Physics 131: tutorial week 6 Waves (1)
Take the speed of sound in air at 0◦ C as 331 ms−1 and the speed of electromagnetic radiation
c = 3, 00 × 108 ms−1 .
1. A person standing in the ocean notices that after a wave crest passes by, ten more
crests pass in a time of 120 s. What is the frequency of the wave?
(0,083 Hz)
Each wave goes by in a time equal to the period of the wave, hence
10 × T = 120 s and
1
1
f =
=
= 0, 083 Hz .
T
12
2. A light wave travels through air at a speed of 3, 0 × 108 ms−1 . Red light has a
wavelength of about 6, 6×10−7 m . What is the frequency of red light? (4, 5×1014 Hz)
f =
3, 0 × 108
v
= 4, 5 × 1014 Hz .
=
λ
6, 6 × 10−7
3. The right-most key on a piano produces a sound wave that has a frequency of
4185,6 Hz . Assuming the speed of sound in air is 343 ms−1 , find the corresponding wavelength.
(8, 19 × 10−2 m)
λ =
v
343
=
= 0, 0819 m .
f
4185, 6
4. One of the Radio Zulu broadcasting frequencies is 90,8 MHz . What is the corresponding wavelength?
(3,30 m)
λ =
3, 00 × 108
v
= 3, 30 m .
=
f
90, 8 × 106
5. A person fishing from a pier observes that four wave crests pass by in 7,0 s and
estimates the distance between two successive crests as 4,0 m . The timing starts with
the first crest and ends with the fourth. What is the speed of the wave? (1,7 ms−1 )
The number of complete waves that pass by is three, hence
3
T =
= 0, 43 s .
7, 0
The estimated distance of 4,0 m between crests is the wavelength, so
v = f λ = 0, 43 × 4, 0 = 1, 7 ms−1 .
6. The wavelength of a sound wave in air is 2,76 m at 0◦ C . What is the wavelength of
this sound in water at 0◦ C ? (Use the data on p 3 of your notes.) (Hint: since the
frequency is dependent on the source, it is the same in both media.)
(12,3 m)
Since the frequency is dependent on the source, it is the same in both air and water.
vwater
vair
=
.
f =
λair
λwater
vwater
1480
Hence λwater = λair ·
= 2, 76 ×
= 12, 3 m .
vair
331
7. At a height of ten metres above the surface of a lake, a sound pulse is generated. The
echo from the bottom of the lake returns to the point of origin 0,140 s later. The air
and water temperatures are 0◦ C . How deep is the lake? (Refer to the table on page
3 of your notes.)
(59 m)
The distance travelled by the pulse in air is 2 × 10 = 20 m. Let d be the depth of the lake, then
the distance travelled by the pulse in water is 2d. If the time the pulse travels in air is ta , then
the time the pulse travels through the water is (0, 140 − ta ) . Since waves travel at constant
speed in a medium, we can use the relation s = vt . Hence
In air :
20 = 331 × ta .
In water : 2h
Hence
h
= 1480 × (0, 140 − ta ) .
20
= 12 × 1480 × 0, 140 −
=
331
59 m .
8. On a day when the temperature is 0◦ C , a man is watching as spikes are being driven
to hold a steel rail in place. The sound of each sledgehammer blow reaches him in
0,14 s through the rail and in 2,0 s through the air, after he sees the blow fall. Find
the speed of sound in the rail.
(4, 7 × 103 ms−1 )
The distance travelled by the sound is the same in air and through the rail. Hence
s = vair tair = vrail trail .
331 × 2, 0
Hence vrail =
= 4, 7 × 103 ms−1 .
0, 14
9. If the speed of sound in air at 0◦ C is 331 ms−1 , what will it be at 27◦ C ? (347 ms−1 )
VT
=
V0
Hence
r
273 + T
273
V27
where T is in degrees celsius .
r
273 + 27
= 331
= 347 ms−1 .
273
10. Calculate (i) the speed (ii) the wavelength of a sound wave in air at 45◦ C produced
by a tuning fork of frequency 400 Hz .
(357 ms−1 ; 0,893 m)
r
273 + 45
= 357 ms−1 .
273
357
V
=
= 0, 893 m .
(ii) λ =
f
400
(i) V45 = 331
11. Standing waves are set up on a string with a frequency of 20 Hz and a distance
between succesive nodes of 0,070 m . Determine the speed with which the separate
waves travel along the string.
(2,8 ms−1 )
The nodes of a standing wave are separated by a distance of half the wavelength, so λ =
2 × 0, 070 = 0, 14 m .
v = f × λ = 20 × 0, 14 = 2, 8 ms−1 .
12. With a source of sound of frequency 5000 Hz , stationary waves are formed in air at
0◦ C . The distance between successive antinodes is 3,31 cm . At a higher temperature
it is found that the separation of the antinodes with the same source is 3,43 cm .
Calculate (i) the speed of sound in air at 0◦ C (ii) the temperature at which the
second observation was made.
(331 ms−1 ; 20,2◦ C)
(i) The wavelength is obtained by taking twice the distance between the antinodes.
V = f λ = 5000 × 2 × 3, 31 = 3, 31 × 102 ms−1 .
r
V
273 + T
=
.
(ii)
V0
273
2
2
V
343
− 273 = 20, 2◦ C .
Hence T = 273
− 273 = 273 ×
V0
331
13. A tuning fork vibrates at a frequency of 521 Hz . An out-of-tune piano string vibrates
at 519 Hz . How much time separates successive beats?
(0,5 s)
Beat frequency = f1 − f2 = 521 − 519 = 2 Hz ,
1
hence
T =
= 0, 5 s .
f
14. Two guitars are slightly out of tune. When they play the ‘same’ note simultaneously,
the sounds they produce have wavelengths of 0,776 m and 0,769 m . On a day when
the speed of sound is 343 ms−1 , what beat frequency is heard?
(4 Hz)
Beat frequency = f2 − f1 =
V
343
343
V
−
=
−
= 4 Hz .
λ2
λ1
0, 769 0, 776
15. A source of frequency 256 Hz is emitting sound energy in air at 0◦ C . The velocity of
sound in air at 0◦ C is 330 ms−1 .
(i) If an observer sets up a stationary wave system by intercepting the original sound
wave with a reflecting barrier, what would be the distance between successive
nodes?
(ii) If the air temperature rises to 20◦ C , what would be the new sound wave velocity?
(iii) If the observer moves steadily towards the surce, would the apparent frequency
be higher than, the same as, or lower than the true frequency? Explain your
answer carefully but without the use of formulae.
(iv) What beat frequency would be heard if a second source is sounded, of frequency
260 Hz (the observer being stationary)?
(0,645 m ; 342 ms−1 ; higher; 4 Hz)
330
1 V
(i) We require the distance between nodes: 12 λ = ·
=
= 0, 645 m .
2
f
2
×
256
r
r
273 + T
293
(ii) V20 = V0
= 330
= 342 ms−1 .
273
273
(iii) The observer encounters more crests and troughs per unit time when moving towards the
source, hence the frequency would be higher.
(iv) f = f2 − f1 = 260 − 256 = 4 Hz .
16. At a football game, a stationary spectator is watching the halftime show. A trumpet
player in the band is playing a 784 Hz tone while marching directly towards the
spectator at a speed of 0,900 ms−1 . On a day when the speed of sound is 343 ms−1 ,
what frequency does the spectator hear?
(786 Hz)
The source is moving towards the listener so the speed must be taken as positive, hence
V
343
f′ =
·f =
× 784 = 786 Hz .
V − vs
343 − 0, 900
17. A hawk is flying directly away from a bird watcher at a speed of 11,0 ms−1 . The hawk
produces a shrill cry whose frequency is 865 Hz . The speed of sound is 343 ms−1 .
What is the frequency that the bird watcher hears?
(838 Hz)
The source is moving away from the listener so the speed must be taken as negative, hence
343
V
·f =
× 865 = 838 Hz .
f′ =
V − vs
343 + 11, 0
18. On a day when the speed of sound in air is 340,0 ms−1 a siren mounted on a car emits
a note whose frequency is 1000 Hz.
(i) Determine the frequency of the sound heard by a stationary observer when the
car approaches him with a speed of 40,00 ms−1 .
(ii) Determine the frequency heard by an observer moving toward the car with a
speed of 40,00 ms−1 while the car remains stationary.
(1133 Hz ; 1118 Hz)
V
340, 0
·f =
× 1000 = 1133 Hz .
V − vs
340, 0 − 40, 00
340, 0 + 40, 00
V + vo
·f =
× 1000 = 1118 Hz .
(ii) f ′ =
V
340, 0
(i) f ′ =
19. A speeding motor-cyclist passes a stationary police car. The car siren is started
(the car still being stationary) and the motor-cyclist hears a frequency of 900 Hz . If
the true siren frequency is 1000 Hz and the speed of sound in the air is 343 ms−1 ,
(i) calculate the speed of the motor-cyclist. The police car starts up in pursuit and
soon reaches a speed of 43,0 ms−1 , still sounding its siren. (ii) what frequency does
the motor-cyclist no hear?
(34,3 ms−1 ; 1029 Hz)
(i) The observer, being the motor-cyclist, is moving away from the source, so we must expect
the calculated velocity to be negative.
V + vo
f′ =
·f ,
V
′
f
900
hence vo =
−1 V =
− 1 343 = −34, 3 ms−1 .
f
1000
(ii) Both the source and the observer are now moving. The observer is still moving away from
the source, whilst the source is moving towards the observer. The motor-cyclist’s speed
must therefore be taken as negative and the police car’s speed as positive.
343 − 34, 3
V + vo
·f =
× 1000 = 1029 Hz .
f ′′ =
V − vs
343 − 43, 0
20. The security alarm on a parked car goes off and produces a frequency of 1000 Hz . As
you drive towards this parked car, pass it and drive away, you observe the frequency
to change by 99 Hz . Calculate your own speed. (Take the speed of sound in the air
as 343 ms−1 .)
(17 ms−1 )
For this problem it is convenient to change the formula for the doppler effect to reflect the
direction of motion of the observer and take the absolute value of the observer’s speed. We
denote the apparent frequencies when moving towards and away from the source as ft′ and fa′
respectively. Thus
V − vo
V + vo
·f
and
fa′ =
·f .
ft′ =
V
V
We know that the difference ft′ − fa′ = 99 Hz (since ft′ > fa′ ) . Now
2vo
V − vo
V + vo
′
′
·f =
−
·f ,
ft − fa =
V
V
V
hence
vo =
(ft′ − fa′ ) V
99 × 343
=
= 17 ms−1 .
2f
2 × 1000
21. A tuning fork of frequency 400 Hz is moved away from an observer and towards a flat
wall with a speed of 2 ms−1 , the air temperature being 27◦ C . What is the apparent
frequency (i) of the unreflected sound wave coming directly towards the observer and
(ii) of the sound waves coming to the observer after reflection? (iii) How many beats
are heard?
(398 Hz ; 402 Hz ; 4 Hz)
(i) The source is moving away from the observer so
V
·f
f =
V − vs
′
−1
where vs = −2 ms
and V = 331 ×
r
300
= 347 ms−1 .
273
347
× 400 = 398 Hz .
347 + 2
(ii) The source is now moving towards the observer, so
347
f′ =
× 400 = 402 Hz .
347 − 2
(iii) Beat frequency = 402 − 398 = 4 Hz .
Hence
f′ =
22. On a hot night a bat emitting ‘squeaks’ of frequency 100 kHz flies with a speed of
20 ms−1 towards a stationary bat which is emitting squeaks of the same frequency.
When the squeaks of the two bats occur together, a beat note of frequency 6 kHz is
produced, as recorded by a suitable detector near the stationary bat. Calculate the
night temperature. (Take the speed of sound in air at 0◦ C as 330 ms−1 .)
(40◦ C )
The recorder is the observer which hears f ′ =
V
V
·f =
× 105 Hz .
V − vs
V − 20
V
× 105 − 105 Hz .
V − 20
Solving the above gives V = 353, 3 ms−1 . The temperature is calculated from
r
273 + T
V
=
. Thus
V0
273
2
353, 3
T =
× 273 − 273 = 40◦ C .
330
Beat frequency = 6 × 103 = f ′ − f =
True or false questions
TRUE
1. The expression v = f λ applies to both transverse waves and
also to longitudonal ones.
2. When a sound wave passes from air into water, the same number
of waves per second must enter the water as leave the air. So the
frequency stays the same. The wavelength however, increases.
3. The ratio of the speed of sound in air at 35◦ C to that in air at
15◦ C is given by
r
V35
35
=
.
V15
15
4. Since water is denser than air, sound travels more slowly
through it.
5. At the antinodes in a stationary wave system, the resultant
displacement is always a maximum.
6. In the doppler effect, the speed of the waves emitted by the
source depends on the speed of the source.
7. The Doppler ‘shift’ for a source moving towards a stationary observer with velocity v is not the same as for an observer moving
towards a stationary source with velocity v .
FALSE
X
X
X
X
X
X
X