IECEP II GEAS S.docx

February 25, 2019 | Author: Jerry Lubi | Category: Sound, Waves, Quantity, Physical Phenomena, Physical Sciences
Share Embed Donate


Short Description

Download IECEP II GEAS S.docx...

Description

GEAS SOLUTION 8. A tornado warning siren on top of a tall pole radiates sound waves uniformly in all diret diretion ions. s. At a distan distane e of !".# !".# m t$e intensity of t$e sound is #.%"# &'m%. At w$at $at dist distan an e from from t$e t$e sir siren is t$e t$e intensity #.#!# &'m%(

r 2  = r1



I2

r 2  = 75 m



0.250 0.010

W 2 m W m

2

ω 2πf k= = v v

A=

m s

10

rad m

Pmax Bk 3×10  Pa

( 1.42×10 5  Pa )

3

m s

!!. ompute t$e speed of sound waves in air at room temperature /T9%#:0 and ;nd t$e range of wavelengt$s in air to w$i$ t$e $uman ear /w$i$ an $ear fre2uenies in t$e range of %#,%#5### 340 is sensitive. T$e mean molar mass for air /a mi*t mi*tur ure e of prin prini ipa pall lly y nitr nitrog ogen en and o*yg o*ygen en00 is %8.8 %8.8 * !#,+ !#,+ 7g'm 7g'mol ol and and t$e t$e ratio of $eat apaities is < 9 !.6#.

v=



v=



 $%& '

(

 (1.4 ) 8.314 ( 28.8×10

-3

m)* +, k# m)*

)(

23, )

m s

!%. onsider onsider an ideali ideali4e 4ed d model model wit$ wit$ a ird /treated as a point soure0 emitting ons onsta tant nt soun sound d powe power5 r5 wit$ wit$ inte intens nsit ity y inversely proportional to t$e s2uare of t$e dist distan ane e from from t$e t$e ir ird. =y $ow $ow many many deie deiels ls does does t$e sound sound intens intensity ity level level drop w$en you move twie as far away from t$e ird(

-8

A=



1."×10  Pa 3 k# 11.3×10 3 m

v =344

 ( 2π rad ) ( 1000 Hz )

k =18.3



 !

v =1.2×10

). In a sinusoidal sound wave of moderate loudness t$e ma*imum pressure variations are of t$e order of +.# * !#,% -a aove and elow atmosp$eri pressure pa /nominally !.#!+ * !#" -a at sea level0. 1ind t$e t$e orresponding ma*imum displaement if t$e fre2ueny is !### 34. In air at normal atmosp$eri pressure and densit density5 y5 t$e speed speed of sound sound is +66 m's and t$e ul7 modulus is !.6% * !#" -a.

344

v=

I1

r 2  = 15 m

k=

v=

(

18.3

rad m

)

-8

A =1.2× 10  m !#. !#. &$at &$at is t$e t$e spee speed d of long longit itud udin inal al waves in a lead rod(

(

2 - 1  = 10dB *)#

I2 I0

- *)# *)#

[

I1 I0

)

2 - 1  = 10dB ( *)# *)# I 2 -*)#I0 ) - ( *)#I1 -*)#I 0 )

( )

2 - 1  = 10dB *)#

I2 I1

]

GEAS SOLUTION

( ) ( )

2 - 1  = 10dB *)#

2 - 1  = 10dB *)#

r1

2

r2

2

r1

2

/2r 1 

!". A stopped organ pipe is sounded near a guitar5 ausing one of t$e strings to virate wit$ large amplitude. &e vary t$e tension of t$e string until we ;nd t$e ma*imum amplitude. T$e string is 8#? as long as t$e stopped pipe. If ot$ t$e pipe and t$e string virate at t$eir fundamental fre2ueny5 alulate t$e ratio of t$e wave speed on t$e string to t$e speed of sound in air.

2

( )

2 - 1  = 10dB *)#

1 4

va 4 a

2 - 1  = -".0dB

vs !6. On a day w$en t$e speed of sound is +6" m's5 t$e fundamental fre2ueny of a stopped organ pipe is %%# 34. a.0 3ow long is t$is stopped pipe( .0 T$e seond overtone of t$is pipe $as t$e same wavelengt$ as t$e t$ird $armoni of an open pipe. 3ow long is t$e open pipe(

s)d =

v 4 f1 345

s)d  = 4

va

 =

 !@. If t$e siren is moving away from t$e listener wit$ a speed of 6" m's relative to t$e air and t$e listener is moving toward t$e siren wit$ a speed of !" m's relative to t$e air5 w$at fre2ueny does t$e listener $ear(

( ) v  v v  6

(  )

( ) 1 220 s

m m   15 s s f   = 300 Hz m m 340  45 s s 340

s)d =0.32 m 1irst Overtone> Seond Overtone>

2 s

 = 0.40

f   = f 6

m s

vs

f + 9 +f ! f " 9 "f !

f   = 277 Hz

f 5  =5 ( 220 Hz ) !. T$e onorde is Bying at Car$ !." at an altitude of 8###m5 w$ere t$e speed of  sound is +%# m's. 3ow long after t$e plane passes diretly over$ead will you $ear t$e soni oom(

f 5  =1100 Hz

( ) 345

f 5  = 3

m s

2 )

(

3 345 )  =

m s

9=s:

)

2 ( 1100 Hz )

-1

1 1.75

9=34.8+

(

vs = (1.75 ) 320

) = 0.470 m vs =5"0

m s

m s

)

GEAS SOLUTION

a9=

=

8000 m vs 

? =-9;& A ? 11 -5 -1 =- ( 0.7×10  Pa ) ( 2.4×10 , ) ( 5.1 , ) A

8000 m m 5"0 ( a 34.8+) s

(

)

? " =-8."×10  Pa A

=20.5s +

!8. A glass Bas7 wit$ volume %## m is ;lled to t$e rim wit$ merury at %#:. 3ow mu$ merury overBows w$en t$e temperature of t$e system is raised to !##:( T$e oeDient of linear e*pansion of t$e glass is #.6# * !# ,"  ,!.

?=A

( ) ? A

?= ( 20×10 m -4

2

) ( -8."×10" Pa )

4

?=-1.7×10 @

#*ass = 39#*ass -5 -1 #*ass =3 ( 0.4×10 , ) -5

#*ass =1.2×10 ,

-1

; #*ass = #*ass  0 ;& ; #*ass = ( 1.2×10 , -5

; #*ass =0.1
View more...

Comments

Copyright ©2017 KUPDF Inc.
SUPPORT KUPDF