide. 340 / 2 = 170.170
- Write the answer. 170
Check. ✓
Method. Think about the direction of travel.
- Recall. The energy travels along the wave, while the particles vibrate.along
- Write the answer. along
Check. ✓
Method. Recall the unit.
- Recall. Frequency is measured in hertz, Hz.hertz (Hz)
- Write the answer. hertz
Check. ✓
Transverse and longitudinal waves
A wave transfers energy from one place to another without transferring matter. There are two main types:
- Transverse: the vibration is at right angles to the direction the wave travels. Examples: water ripples, light, and other electromagnetic waves, and waves on a rope.
- Longitudinal: the vibration is along the direction the wave travels. The wave has compressions (crowded particles) and rarefactions (spread-out particles). Examples: sound waves in air, and some waves in a slinky.
Key ideas
- Transverse: vibration at right angles to the direction of travel.
- Longitudinal: vibration along the direction of travel.
- Examples of transverse: light, water ripples, electromagnetic waves.
- Examples of longitudinal: sound waves in air, waves in a slinky spring.
Method. Recall the definition of a transverse wave.
- Recall. In a transverse wave the vibration is at right angles to the travel.at right angles
- Decide. So the vibrations are perpendicular to the direction of travel.perpendicular
- Write the answer. perpendicular
Check. ✓
Method. Recall the name of the crowded part.
- Look. The crowded regions are where the particles are pushed together.crowded
- Recall. The crowded parts of a longitudinal wave are compressions.compressions
- Write the answer. compressions
Check. The spread-out parts are rarefactions. ✓
Method. Recall the definition.
- Look. In one type of wave, the vibrations are at right angles to the travel.right angles
- Decide. That type is transverse.transverse
- Write the answer. transverse
Check. ✓
Watch out
Watch out: in both types of wave, the particles do not travel with the wave. They vibrate about a fixed position, and the energy moves on.
Exam tip
Exam tip. Say “at right angles” for transverse and “along the direction of travel” for longitudinal. Both phrases score.
Exercise 17A Types of wave
Method. Recall the definition.
- Recall. Transverse waves vibrate at right angles to the direction of travel.transverse
- Write the answer. transverse
Check. ✓
Method. Recall the definition.
- Recall. Longitudinal waves vibrate along the direction of travel.longitudinal
- Write the answer. longitudinal
Check. ✓
Method. Recall an example.
- Recall. Light is a transverse electromagnetic wave.light
- Write the answer. light
Check. Accept water waves, or other electromagnetic waves. ✓
Method. Recall an example.
- Recall. Sound in air is a longitudinal wave.sound
- Write the answer. sound
Check. Accept a wave in a slinky. ✓
Method. Recall the name.
- Recall. The crowded regions are compressions.compressions
- Write the answer. compressions
Check. ✓
Method. Recall the name.
- Recall. The spread-out regions are rarefactions.rarefactions
- Write the answer. rarefactions
Check. ✓
Method. The particles vibrate about fixed positions.
- Recall. The water particles move up and down; the energy moves to the shore.no
- Write the answer. no
Check. ✓
Method. The coils move along the spring.
- Recall. The coils vibrate along the direction of the spring, which is longitudinal.longitudinal
- Write the answer. longitudinal
Check. ✓
Method. The rope moves at right angles to the travel.
- Recall. The rope vibrates up and down, at right angles to the direction of travel.transverse
- Write the answer. transverse
Check. ✓
Method. Think about the particles.
- Particles. The air particles are pushed and pulled along the direction of travel.along
- Waves. So the wave is longitudinal.longitudinal
- Write the answer. Accept any sentence that says the air particles vibrate along the direction the sound travels.
Check. ✓
Method. Think about what moves.
- Matter. The particles only vibrate about their positions.vibrate
- Energy. Waves transfer energy, not matter.energy
- Write the answer. Accept any sentence that says waves transfer energy, not matter.
Check. ✓
Method. Compare the directions.
- Transverse. Transverse: at right angles to the travel.right angles
- Longitudinal. Longitudinal: along the travel.along
- Write the answer. Accept any sentence that says transverse vibrations are at right angles and longitudinal ones are along the travel.
Check. ✓
Amplitude, wavelength, frequency and period
The amplitude is the maximum displacement from the rest position. The wavelength (lambda) is the distance from one point on a wave to the same point on the next wave, such as crest to crest. The frequency is the number of waves passing a point each second, in hertz (Hz). The period is the time for one complete wave, and it is the inverse of the frequency:
T = 1 / f, so f = 1 / T
| Quantity | Symbol | Unit |
|---|---|---|
| amplitude | A | metre (m), the maximum displacement from the rest position |
| wavelength | lambda | metre (m), from one point to the same point on the next wave |
| frequency | f | hertz (Hz), waves per second |
| period | T | second (s), the time for one complete wave |
| wave speed | v | metre per second (m/s) |
Key ideas
- Amplitude: height from rest to crest.
- Wavelength: crest to crest (or any point to the same point on the next wave).
- Frequency f in hertz: waves per second.
- Period T = 1 / f, in seconds.
Method. Use T = 1 / f.
- Take the numbers. f = 50 Hz.f = 50
- Divide. T = 1 / 50 = 0.02 s.1 / 50 = 0.02
- Write the answer. 0.02 s
Check. 1 / 0.02 = 50 Hz. ✓
Method. Use f = 1 / T.
- Take the numbers. T = 0.25 s.T = 0.25
- Divide. f = 1 / 0.25 = 4 Hz.1 / 0.25 = 4
- Write the answer. 4 Hz
Check. ✓
Method. The wavelength is the length of one complete wave.
- Read. One complete wave spans 2 cm, from crest to the next crest.2 cm
- Decide. So the wavelength is 2 cm.2 cm
- Write the answer. 2 cm
Check. ✓
Watch out
Watch out: amplitude is the height from the rest position to a crest, not the whole height from trough to crest.
Exam tip
Exam tip. Give frequency in hertz and period in seconds. A higher frequency gives a shorter period.
Exercise 17B Amplitude, wavelength, frequency and period
Method. Recall the unit.
- Recall. Frequency is measured in hertz, Hz.Hz
- Write the answer. hertz (Hz)
Check. ✓
Method. Recall the definition.
- Recall. The amplitude is the maximum height from the rest position.height
- Write the answer. height
Check. ✓
Method. Frequency is waves per second.
- Recall. A frequency of 10 Hz is 10 waves per second.10
- Write the answer. 10
Check. ✓
Method. T = 1 / f.
- Divide. T = 1 / 4 = 0.25 s.1 / 4 = 0.25
- Write the answer. 0.25 s
Check. ✓
Method. f = 1 / T.
- Divide. f = 1 / 0.5 = 2 Hz.1 / 0.5 = 2
- Write the answer. 2 Hz
Check. ✓
Method. T = 1 / f.
- Divide. T = 1 / 200 = 0.005 s.1 / 200 = 0.005
- Write the answer. 0.005 s
Check. ✓
Method. f = 1 / T.
- Divide. f = 1 / 0.001 = 1000 Hz.1 / 0.001 = 1000
- Write the answer. 1000 Hz
Check. ✓
Method. Each wave is 4 cm long.
- Multiply. Two wavelengths = 2 x 4 = 8 cm.2 x 4 = 8
- Write the answer. 8 cm
Check. ✓
Method. Amplitude and wavelength are separate properties.
- Check. Amplitude does not fix frequency, and different wavelengths give different frequencies at a set speed.no
- Write the answer. no
Check. ✓
Method. Think about what frequency counts.
- Frequency. Frequency counts the waves per second.per second
- Period. So the period is the time for one wave, the inverse of the frequency.inverse
- Write the answer. Accept any sentence that says it is the time for one full wave to pass a point, which is the inverse of the frequency.
Check. ✓
Method. Use T = 1 / f.
- Halve. If f doubles, then 1 / f is halved.halved
- Link. So the period is halved, not doubled.halved
- Write the answer. Accept any answer that says doubling the frequency halves the period, since T = 1 / f.
Check. ✓
The wave equation: v = f lambda
The speed of a wave is the frequency multiplied by the wavelength:
v = f x lambda, with v in m/s, f in Hz and lambda in m
The equation can be rearranged: f = v / lambda and lambda = v / f. For a given wave speed, a higher frequency gives a shorter wavelength.
Key ideas
- v = f x lambda. Speed in m/s, frequency in Hz, wavelength in m.
- lambda = v / f; f = v / lambda.
- For a fixed speed, higher frequency means shorter wavelength.
- Convert cm to m before you calculate.
Method. Use v = f x lambda.
- Take the numbers. f = 5 Hz and lambda = 2 m.f = 5, lambda = 2
- Multiply. v = 5 x 2 = 10 m/s.5 x 2 = 10
- Write the answer. 10 m/s
Check. 10 / 5 = 2 m. ✓
Method. Use lambda = v / f.
- Take the numbers. v = 0.2 m/s and f = 4 Hz.v = 0.2, f = 4
- Divide. lambda = 0.2 / 4 = 0.05 m.0.2 / 4 = 0.05
- Write the answer. 0.05 m
Check. 4 x 0.05 = 0.2 m/s. ✓
Method. Use lambda = v / f.
- Take the numbers. v = 340 m/s and f = 170 Hz.v = 340, f = 170
- Divide. lambda = 340 / 170 = 2 m.340 / 170 = 2
- Write the answer. 2 m
Check. 170 x 2 = 340 m/s. ✓
Watch out
Watch out: convert the wavelength from cm to m before you substitute. 50 cm is 0.5 m.
Exam tip
Exam tip. Write v = f lambda, and then rearrange to the quantity you need. Check that the units work out.
Exercise 17C The wave equation
Method. Recall the wave equation.
- Recall. The wave equation is v = f lambda, so speed = frequency x wavelength.wavelength
- Write the answer. wavelength
Check. ✓
Method. Use v = f lambda.
- Multiply. v = 3 x 4 = 12 m/s.3 x 4 = 12
- Write the answer. 12 m/s
Check. ✓
Method. Use f = v / lambda.
- Divide. f = 20 / 5 = 4 Hz.20 / 5 = 4
- Write the answer. 4 Hz
Check. ✓
Method. Use lambda = v / f.
- Divide. lambda = 30 / 6 = 5 m.30 / 6 = 5
- Write the answer. 5 m
Check. ✓
Method. Use f = v / lambda.
- Divide. f = 100 / 0.5 = 200 Hz.100 / 0.5 = 200
- Write the answer. 200 Hz
Check. ✓
Method. Convert 50 cm to 0.5 m, then v = f lambda.
- Convert. 50 cm = 0.5 m.0.5 m
- Multiply. v = 8 x 0.5 = 4 m/s.8 x 0.5 = 4
- Write the answer. 4 m/s
Check. ✓
Method. Use lambda = v / f.
- Divide. lambda = 340 / 34 = 10 m.340 / 34 = 10
- Write the answer. 10 m
Check. ✓
Method. For the same speed, lambda is proportional to 1 / f.
- Ratio. lambda A = v / 2f and lambda B = v / f, so lambda A / lambda B = 1/2 = 0.5.0.5
- Write the answer. 0.5
Check. ✓
Method. Find lambda in m, then convert to cm.
- Divide. lambda = 340 / 400 = 0.85 m.0.85
- Convert. 0.85 m = 85 cm.85
- Write the answer. 85 cm
Check. ✓
Method. Use v = f lambda in cm and Hz.
- Multiply. v = 5 x 3 = 15 cm/s.5 x 3 = 15
- Write the answer. 15 cm/s
Check. ✓
Method. Look at lambda = v / f.
- Equation. lambda = v / f, with v fixed.v / f
- Result. A larger f makes the divisor bigger, so lambda is smaller.smaller
- Write the answer. Accept any sentence that says lambda = v / f, so a larger f gives a smaller lambda.
Check. ✓
Reflection and refraction of waves (idea)
Reflection: a wave bounces back when it meets a boundary, with the angle of reflection equal to the angle of incidence. Refraction: a wave changes direction when it passes from one medium into another where its speed changes. When the wave slows down (for example, moving from deep to shallow water), it bends towards the normal. The frequency stays the same, but the wavelength changes.
Key ideas
- Reflection: the wave bounces back from a boundary.
- Refraction: the wave changes direction when its speed changes at a boundary.
- Refraction: the frequency stays the same; the wavelength and speed change.
- A wave slowing down bends towards the normal.
Method. The frequency is set by the source.
- Recall. The frequency of a wave stays the same at a boundary.no
- Change. The wave speed and wavelength change instead.speed and wavelength
- Write the answer. no
Check. ✓
Method. The angle of reflection equals the angle of incidence.
- Look. The boundary is flat, so the law of reflection applies.law of reflection
- Recall. Angle of reflection = angle of incidence = 30 degrees.30
- Write the answer. 30 degrees
Check. ✓
Method. Use the law of reflection.
- Look. The angle is measured from the normal, which is at right angles to the surface.normal
- Recall. The angle of reflection equals the angle of incidence: 40 degrees.40
- Write the answer. 40 degrees
Check. ✓
Watch out
Watch out: at a boundary, the frequency does not change. Only the speed and the wavelength change in refraction.
Exam tip
Exam tip. When describing refraction, say which way the wave bends, and whether it is slowing down or speeding up.
Exercise 17D Reflection and refraction
Method. Recall the process.
- Recall. Bouncing back from a boundary is reflection.reflection
- Write the answer. reflection
Check. ✓
Method. Recall the process.
- Recall. Changing direction when the speed changes is refraction.refraction
- Write the answer. refraction
Check. ✓
Method. The law of reflection.
- Recall. The angle of reflection equals the angle of incidence.equal
- Write the answer. equal
Check. ✓
Method. Shallow water slows waves.
- Recall. The wave slows down in shallow water.decrease
- Write the answer. decrease
Check. ✓
Method. Frequency is set by the source.
- Recall. The frequency stays the same; the speed and wavelength change.no
- Write the answer. no
Check. ✓
Method. A slower wave bends towards the normal.
- Recall. When a wave slows down, it bends towards the normal.towards
- Write the answer. towards
Check. ✓
Method. Angle of reflection equals angle of incidence.
- Recall. The angle of reflection is 45 degrees.45
- Write the answer. 45 degrees
Check. ✓
Method. Use lambda = v / f with the new speed.
- Divide. lambda = 0.2 / 4 = 0.05 m.0.2 / 4 = 0.05
- Write the answer. 0.05 m
Check. ✓
Method. Think about the speed.
- Speed. The wave speed changes at the boundary.speed changes
- Direction. So one side of the wave changes speed first, and the wave changes direction.direction
- Write the answer. Accept any sentence that says the speed of the wave changes in the new material, so it changes direction.
Check. ✓
Method. Think about the source.
- Source. The frequency is set by the source of the wave.source
- Boundary. It does not change at a boundary, only the speed and wavelength do.does not change
- Write the answer. Accept any sentence that says the frequency is set by the source and does not change at a boundary.
Check. ✓
Method. The wave bounces off the barrier.
- Recall. Bouncing off a barrier is reflection.reflection
- Write the answer. reflection
Check. ✓
Sound, ultrasound and measuring the speed of sound
Sound is a longitudinal wave. In air, sound travels at about 340 m/s. Sound with a frequency above about 20 000 Hz is called ultrasound. Ultrasound is used in medicine (scans), in sonar and depth finding, and in cleaning. It is used because it reflects from boundaries inside objects and does not use harmful ionising radiation.
To measure the distance to a wall by echo, measure the time between the sound being made and the echo returning. The sound travels to the wall and back, so the total distance is twice the distance to the wall:
distance to the wall = speed x time / 2
| Use | Wave | Why |
|---|---|---|
| ultrasound scan of a baby | ultrasound (above 20 kHz) | reflects from boundaries inside the body; no harmful ionising radiation |
| sonar or depth finding | ultrasound or sound | echo time gives the distance |
| cleaning jewellery | ultrasound | vibrates dirt off the surface |
| ripple tank | water waves | shows reflection and refraction of waves |
Key ideas
- Sound is longitudinal; its speed in air is about 340 m/s.
- Ultrasound: sound above 20 000 Hz.
- Echo: distance to the reflector = speed x time / 2.
- Ultrasound is used in scans because it reflects from boundaries inside the body.
Method. The sound travels there and back, so use distance = speed x time / 2.
- Total. Total distance = 340 x 2 = 680 m.340 x 2 = 680
- Halve. Distance to the wall = 680 / 2 = 340 m.680 / 2 = 340
- Write the answer. 340 m
Check. ✓
Method. Use distance = speed x time / 2.
- Multiply. Total = 1500 x 0.0001 = 0.15 m.1500 x 0.0001 = 0.15
- Halve. Distance = 0.15 / 2 = 0.075 m.0.15 / 2 = 0.075
- Write the answer. 0.075 m
Check. ✓
Method. Use distance = speed x time / 2.
- Total. Total path = 340 x 4 = 1360 m.1360
- Halve. Distance to the wall = 1360 / 2 = 680 m.680
- Write the answer. 680 m
Check. ✓
Watch out
Watch out: the sound goes to the wall and back. Always halve the total distance to find the distance to the wall.
Exam tip
Exam tip. Write the total path (speed x time), then halve it. A missing halving is the most common mistake in echo questions.
Exercise 17E Sound, ultrasound and echoes
Method. Recall the value.
- Recall. The speed of sound in air is about 340 m/s.340
- Write the answer. 340 m/s
Check. ✓
Method. Recall the limit.
- Recall. Ultrasound is sound above about 20 000 Hz.20 000
- Write the answer. 20 000 Hz
Check. ✓
Method. Recall the type.
- Recall. Sound in air is a longitudinal wave.longitudinal
- Write the answer. longitudinal
Check. ✓
Method. Distance = speed x time / 2.
- Multiply. Total = 340 x 1 = 340 m.340
- Halve. Distance = 340 / 2 = 170 m.170
- Write the answer. 170 m
Check. ✓
Method. Distance = speed x time / 2.
- Multiply. Total = 340 x 0.5 = 170 m.170
- Halve. Distance = 170 / 2 = 85 m.85
- Write the answer. 85 m
Check. ✓
Method. Think about what ultrasound does in the body.
- Reflects. Ultrasound reflects from boundaries inside the body.reflects
- Safe. It does not use ionising radiation.no ionising radiation
- Write the answer. Accept any reason such as reflects from body boundaries or no ionising radiation.
Check. ✓
Method. Compare the two waves.
- Ultrasound. Ultrasound is a sound wave with no ionising radiation.no ionising
- Safety. So it is safer for the baby than X-rays.safer
- Write the answer. Accept any sentence that says ultrasound does not use ionising radiation, so it is safer.
Check. ✓
Method. Total path = speed x time, then halve it.
- Total. Total = 1500 x 0.6 = 900 m.900
- Halve. Depth = 900 / 2 = 450 m.450
- Write the answer. 450 m
Check. ✓
Method. Think about the path.
- Path. The sound goes to the reflector and returns.there and back
- Half. So the distance to the reflector is half of the total path.half
- Write the answer. Accept any sentence that says the sound travels to the reflector and back.
Check. ✓
Method. Total path first, then halve.
- Total. Total path = 340 x 0.2 = 68 m.68
- Halve. Distance = 68 / 2 = 34 m.68 / 2 = 34
- Write the answer. 34 m
Check. ✓
Chapter summary
| Skill | Method |
|---|---|
| Transverse and longitudinal | Transverse: vibration at right angles; longitudinal: vibration along the travel |
| Describe a wave | Amplitude (height); wavelength (repeat distance); frequency (Hz); period T = 1 / f |
| Wave equation | v = f lambda; lambda = v / f; f = v / lambda |
| Reflection and refraction | Reflection bounces back; refraction changes direction when speed changes; frequency unchanged |
| Sound and echoes | Sound about 340 m/s in air; distance = speed x time / 2; ultrasound above 20 000 Hz |
Can you…?
- I can tell transverse and longitudinal waves apart
- I can describe amplitude, wavelength, frequency and period
- I can calculate speed, frequency and wavelength using v = f lambda
- I can explain reflection and refraction at a boundary
- I can calculate the distance from an echo time, and explain ultrasound uses
Key terms
- transverse wave
- a wave whose vibration is at right angles to the direction of travel
- longitudinal wave
- a wave whose vibration is along the direction of travel
- amplitude
- the maximum displacement from the rest position
- wavelength
- the distance from one point on a wave to the same point on the next wave
- frequency
- the number of waves passing a point each second, in hertz (Hz)
- period
- the time for one complete wave, T = 1 / f
- ultrasound
- sound with a frequency above about 20 000 Hz
Exercise 17Z Check-up
Method. Recall.
- Recall. Longitudinal waves vibrate along the direction of travel.longitudinal
- Write the answer. longitudinal
Check. ✓
Method. T = 1 / f.
- Divide. T = 1 / 25 = 0.04 s.1 / 25 = 0.04
- Write the answer. 0.04 s
Check. ✓
Method. v = f lambda.
- Multiply. v = 6 x 2 = 12 m/s.6 x 2 = 12
- Write the answer. 12 m/s
Check. ✓
Method. lambda = v / f.
- Divide. lambda = 10 / 5 = 2 m.10 / 5 = 2
- Write the answer. 2 m
Check. ✓
Method. Recall.
- Recall. The amplitude is the maximum height from the rest position.height
- Write the answer. height
Check. ✓
Method. Recall.
- Recall. Bouncing off a surface is reflection.reflection
- Write the answer. reflection
Check. ✓
Method. Distance = speed x time / 2.
- Multiply. 340 x 3 = 1020 m.1020
- Halve. 1020 / 2 = 510 m.510
- Write the answer. 510 m
Check. ✓
Method. Recall.
- Recall. The speed of sound in air is about 340 m/s.340
- Write the answer. 340 m/s
Check. ✓
Method. Frequency is set by the source.
- Recall. Refraction changes the speed and wavelength, not the frequency.no
- Write the answer. no
Check. ✓
Method. Think about the particles.
- Push. The vibration pushes particles together and then apart.together and apart
- Write the answer. Accept any sentence that says the particles are pushed together and spread apart along the direction of travel.
Check. ✓
Total 29 marks. Allow about 29 minutes.
Exercise 17X Exam-style questions
Method. f = 1 / T.
- Divide. f = 1 / 0.02 = 50 Hz.1 / 0.02 = 50
- Write the answer. 50 Hz
Check. ✓
Method. v = f lambda.
- Multiply. v = 8 x 1.5 = 12 m/s.8 x 1.5 = 12
- Write the answer. 12 m/s
Check. ✓
Method. Recall each type.
- Light. Light is transverse.transverse
- Sound. Sound in air is longitudinal.longitudinal
- Rope. A rope shaken up and down is transverse.transverse
- Write the answer. transverse, longitudinal, transverse
Check. ✓
Method. Compare the two types.
- Transverse. Transverse: the vibration is at right angles to the direction of travel, for example light.right angles
- Longitudinal. Longitudinal: the vibration is along the direction of travel, for example sound.along
- Energy. Both types transfer energy without transferring matter.energy
- Write the answer. Accept any answer that says transverse vibrations are at right angles (e.g. light) and longitudinal vibrations are along the direction of travel (e.g. sound).
Check. ✓
Method. Use v = f lambda.
- Multiply. v = 25 x 0.04 = 1 m/s.25 x 0.04 = 1
- Write the answer. 1 m/s
Check. ✓
Method. Use lambda = v / f.
- Divide. lambda = 340 / 440 = 0.7727... m, which is 0.77 m to 2 decimal places.340 / 440 = 0.77
- Write the answer. 0.77 m
Check. ✓
Method. Total path first, then halve.
- Total. Total path = 340 x 4 = 1360 m.1360
- Halve. Distance = 1360 / 2 = 680 m.680
- Write the answer. 680 m
Check. ✓
Method. Describe the method.
- Pulse. An ultrasound pulse is sent down to the sea floor.pulse
- Echo. The time for the echo to return is measured.time
- Calculate. Depth = speed x time / 2, using the speed of ultrasound in water.speed x time / 2
- Reason. Halving is needed because the pulse travels down and back.halve
- Write the answer. Accept any answer that says a pulse is sent down, the echo time is measured, and the depth is half of speed times time.
Check. ✓
Method. Use the wave equation.
- Equation. v = f lambda, so lambda = v / f.lambda = v / f
- Doubling. If f doubles with v fixed, lambda halves.halves
- Write the answer. Accept any answer that says lambda = v / f, so doubling f halves the wavelength.
Check. ✓
Method. Think about each quantity.
- Speed. The speed decreases as the wave slows.decreases
- Wavelength. The wavelength decreases because lambda = v / f with f fixed.decreases
- Frequency. The frequency stays the same, as it is set by the source.same
- Write the answer. Accept any answer that says the speed and wavelength decrease, and the frequency stays the same.
Check. ✓