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Unformatted text preview: A skyrocket explodes 108 In above the ground.
Three observers are spaced 87 n1 apart, with
observer A directly under the point of the
explosion. 1% it t t h—STm—+—87m—1 Find the ratio of the sound intensity heard
by observer A to that heard by observer B.
Correct answer: 1.64892 . Explanation: Given: h: 108m and
d287n1. E ‘9...
co \ h.“
\ II‘Iu.
I: \ “HM
1—I \\ Marc
II x a“
q: \\TB “an
5 \\ B “H...“
— ———————— ‘I———————:=I
A' C The intensity at a distance a“ from the
source is P
I— _ 413T r2 1
and distances from the scurce to points A and
B are TA : h and TB: VhEI—dg, SD c_%_m+£ E—TEI— .312
__(1081n)24—(871n)2
“ (1081n)2 Find the ratiouof the intensity heard by ob— server A to that heard by observer C.
Correct answer: 3.59568 . Explanation:
The distance from the source to C is To: h2+(2d)2, SO I"; T3 #312
_ (108 m)2 + 4 (87 m)?
(108 m)2 The sound level produced by one singer is
85.5 dB. What would be the sound level produced
by a chorus of 33 such singers (all singing at the sarne intensity at approximately the sarne
distance as the original singer)?
Correct answer: 100.685 dB. Explanation:
The total sound intensity is the surn of the sound intensities produced by each individual
singer.
Note: This is NOT true of sound levels measured on the decibel scale.
Therefore In I T111 .
The resulting sound level in decibels is In
ﬁn — ll] log 1—0 I
2 ll] log?1 —I— ll] logo.
u = E31 + $5
2 (85.5 dB) + (15.1851 dB)
2 100.585 dB. Everyone at a party is talking equally loudly. If only one person were talking, the sound level would be
72 dB. Find the sound level when all 38 people are talking. I... = 3811; 5.0. = [10 log (38) + 72] dB 2 (15.8 + 72) dB 2 88.8 dB. A bat, rnoving at 3.2 n1/s, is chasing a
ﬂying insect. The bat emits a 45 kHz chirp
and receives back an echo at 45.33 kHz. The speed of sound in air is 338 n1/s.
At what speed is the bat gaining on its prey?
Correct ansvver: 1.23474 n1/s (tolerance ﬂ: 1 %). Explanation: Let: u : 338 rn/s.
cg. : 3.2 rn/s.
or : 1.96526 n1/s,
f = 45 kHz. and
m = 45.33 kHz. The prey receives a frequency (“H — Use)
(a — uh) f=f
and the bat receives an echo with frequency ﬂ=f————.SO (e —— eb)(e — e1.) f1 : f (e —eb)(e +15.) '. where m, is the speed of the bet and ex is the
speed of the prey.
Solving fer “LII, we obtain fr (“U _ vb)
f (1’ + it) ”I [3535) + 1i Z “ i1 ‘ MU Wm] (n+em):e—ex Since f3: (1.! — e5) _ (45.33 kHz) (338 m/s — 3.2 rn/e) f (e —— vb) _ (45 kHz) (338 m/s —— 3.2 rn/e)
: [3.988438 . then 1 — 0.988438 1 + 0.988438
2 1.96526 m/s, rum 2 (338 ni/e) £1: : 1.3g, — 111
= 3.2 m/s — 1.96526 m/s 2 1.23474111/5 . Two loudspeakers are placed en a wall
4.19 1n apart. A listener stands directly in front ef one of the speakers, 81.1 n1 frern the
wall. The speakers are being driven by the
same electric signal generated by a harrnenic
escillater of frequency 4400 Hz. The speed of the seund in air is 343 rn/s. What is the phase diiference A‘E between
the twe waves (generated by each speaker) when the}r reach the listener?
Cerrect answer: 8.71819 rad (tolerance :I: 1 %). Explanation:
Each speaker prcduces a spherical wave cf the ferrn
5P 0: sin(<p), (,0 : k?" — wt — qblﬂl. where r is the straight—line distance from the
speaker to the listener. The twe speakers
have equal frequencies all = M2 = a! and equal initial phases ail”) 2 (E5?) = Q5”), but the}r are
at different distances fren1 the listener: The
distance te the ﬁrst speaker is r1 : 81.1 n1 but the distance to the second speaker is r2 = (81.11'n)2+(4.19rn)2 = 81.2082 rn. Censequently, there is a phase difference be—
tween the twe waves £90 9'32 — 9'91
(km — wt — gblﬂl) — (krl — wt — gall”) k X (T2 —’i"'1). rThe wave—number of each sound wave fol—
lows from wavelength which in turn follows
from the frequency: 2:0 20f kzk 10 = 80.0000 rad/In. Consequently, the phase difference is £00 : kx (”Pg—T1)
= (80.6006 rad/n1) X
X ((81.2082 111) — (81.1 01))
8.71819 rad. Question 11
part 2 of 2 10 points Suppose the waves generated by each
speaker have equal pressure amplitudes 6P1mm‘ : 513an : 5PM : c.0013 Pa at the listener’s location. What is the pressure amplitude 6Pﬂa§ of
the combined sound of the two speakers?
Correct answer: 0.00089958 Pa (tolerance :I: 1%). Explanation:
The two individual waves have pressures 5P1 : 6PM“ >< sin(<,01),
5P2 2 6pm” >< sin(<,02). Adding them up to obtain the cornbined
sound wave, we ﬁnd 5P1+2 Z 5P1 — 5P2
: 5pm” >< [sinﬁplj —— sin(r,02)] 513m” X 2 cos (log ; {’01) X X sin (—902: ['01) 5Pmax X 2 cos (%) >< E 5Pﬂa§ >< sin(eenst — wt) (s)! : 0.00089958 Pa where 5pm: — 2 5PM >< is the pressure smplidude ef the combined
sound of the two speakers. Two loudspeakers are placed a distance of
5 m above and below one another, as shown,
and driven by the same source. A detector is positioned a perpendicular
distance of 12 m from the two speakers and a vertical height 4 m from the lower speaker.
This results in constructive interference at
the ﬁrst order maximum. The velocityr of sound is 345 m / s . H—lgﬂl—H Kirsua my What minimum distance (in an}r direction) should the top speaker he moved in order
to create destructive interference between the two speakers?
Correct answer: 0.303758 m (tolerance j: 1 %). Explanation: Let h24rn
d=5rn
L212rn,
_ if
3’— 2
awn?
21.5m,
y———[1.5m)—%
:—1n1, and
d (5H1)
y+§—[1.5rn]+T
24m
1132345rn/s. Constructive and destructive interference.
Phase difference: Maxims: (15 2 2111s.
Minima: gab : (2111+ 1) 1T. gbzkd, and kngll.
5 is difference in path length. Deuble Slit interference.
The rules fer determining interference max—
imum or minimum are the same for sound waves and light waves.
Thus, the path length difference is 5=a2_d1=m, (1) where n = 1 for the ﬁrst maximum. We cannot use the formula sinfi' : d since the detector is not extremelyr far from the
speakers compared to the distance between the speakers! This formula is onl},r valid when
the detector is far enough from the sources that the signals are traveling nearlyr parallel
to each other. We must go back to the deﬁni— tions and basic concepts of constructive and
destructive interference, since we know how
to relate the difference in path lengths to the
wavelength. We can start by computing the
distances the two signals travel. FIom the pic—
ture and using the Pythagorean Theorem. the
north antenna’s signal travels a distance d 2
d1: L2+(y—§)l and the south antenna’s signal travels a dis— tance
d 2 Thus. the path length difference is
5 : d2 — d1 .
For the second maximum. 521A. (f10m5 : 111A, 111 :1). Substituting the values of = 5 111, L =
12111, y=1.5111,weget 2
d1: LZl(y—g) 2
5
= (12 m)2 + (1.5 111 — in) = 12.0416 111, d 2
L2 _
+(y+2)
5111 2 = 12.6491 111 . and d2 Sc, .5 : d2 — d1
2 (12.5491 m) _ (12.0416 m)
: [3.607516 m. In order to he at the ﬁrst maximum, we need tc have 5 = (1) )1. Hence, Azgz m 20.607516m. Te prcdnce destructive interference the speaker should be moved a half—wavelength
distance farther from the detecter )1
dB : —
2
_ [1607516 m
_ 2
: [1303758 m ,
with a frequencyr 1:13 (345 m/s) :_:—:567.886H.
f )1 (0.303758 m) E The human ear canal is about 2.44 cm longr
and can be regarded as a tube open at one
end and closed at the eardrum. What is the fundamental frequencyr around
which we would expect hearing to be most sensitive? Assume the speed of sound in air
to be 345 m/ s. 1. 2.9385 kHz
2. 3.03004 kHz
3. 3.125 kHz 4. 3.22243 kHz
5. 3.32375 kHz 6. 3.42857 kHz
7. 3.53484 kHz correct Explanation: Given : L = 2.44 cm and
w = 345 m/s. Hearing would be best at the fundamental
resonance of A = 4L, so “0
f — x
f _ 1 (345 m/s) 100 cm 1 kHz
1 _ 49.44 cm) 1 m 1000 Hs : 3.53484 kHz . Given: The speed of sound in air is 344 n1/s.
151 student uses an audio oscillator of ad—
justable frequencyr to measure the depth of a water well. rI‘wo successive resonant frequen—
cies are heard at 92.4 Hz and 109.2 Hz. What is the depth of the well? Correct answer: 10.2381 1n (tolerance ﬂ: 1 %). Explanation:
Basic Concept: Fer a tube closed at one
end, resonances occur at _4L
‘2n—1’ Solution: Call L the depth of the well and 113 the speed of sound. Then for some integer
n )1 where n: 1,2,3,«~~. L: (211—1)?
= (2n— 11:3, (1) then Preeurning that f2 3:) f1, let ta : ta —— 1 for the
next reennant frequene},r f2 L:[2(a+1)—1]%
: (2m 1) 4:, (3)
then 2f2 “Us
a: as (L—4f2) (4) from where we ean obtain L
a
L = —3
2 [f2 — f1]
(344 rn/e)
2 [(109.2 HZ) — (92.4 Haj]
: 10.2381 In. me Eq. (4), we have 2
n: hL_l
“Us 2
_ 2 (11192 Hz) (10.2381 111) 1
_ 344 m/s 2
: 5. Te eheek this result, using Eq. (1) and (3), we have Us 4f1 (344 m/s]
[2 (6) _ 1] 4 (92.4 Hz)
2 102381111, and L: [214+ 1] ”3 L:[2n—1] :[2(6)+1]W rI‘he wavelengths are 21 = ”—3 = 3.72294 111
f1 22 : ”—3 : 3.15018 111.
f2 A11 aluminum rod 1.85 H] long is held at its
center. It is stroked with a rosin—coated cloth
to set up longitudinal vibrations in the fun— damental mode. The speed of sound in alu—
n1inun1is 5040 1n/s. What is the frequency of the fundamental
vibrational n1ode established in the rod?
Correct answer: 1362.16 Hz. Explanation: Since the rod is held at its center, the wave—
length }. of the fundamental vibrational mode
in the rod is equal to twice the length L of the
rod )1 = 2 L . This relation together with the formula 1.:
)1on:—
f gives us the fundamental frequency _ 2L _ 504D n1/s
_ 2 (1.85 1n)
= 1362.16 Hz. ...
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 Fall '08
 RobertR.Criss
 Physics

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