Waves and Radiation

Waves and Radiation

Frequency

Speed Distance and Time

Physics

National 4/5

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Speed Distance and Time

In this section you can use the equation:

speed =

also written as

wherev = average speed in metres per second (m/s)

d = distance travelled in metres (m)

t= time taken in seconds (s).

1.Find the missing values in the following table:

Speed (m/s) / Distance (m) / Time (s)
(a) / 15 000 / 5
(b) / 38 / 0·02
(c) / 1 500 / 0·25
(d) / 5 200 / 0·01
(e) / 340 / 17
(f) / 330 / 3 465

2.The speed of sound in tissue is 1 500 metres per second. How far would sound travel in tissue in a time of 0·000 2 seconds?

3. Sound in jelly can travel a distance of 0·435 metres in a time of 0·000 3 seconds. What is the speed of sound in jelly?

4.How long would it take for sound to travel 0·435 m through air if the speed of sound in air is 340 m/s?

5.The speed of sound in muscle is 1 600 m/s. How far would sound travel in muscle in a time of 0·5 milliseconds?

6.Calculate the speed of sound in bone given that it takes 0·05 ms for sound to travel 0·15 m through bone.

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Echoes

1.A scientist is testing some sonar equipment in a valley.

The equipment operates at 40 kHz, giving a wavelength in air of 0·83 cm, and consists of an ultrasound transmitter / receiver and a sensitive timing device. There are two horns on the transmitter which direct the ultrasound in two opposite directions, as shown above.

The scientist positions the equipment 60 m from the nearby valley wall and transmits a pulse of ultrasound. The time between transmission and detection of each pulse echo is then displayed on the timer.

(a)Calculate the speed of sound in the valley.

(b)What time elapsed, after transmission, before the first echo was detected?

(c)The second echo was detected at a time of 0·675 seconds after transmission. What was the distance between the scientist and the faraway valley wall

(d)How wide was the valley?

2.A boy is standing at a distance of 100 m from a large building. He shouts loudly and hears an echo.

(a)How far did the sound travel between leaving the boy and returning to him as an echo?

(b)If the speed of sound in air is 340 m/s, how long did it take for the sound to cover this distance?

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Ultrasound

  1. An ultrasound pulse of frequency 8 MHz is transmitted into an expectant mother’s womb and reflects from baby’s bottom. The pulse echo is detected 0·08 milliseconds after being transmitted. The speed of sound through the body tissue and fluid is 1500 m/s.

(a)How far does the pulse travel?

(b)How far from the transmitter is the baby’s bottom?

(c)Another pulse is reflected from the foot of the baby. If this reflected pulse is detected 0·15 milliseconds after being transmitted, how far from the transmitter is the baby’s foot?

  1. During an ultrasound scan, a baby’s forehead is situated 7·5 cm from the transmitter. The ultrasound pulse travelling at 1500 m/s is reflected from the baby’s forehead.

(a)What is the total distance travelled by the pulse?

(b)What time elapses between the transmission of the pulse and the detection of the pulse echo?

  1. An ultrasound pulse is transmitted into the womb of an expectant mother and the pulse echo is detected after a time of 0·38 milliseconds. The pulse was reflected by one of the baby’s knees situated 28·5 cm from the transmitter. Show that the speed of sound in the womb is 1 500 m/s.

4.Ultrasound can be used in medicine to monitor the growth of unborn babies.

On a visit to hospital an expectant mother is scanned with ultrasound of frequency 8 MHz, which has a wavelength of 0·19 mm in the body.

One pulse of ultrasound leaves the transmitter, is reflected by the baby’s hand and returns to the detector in a time of 0·15 milliseconds.

(a)What is meant by ultrasound?

(b)Why is a layer of jelly placed between the mother and the transmitter / detector?

(c)How far from the transmitter / detector is the baby’s hand positioned?

(d)Explain how the image of a baby is produced during an ultrasound scan.

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Speed, Frequency and Wavelength

In this section you can use the equation:

speed = frequency x wavelength

also written as

wherev = average speed in metres per second (m/s)

f = frequency in Hertz (Hz)

 = wavelength in metres (m)

1.Find the missing values in the following table:

Speed (m/s) / Frequency (Hz) / Wavelength (m)
(a) / 7,000,000 / 0·0005
(b) / 80,000 / 0·02
(c) / 1,360 / 6,800,000
(d) / 330 / 660
(e) / 1,500 / 0·0025
(f) / 5,200 / 1·3
Helpful Hint
1 kHz = 1 000 Hz = 1 x 103 Hz
1 MHz = 1 000 000 Hz = 1 x 106 Hz

2.Calculate the wavelength of sound with frequency 1 000 Hz which is passing through carbon dioxide gas. (Speed of sound in carbon dioxide = 270 m/s.)

3.What is the speed of ultrasound in Glycerol given that a 40 kHz ultrasound pulse has a wavelength of 4·75 cm in Glycerol?

4.An 8 MHz ultrasound pulse is transmitted into water. It has a wavelength of 1·87 x 10-4 m in water. Calculate its speed.

5.A buzzer emitting sound of frequency 12 kHz is switched on. What is the wavelength of the sound waves in air where the speed of sound is 340 m/s?

6.An ultrasound pulse of frequency 7 MHz is transmitted through 8 cm of muscle. The wavelength of the ultrasound in muscle is 2·29 x 10-4 m.

(a)Calculate the speed of sound in muscle.

(b)Calculate the time taken for the ultrasound to pass through the muscle.

7.Sound waves of frequency 4 kHz travel along a 2·6 m length of aluminium in a time of 0·5 milliseconds.

(a)What is the speed of sound in aluminium?

(b)Calculate the wavelength of this sound in aluminium.

8.An ultrasound pulse, of wavelength 3·75 x 10-4 m, is transmitted into the womb of an expectant mother. It is reflected by the head of her baby and the reflected pulse is detected 0·2 ms after transmission. The baby’s head is positioned 15 cm from the transmitter / detector.

(a)Show that the speed of ultrasound is 1 500 m/s inside the woman.

(b)What frequency of ultrasound was used?

9.

/ It takes 0·02 milliseconds for a 15 kHz sound vibration to travel 6 cm through the humerus bone in your arm.
What is the wavelength of this sound in bone?

10.Assuming no energy losses, how far would sound travel in 1 second through a material in which a 2 kHz sound vibration has a wavelength of 95 cm?

Sound Level

1.David and Majhid were doing a project on sound in their physics class. They used a sound level meter to measure the noise level in decibels of various situations and then did some research using material in the school library.

(a) The list below shows the different situations the boys investigated.

Quiet lane.

Heavy lorry passing on the main road.

Noise in the class during a practical lesson.

Pneumatic drill.

Give a sound level reading in decibels for each situation.

(b)In their report the boys mentioned ‘noise pollution’. Give two examples of noisepollution.

(c)Why should workmen using pneumatic drills wear ear protectors?

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Mixed questions

  1. A sound wave is generated and the amplitude is traced against time as shown below

A second sound wave is produced and shown below

a)On hearing the sounds, what would you hear to be different?

b)Draw a wave form that would be just as loud as the first wave but of a higher pitch (a higher frequency)

c)Sound waves are examples of longitudinal waves but the waveforms shown above are transverse waves.

The diagram from the oscilloscope shows how the voltage output from a microphone would vary when a sound is made.

When a sound wave moves through air what is happening to the air molecules.

  1. soundCopy the diagram, label the equipment and explain how you would use the following set up to measure the speed of sound in air
  1. a) On a cold winters day the temperature was 0oC. A pile driver was at work. James was watching the pile driver from 500m away and noticed the time delay between seeing the pile being driven and the hearing the sound. Given that the speed of sound in air is 330m/s at 0oC what was the time delay in seconds that James experienced?

b) 6 months later Helen (using sophisticated timers) measured the time delay between seeing and hearing the pile driver to be 2.94seconds from 1km away. Calculate the speed of sound, on that day, when the air temperature was 20oC.

  1. Copy the table into your jotter, adding appropriate headings, then match the following speeds of sounds to the medium that the sound is travelling through.

340m/s / Glycerol
355m/s / Air at 20oC
6320m/s / Air at 40oC
1500m/s / Diamond
1900m/s / Aluminium
12 000m/s / Carbon Dioxide
270m/s / Water
  1. What is the definition of Noise pollution?
  2. How do ear defenders work? Include a labelled diagram to help your explanation.
  3. Sound levels are measured in decibels (dB). Suggest reasonable sound level for the following experiences:
  • Normal Conversation at 1m
  • Quiet Classroom
  • Threshold of Hearing
  • Lowest prolonged sound level at which ear defenders are recommended
  • Jet engine (close up).
  1. Explain how bats can “see” with sound.
  2. How is ultrasound used for medical applications to “see” inside the body?
  3. A fishing boat was using sonar to track a shoal of fish. The time between sending and receiving the sound pulse was 0.36 seconds. Given that the speed of sound in water is 1500m/s, what depth are the fish swimming at?

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Mixed questions - 2

  1. Thunder is heard 20 seconds after a lightning flash. If the speed of sound is 340 m/s, how far away is the storm?
  1. Explain why, during a thunder storm, you see the lightning before you hear the thunder.
  1. On a day when the speed of sound in air is 330 m/s, how long would sound take to travel a distance of 1.6 km?
  1. During a thunder storm it is noticed that the time interval between the flash of lightning and the clap of thunder gets less. What does this tell you about the storm?
  1. Describe a method of measuring the speed of sound in air giving:

a) the apparatus used

b) the measurements taken

c) any equations used in the calculation.

  1. Ten pupils are standing on Calton Hill, looking at Edinburgh Castle. They measure the time difference between seeing the smoke from the one o’clock gun and hearing the bang. The measured times are 3.8 s, 4.2 s, 4.0 s, 3.8 s, 4.4 s, 3.8 s, 4.0 s, 4.2 s, 3.6 s, and 4.2 s.

a)Calculate the average time for the group.

b)Calculate the distance from the Castle to Calton Hill if the speed of sound is 330 m/s.

  1. An explosion in Grangemouth could be heard in South Queensferry one minute later . Given they are 20 km apart, calculate the speed of sound in air.
  1. On a day when the speed of sound is 330 m/s, how long would the sound take to travel a distance of 19.8 km?
  1. In a race the runners are at different distances away from the starter. They will hear the starting horn at different times. Using the speed of sound as 340 m/s, calculate the time difference in hearing the horn for two runners who are 5 m and 15 m from the starter.
  1. a) Explain, using a diagram, the difference between a transverse and longitudinalwave.

b)What type of waves are the following:

i) sound waves

ii) water waves

iii) light waves.

  1. Explain, using the particle model, why sound travels quicker in metals than gases.
  1. Explain why sound cannot travel through a vacuum.

Waves and Optics (Int 2) – Student Material1

Numerical Solutions

Speed Distance and Time (P 2)

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Waves and Optics (Int 2) – Student Material

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Numerical Solutions

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Waves and Optics (Int 2) – Student Material

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Numerical Solutions

1a. 3000 m/s

b. 1900 m/s

c. 375 m

d. 52 m

e. 0.05 s

f. 10.5 s

20.3 m

31450 m/s

41.3 ms

50.8 m

63000m/s

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Waves and Optics (Int 2) – Student Material

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Numerical Solutions

Echoes P4

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Waves and Optics (Int 2) – Student Material

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Numerical Solutions

1a. 332 m/s

b. 0.36 s

c. 112 m

d. 172 m
2 a. 200 m

b 0.59 s

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Waves and Optics (Int 2) – Student Material

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Numerical Solutions

Ultrasound P5

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Waves and Optics (Int 2) – Student Material

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Numerical Solutions

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Waves and Optics (Int 2) – Student Material

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Numerical Solutions

1a. 0.12 m

b. 0.06 m

c. 0.1125 m
2 a. 0.15m

b. 0.01s

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Waves and Optics (Int 2) – Student Material

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Numerical Solutions

3Total distance = 2 x 0.285 = 0.57m

Use speed = distance/time = 0.57/0.00038 = 1500m/s

4a. High frequency sound waves above the range of human hearing /above 20,000 Hz

b The layer of jelly provides a good contact between the transmitter and the mother’s skin. This prevents ‘false echoes’ caused by air bubbles.

c. 0.114m

d. A high frequency sound signal is sent out from the transmitter. The signal passes through the skin and tissue until it bounces off the babies bones. The time it takes to go from the transmitter to the baby and back to the receiver allows the doctor to calculate the depth of the baby. The doctors use the image to estimate the delivery date, to monitor the growth of the baby and to check that the baby is developing normally.

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Waves and Optics (Int 2) – Student Material

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Numerical Solutions

Speed, Frequency and Wavelength P7

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Waves and Optics (Int 2) – Student Material

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Numerical Solutions

1 a.3500 m/s

b.1600 m/s

c.0.0002 m

d.0.5 m

e.600,000 Hz

f.4000 Hz

2. 0.27 m

3. 1900 m/s

4. 1496 m/s

5. 0.028 m

6 a.1603 m/s

b.50 x 10-6 s
7 a. 5200 m/s

b. 1.3 m

8 a. Distance =

0.3m,

time =0.0002s

Speed = distance/time 0.3/0.002 = 1500m/s

b.4MHz

9. speed = 3000m/s

Wavelength = 0.2m

10. speed = 1900m/s

distance = 19m

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Waves and Optics (Int 2) – Student Material

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Numerical Solutions

Sound Level P9

1a.Quiet lane

Heavy lorry passing on the main road

Noise in the class during a practical lesson

Pneumatic drill

b.Noise pollution is unwanted noise. Examples of noise pollution are traffic noise from a motorway, the noise of a football crowd when a goal is scored or noise from an airport.

c.Workmen using pneumatic drills should wear ear protectors because the volume of the drill is above the danger level. If the workmen do not wear protection they risk permanent damage to their hearing if they are exposed to the loud sound for a long period of time.

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Waves and Optics (Int 2) – Student Material

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Numerical Solutions

Mixed Questions P 10

  1. a) The second sound will be quieter but will be the same note because the frequencies (pitch) are the same.

b)

c) The air molecules tend to bunch up and spread out along the direction of the sound wave.

  1. You would measure the straight line distance between the microphones. A loud sound is made by the bell this starts the timer, when the sound reaches the first microphone the timer starts when the sound reaches the second microphone the timer is stopped and displayed on the screen. Finally you would use the equation speed = distance divided by time to work out the speed of sound.
  2. a) Time = Distance ÷ Speed

Time = 500 ÷ 330 = 1.51 seconds

b) Speed = Distance ÷ Time

Speed = 1000 ÷ 2.94 = 340m/s

4.

340m/s / Air at 20oC
355m/s / Air at 40oC
6320m/s / Aluminium
1500m/s / Water
1900m/s / Glycerol
12 000m/s / Diamond
270m/s / Carbon Dioxide
  1. Noise is a sound that is loud, unpleasant, unexpected, or undesired. Noise gets in the way of the wanted signal.
  2. Ear defenders should have a hard outside shell to reflect away sound. The shell is usually made of plastic. On the inside they should be foam like to absorb any sound that was not reflected away. The ear defenders should fit snugly so as to prevent sound entering in-between the ears and the defenders and be comfortable.
  3. Normal conversation at 1m – 60dB

Quiet classroom 30dB

Threshold of Hearing 0dB

Lowest prolonged sound level at which ear defenders are recommended- 80db

Jet engine (close up) 150dB

  1. They don't actually "see" using sound, but yes, they use a sonar-like ability called echolocation to determine distance from objects. They emit ultrasonic sound that only they can hear. When the sound hits an object such, it bounces back and the bat can determine the distance and direction to the object. Also, contrary to popular belief, bats are not blind. They have eyes and limited vision. They primarily use their vision to travel long distances.
  1. Ultrasound is passed into the body from a transducer which contains a transmitter and receiver of ultrasound. A gel is placed on the skin to expel air and provide a good match to the skin so that unwanted reflections at the probe/skin contact are reduced. The ultrasound is strongly reflected off hard surfaces such as bone or denser tissue. The returning pattern of echoes is translated by a computer into images and videos.
  2. This is an echo, so the distance calculated will be twice the depth.

Distance = Speed x Time

Distance = 1500 x 0.36 = 540m

So the fish are at a depth of 540÷2 = 270m

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Waves and Optics (Int 2) – Student Material

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Numerical Solutions

Mixed Questions – 2 P 13

  1. 6800m
  2. The speed of light is much greater than the speed of sound.
  3. 4.85s
  4. The storm is getting closer (the distance is decreasing)
  5. Equipment – two microphones connected to a fast timer, a hammer and something to strike and a metre stick.

When the hammer hits the object sound travels to the first microphone, starting the timer. When the sound reaches the second microphone it switches off the timer. The distance between the microphones is measured using a metre stick.

Calculation – speed = distance between microphones

Time on timer

  1. 4s
  2. 1320 m
  3. 333 m/s
  4. 60 s
  5. 0.029s
  6. In longitudinal waves the direction of wave travel and the direction of particle travel are along the same axis.

In transverse waves the direction of wave travel and the direction of particle travel are at right angles to one another.