GEAS SOLUTION 8. A tornado warning siren on top of a tall pole radiates sound waves uniformly in all diret diretion ions. s. At a distan distane 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 $ear fre2uenies in t$e range of %#,%#5### 340 is sensitive. T$e mean molar mass for air /a mi*t mi*tur ure e of prin prini ipa pall lly y nitr nitrog ogen en and o*yg o*ygen en00 is %8.8 %8.8 * !#,+ !#,+ 7g'm 7g'mol ol and and t$e t$e ratio of $eat apaities is < 9 !.6#.
v=
√
v=
√
$%& '
(
(1.4 ) 8.314 ( 28.8×10
-3
m)* +, k# m)*
)(
23, )
m s
!%. onsider onsider an ideali ideali4e 4ed d model model wit$ wit$ a ird /treated as a point soure0 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 ane e from from t$e t$e ir ird. =y $ow $ow many many deie deiels ls does does t$e sound sound intens intensity ity level level drop w$en you move twie as far away from t$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 aove and elow atmosp$eri pressure pa /nominally !.#!+ * !#" -a at sea level0. 1ind t$e t$e orresponding ma*imum displaement if t$e fre2ueny 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 virate 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 virate at t$eir fundamental fre2ueny5 alulate t$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 fre2ueny of a stopped organ pipe is %%# 34. a.0 3ow long is t$is stopped pipe( .0 T$e seond 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 fre2ueny does t$e listener $ear(
( ) v v v 6
( )
( ) 1 220 s
m m 15 s s f = 300 Hz m m 340 45 s s 340
s)d =0.32 m 1irst Overtone> Seond Overtone>
2 s
= 0.40
f = f 6
m s
vs
f + 9 +f ! f " 9 "f !
f = 277 Hz
f 5 =5 ( 220 Hz ) !. T$e onorde is Bying at Car$ !." at an altitude of 8###m5 w$ere t$e speed of sound is +%# m's. 3ow long after t$e plane passes diretly over$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
a9=
=
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$e rim wit$ merury at %#:. 3ow mu$ merury overBows w$en t$e temperature of t$e system is raised to !##:( T$e oeDient of linear e*pansion of t$e glass is #.6# * !# ," ,!.
?=A
( ) ? A
?= ( 20×10 m -4
2
) ( -8."×10" Pa )
4
?=-1.7×10 @
#*ass = 39#*ass -5 -1 #*ass =3 ( 0.4×10 , ) -5
#*ass =1.2×10 ,
-1
; #*ass = #*ass 0 ;& ; #*ass = ( 1.2×10 , -5
; #*ass =0.1
-1
) ( 200
%#. ou are designing an eletroni iruit element made of %+ mg of silion. T$e eletri urrent t$roug$ it adds energy at t$e rate of .6 m& 9 .6 * !#,+ H's. If your design doesnt allow any $eat transfer out of t$e element5 at w$at rate does its temperature inrease( T$e spei; $eat of silion is #" H'7gJ.
;&=
3
m< -3
;&=
; mr<r> = mr<r> 0 ;& ; mr<r> = ( 18×10 , -5
; mr<r> =2.
-1
) ( 200
3
7.4×10 (
( 23×10-" k# ) ;&=0.4"
3
; mr<r> - ; #*ass =2.
- ; #*ass = 2.7
(
705
( k#,
)
, s
3
3
!). An aluminium ylinder !# m long5 wit$ a ross,setional area of %# m%5 is to e used as a spaer etween two steel walls. At !.%: it Fust slips in etween t$e walls. &$en it warms to %%.+:5 alulate t$e stress in t$e ylinder and t$e total fore it e*erts on ea$ wall5 assuming t$at t$e walls are perfetly rigid and a onstant distane apart.
%!. A geologist wor7ing in t$e ;eld drin7s $er morning oKee out of an aluminium up. T$e up $as a mass of #.!%# 7g and is initially at %#.# : w$en s$e pours in +.## 7g of oKee initially at #.# :. &$at is t$e ;nal temperature after t$e oKee and t$e up attain t$ermal e2uilirium( /Assume t$at oKee $as t$e same spei; $eat as water and t$at t$ere is no $eat e*$ange wit$ t$e surroundings.0
<)ff = m <)ff < Car ; & <)ff
(
<)ff = ( 0.300 k# ) 410
( k#,
)(
&-70.0 ℃ )
GEAS SOLUTION /inluding t$e lid0 of #.8# m% and wall t$i7ness %.# m. It is ;lled wit$ ie5 water5 and ans of Omni,ola at #:. &$at is t$e rate of $eat Bow into t$e o* if t$e temperature of t$e outside wall is +#:(
a*m:m = ma*m:m
(
a*m:m = ( 0.120 k# ) 10
( k#,
)(
&-20.0 ℃)
<)ff a*m:m =0
(
H=kA
)(
( 0.300 k# ) 410 (
k#,
&-70.0 ℃ ) ( 0.120
(
H=12 %%. A p$ysis student wants to ool #.%" 7g of iet Omni,ola /mostly water05 initially at %":5 y adding ie initially at ,%#:. 3ow mu$ ie s$ould s$e add so t$at t$e ;nal temperature will e #: wit$ all t$e ie melted if t$e $eat apaity of t$e ontainer may e negleted(
4
1 = m:< 4.2×10
( k#
(
5
)(
0 ℃ -25.0 ℃ )
)(
(
4
H= ( 0.020 m
0 ℃ - ( -20.0 ℃ ) )
2
) ( 0."0 )
(
5."7×10
-8
W 2
m ,
4
)
( 1073, )4
H=00W %". T$e ondition alled standard temperature and pressure /ST-0 for a gas is de;ned to e a temperature of #: 9 %+.!" and a pressure of ! atm 9 !.#!+ * !#" -a. If you want to 7eep a mole of an ideal gas in your room at ST-5 $ow ig a ontainer do you need(
)
)m: 1 2 = 0 -2"000 ( m :< 42000
)
%6. A t$in s2uare steel plate5 !# m on a side5 is $eated in a la7,smit$s forge to a temperature of 8##:. If t$e emissivity is #.@#5 w$at is t$e total rate of radiation of energy(
)
( k#
30 ℃ -0 ℃ 0.020m
"
2 = m :< f 2 = m:< 3.34×10
(
=1.04×10 (
( k#,
( k#,
)
( ( 8"D400s ) s
H=AF&
3
2
( )
1 = m :< <:< ; & :<
( (
0.80m
( s
= 12
)m: =-2"D000 (
1 = m :< 2.1×10
W m,
) )(
=H
)m: = m)m:
(
H= 0.010
&="".0 ℃
)m: = ( 0.25 k# ) 410
(
&H - &E
)
(
( ( m:< 334000 k# k#
m :< = "# %+. A Styrofoam o* used to 7eep drin7s old at a pini $as total wall area
)
=
%&
(
( 1m)* ) 8.314 ( =
m)*,
5
1.013×10 Pa
=0.0224 m
3
)(
273.15, )
GEAS SOLUTION instant5 t$e t$rower is spinning at an angular speed of !#.# rad's and t$e angular speed is inreasing at "#.# rad's%. At t$is instant5 ;nd t$e tangential and entripetal omponents of t$e aeleration of t$e disus and t$e magnitude of t$e aeleration.
=22.4 %@. 1ive gas moleules $osen at random are found to $ave speeds of "##5 @##5 ##5 8##5 and )## m's. 1ind t$e rms speed.
v
2 av
=
[(
500
)( 2
m s
"00
m s
) ( 2
700
m s
)( 2
800
5
m v av =5.10× 10 2 s 5
v rms = √ v
a a = 40.0
m
2
2
s
m)
2
s
(
a=
√ a a
a=
√(
rad
2
s
2
)(
0.800m )
m 2
s
a rad
40.0
a = 8.4
dm 120 =d 1s
m s
2
2
)( 2
80.0
m s
2
)
2
m s
2
"!. A fore of "## N forms angles of @##5 6"#5 and !%##5 respetively wit$ t$e *5 y and 4 a*es. Otained t$e salar omponent of 1.
dm m 0 =d 120
( )
? x =500@ ( <)s"0+ ) =250@
v x dm m 0 d
m0
)
m
a rad = 10.0
a rad = 80.0
%. A ro7et is in outer spae5 far from any planet5 w$en t$e ro7et engine is turned on. In t$e ;rst seond of ;ring5 t$e ro7et eFets !'!%# of its mass wit$ a relative speed of %6## m's. &$at is t$e ro7ets initial aeleration(
m s
2
2
m s
2400
s
a rad = ω r 5
v rms =714
rad
2
av
√
a = 20
(
2
v rms = 5.10×10
a=-
a a = r9 a a = ( 0.800m ) 50.0
2
a=-
m s
? > =500@ ( <)s 45+ ) =354@
( ) -
m0 120s
m
?z =500@ ( <)s120+ ) =-250@ T$ereforeM
?= ( 250@ ) : ( 354@ ) G- ( 250@ ) k
2
s
%8. A disus t$rower moves t$e disus in a irle of radius 8#.# m. At a ertain
@6. In -$ysis5 w$at s$ould e t$e speed at w$i$ an oFet must travel so t$at its mass will inrease y !#? elativisti Cass
GEAS SOLUTION
m=
m0
√
1.21=
( )
1<
1.1m 0 =
2
√
( )
[ ( )] [ ( )] 1- 2 <
m0 1<
1
2
1.21 1- 2 < 2
2
2
=1
0.21< =1.21 =0.41"<
2
