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PHYS	
  101	
  Learning	
  Object:	
  
Sound	
  Wave	
  Speed	
  &	
  Phase	
  of	
  
Medium	
  
Damian	
  Feldman-­‐Kiss	
  
12036133	
  
Sound	
  Wave	
  Speed	
  &	
  Phase	
  of	
  
Medium	
  
Do	
  sound	
  waves	
  travel	
  faster	
  through	
  water	
  or	
  
through	
  air?	
  Make	
  a	
  predicLon.	
  
	
  
Now	
  that	
  you	
  have	
  your	
  predicLon,	
  lets	
  solve	
  a	
  
simple	
  physics	
  problem	
  and	
  explore	
  the	
  science	
  
behind	
  the	
  phenomenon!	
  
QuesLon	
  
Knowing	
  that	
  your	
  physics	
  midterm	
  is	
  coming	
  up,	
  you	
  
want	
  to	
  have	
  some	
  fun	
  with	
  science	
  to	
  keep	
  your	
  brain	
  
acLve.	
  You	
  decide	
  to	
  see	
  how	
  fast	
  sound	
  travels	
  in	
  the	
  
dry	
  air	
  outside	
  the	
  pool	
  versus	
  in	
  the	
  water.	
  You	
  and	
  your	
  
pal	
  shout	
  at	
  each	
  other	
  from	
  50	
  m	
  apart	
  outside	
  the	
  pool.	
  
You	
  then	
  head	
  inside,	
  get	
  changed	
  and	
  try	
  again,	
  but	
  this	
  
Lme	
  under	
  water.	
  Calculate	
  the	
  Lme	
  it	
  takes	
  for	
  the	
  
sound	
  waves	
  to	
  travel	
  to	
  your	
  pal	
  in	
  each	
  case.	
  Assume	
  
temperature	
  is	
  20˚C	
  for	
  both	
  scenarios.	
  Some	
  helpful	
  
values	
  follow.	
  Report	
  your	
  answer	
  with	
  3	
  sig.	
  figs.	
  to	
  
emphasize	
  the	
  difference	
  in	
  Lmes.	
  
Useful	
  Info	
  
	
  
Approximate	
  velocity	
  of	
  a	
  sound	
  wave	
  in	
  dry	
  air:	
  	
  
v	
  =	
  331	
  ms-­‐1	
  +	
  (0.6	
  m(s	
  C)-­‐1)	
  x	
  T	
  
(T	
  measured	
  in	
  Celsius)	
  
	
  
v	
  =	
  √(B	
  /	
  ρ)	
  
	
  
See	
  below	
  for	
  soluLon	
  
Bulk	
  modulus	
  
(Pa)	
  
	
  
Density	
  (kg/
m3)	
  
Water(l)	
  20˚C	
  
	
  
	
  
2.2	
  x	
  109	
  
	
  
1	
  000	
  
	
  
SoluLon	
  
Strategy	
  
•  What	
  are	
  we	
  given	
  and	
  what	
  do	
  we	
  need	
  to	
  
solve?	
  	
  
•  Visualize	
  problem	
  
•  Manipulate	
  equaLons	
  and	
  solve	
  
Step	
  1:	
  Given	
  and	
  unknown	
  values	
  
BWater(l)	
  =	
  2.2	
  x	
  109	
  Pa	
  
ρWater(l)	
  =	
  1	
  000	
  kg/m3	
  
vWater(l)	
  =	
  ?	
  
T	
  =	
  20˚C	
  
vAir	
  =	
  ?	
  
d	
  =	
  50	
  m	
  
tAir	
  	
  =	
  ?	
  
tWater(l)	
  =	
  ?	
  
Step	
  2:	
  Draw	
  a	
  Picture	
  
Air:	
  
Water(l):	
  
Hey!!	
  
50	
  m	
  
50	
  m	
  
Hey!!	
  
Step	
  3:	
  EquaLons	
  
Velocity	
  of	
  a	
  sound	
  wave:	
  v	
  =	
  √(B/ρ)	
  
Approximate	
  velocity	
  of	
  a	
  sound	
  wave	
  in	
  dry	
  air:	
  
v	
  =	
  331	
  ms-­‐1	
  +	
  (0.6	
  m(s	
  C)-­‐1)	
  x	
  T	
  
	
  
Time:	
  v	
  =	
  d/t	
  =>	
  t	
  =	
  d/v	
  
Step	
  4:	
  Solve	
  Unknown	
  Values	
  
vWater(l)	
  =	
  √(B/ρ)	
  
tWater(l)	
  	
  =	
  d/v	
  =>	
  d/(√(B/ρ))	
  
=	
  50m(√(2.2	
  x	
  109	
  Pa	
  /	
  1	
  000	
  kg/m3))	
  
=	
  0.0337	
  s	
  
	
  
vsound	
  wave	
  in	
  dry	
  air	
  =	
  331	
  ms-­‐1	
  +	
  (0.6	
  m(s	
  C)-­‐1)	
  x	
  T	
  
tAir	
  	
  =	
  d/v	
  =>	
  d/(331	
  ms-­‐1	
  +	
  (0.6	
  m(s	
  C)-­‐1)	
  x	
  20	
  C)	
  
=	
  0.146	
  s	
  
	
  
Step	
  5:	
  Final	
  Answer	
  
At	
  20˚C,	
  it	
  takes	
  approximately	
  0.0337	
  s	
  for	
  
sound	
  waves	
  to	
  travel	
  50	
  m	
  in	
  liquid	
  water,	
  and	
  
approx.	
  0.146	
  s	
  in	
  dry	
  air.	
  In	
  other	
  words,	
  sound	
  
waves	
  travel	
  approx.	
  4	
  Lmes	
  faster	
  in	
  liquid	
  
water	
  than	
  in	
  air	
  under	
  these	
  condiLons.	
  
	
  
	
  
Learning	
  Goal	
  1	
  
As	
  we	
  just	
  observed,	
  the	
  propagaLon	
  speed	
  of	
  a	
  
sound	
  wave	
  depends	
  on	
  the	
  properLes	
  of	
  the	
  
medium.	
  
	
  
Let’s	
  examine	
  this	
  in	
  more	
  detail.	
  
	
  
	
  
(Dr.’s	
  Bates	
  &	
  Ropler,	
  2015,	
  Lecture	
  17)	
  
	
  
Learning	
  Goal	
  2	
  –	
  Why?	
  
A	
  sound	
  wave	
  is	
  a	
  longitudinal	
  wave—as	
  the	
  wave	
  propagates	
  
through	
  an	
  elasLc	
  medium,	
  the	
  medium	
  alternates	
  between	
  
regions	
  of	
  compression	
  and	
  rarefacLon.	
  
	
  
The	
  wave	
  speed	
  (v)	
  equals	
  the	
  square	
  root	
  of	
  the	
  bulk	
  
modulus	
  (B)	
  divided	
  by	
  the	
  density	
  of	
  the	
  medium	
  (p):	
  
	
  
v	
  =	
  √(B/ρ)	
  
	
  
The	
  bulk	
  modulus	
  is	
  defined	
  as	
  “the	
  raLo	
  of	
  the	
  change	
  in	
  
pressure	
  divided	
  by	
  the	
  fracLonal	
  change	
  in	
  volume.”	
  
	
  
(Hawkes	
  et	
  al.,	
  2014,	
  p.	
  425)	
  
	
  
Learning	
  Goal	
  2	
  –	
  Why?	
  
The	
  bulk	
  modulus	
  describes	
  how	
  resistant	
  a	
  medium	
  is	
  to	
  
compression.	
  	
  
	
  
Liquids	
  are	
  nearly	
  incompressible,	
  while	
  gases	
  are	
  compressible.	
  
Liquids	
  are	
  denser	
  than	
  gases.	
  Sound	
  waves	
  generally	
  propagate	
  faster	
  
through	
  a	
  liquid	
  medium	
  than	
  a	
  gaseous	
  medium	
  because	
  of	
  the	
  
difference	
  in	
  the	
  compressibility	
  and	
  density	
  between	
  the	
  two	
  phases.	
  	
  
	
  
But	
  wait,	
  what	
  about	
  the	
  concomitant	
  increase	
  in	
  density	
  from	
  gases	
  
to	
  liquids?	
  Would	
  this	
  not	
  compensate	
  for	
  the	
  difference	
  in	
  bulk	
  
modulus	
  between	
  the	
  two	
  phases?	
  It	
  turns	
  out	
  that	
  liquids	
  are	
  indeed	
  
so	
  resistant	
  to	
  compression	
  that	
  this	
  is	
  not	
  the	
  case.	
  Lets	
  look	
  at	
  
another	
  quesLon	
  to	
  drive	
  this	
  point	
  home.	
  
	
  
(Hawkes	
  et	
  al.,	
  2014,	
  p.	
  425-­‐426)	
  
	
  
QuesLon	
  2	
  
Mercury	
  is	
  a	
  very	
  dense	
  liquid	
  	
  
(ρ	
  =	
  13	
  534	
  kg/m3).	
  Its	
  bulk	
  modulus	
  is	
  2.85	
  x	
  
1010	
  Pa.	
  Determine	
  the	
  speed	
  of	
  sound	
  in	
  liquid	
  
mercury.	
  	
  
SoluLon	
  
v	
  =	
  √(B/ρ)	
  
v	
  =	
  √(2.85	
  x	
  1010	
  Pa	
  	
  /	
  13	
  534	
  kg/m3)	
  
v	
  =	
  1	
  450	
  m/s	
  
	
  
The	
  speed	
  of	
  sound	
  in	
  liquid	
  mercury	
  is	
  	
  
1	
  450	
  m/s.	
  
	
  
References	
  
Dr.’s	
  Bates	
  and	
  Ropler.	
  2015.	
  PHYS	
  101,	
  202	
  
Lecture	
  Slides.	
  	
  
	
  
Hawkes	
  et	
  al.	
  2014.	
  Physics	
  for	
  ScienLsts	
  and	
  
Engineers:	
  An	
  interacLve	
  Approach.	
  Revised	
  
Custom	
  Volume	
  1:	
  PHYS	
  101.	
  Custom	
  Ed	
  UBC.	
  
	
  

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Phys 101 learning object

  • 1. PHYS  101  Learning  Object:   Sound  Wave  Speed  &  Phase  of   Medium   Damian  Feldman-­‐Kiss   12036133  
  • 2. Sound  Wave  Speed  &  Phase  of   Medium   Do  sound  waves  travel  faster  through  water  or   through  air?  Make  a  predicLon.     Now  that  you  have  your  predicLon,  lets  solve  a   simple  physics  problem  and  explore  the  science   behind  the  phenomenon!  
  • 3. QuesLon   Knowing  that  your  physics  midterm  is  coming  up,  you   want  to  have  some  fun  with  science  to  keep  your  brain   acLve.  You  decide  to  see  how  fast  sound  travels  in  the   dry  air  outside  the  pool  versus  in  the  water.  You  and  your   pal  shout  at  each  other  from  50  m  apart  outside  the  pool.   You  then  head  inside,  get  changed  and  try  again,  but  this   Lme  under  water.  Calculate  the  Lme  it  takes  for  the   sound  waves  to  travel  to  your  pal  in  each  case.  Assume   temperature  is  20˚C  for  both  scenarios.  Some  helpful   values  follow.  Report  your  answer  with  3  sig.  figs.  to   emphasize  the  difference  in  Lmes.  
  • 4. Useful  Info     Approximate  velocity  of  a  sound  wave  in  dry  air:     v  =  331  ms-­‐1  +  (0.6  m(s  C)-­‐1)  x  T   (T  measured  in  Celsius)     v  =  √(B  /  ρ)     See  below  for  soluLon   Bulk  modulus   (Pa)     Density  (kg/ m3)   Water(l)  20˚C       2.2  x  109     1  000    
  • 5. SoluLon   Strategy   •  What  are  we  given  and  what  do  we  need  to   solve?     •  Visualize  problem   •  Manipulate  equaLons  and  solve  
  • 6. Step  1:  Given  and  unknown  values   BWater(l)  =  2.2  x  109  Pa   ρWater(l)  =  1  000  kg/m3   vWater(l)  =  ?   T  =  20˚C   vAir  =  ?   d  =  50  m   tAir    =  ?   tWater(l)  =  ?  
  • 7. Step  2:  Draw  a  Picture   Air:   Water(l):   Hey!!   50  m   50  m   Hey!!  
  • 8. Step  3:  EquaLons   Velocity  of  a  sound  wave:  v  =  √(B/ρ)   Approximate  velocity  of  a  sound  wave  in  dry  air:   v  =  331  ms-­‐1  +  (0.6  m(s  C)-­‐1)  x  T     Time:  v  =  d/t  =>  t  =  d/v  
  • 9. Step  4:  Solve  Unknown  Values   vWater(l)  =  √(B/ρ)   tWater(l)    =  d/v  =>  d/(√(B/ρ))   =  50m(√(2.2  x  109  Pa  /  1  000  kg/m3))   =  0.0337  s     vsound  wave  in  dry  air  =  331  ms-­‐1  +  (0.6  m(s  C)-­‐1)  x  T   tAir    =  d/v  =>  d/(331  ms-­‐1  +  (0.6  m(s  C)-­‐1)  x  20  C)   =  0.146  s    
  • 10. Step  5:  Final  Answer   At  20˚C,  it  takes  approximately  0.0337  s  for   sound  waves  to  travel  50  m  in  liquid  water,  and   approx.  0.146  s  in  dry  air.  In  other  words,  sound   waves  travel  approx.  4  Lmes  faster  in  liquid   water  than  in  air  under  these  condiLons.      
  • 11. Learning  Goal  1   As  we  just  observed,  the  propagaLon  speed  of  a   sound  wave  depends  on  the  properLes  of  the   medium.     Let’s  examine  this  in  more  detail.       (Dr.’s  Bates  &  Ropler,  2015,  Lecture  17)    
  • 12. Learning  Goal  2  –  Why?   A  sound  wave  is  a  longitudinal  wave—as  the  wave  propagates   through  an  elasLc  medium,  the  medium  alternates  between   regions  of  compression  and  rarefacLon.     The  wave  speed  (v)  equals  the  square  root  of  the  bulk   modulus  (B)  divided  by  the  density  of  the  medium  (p):     v  =  √(B/ρ)     The  bulk  modulus  is  defined  as  “the  raLo  of  the  change  in   pressure  divided  by  the  fracLonal  change  in  volume.”     (Hawkes  et  al.,  2014,  p.  425)    
  • 13. Learning  Goal  2  –  Why?   The  bulk  modulus  describes  how  resistant  a  medium  is  to   compression.       Liquids  are  nearly  incompressible,  while  gases  are  compressible.   Liquids  are  denser  than  gases.  Sound  waves  generally  propagate  faster   through  a  liquid  medium  than  a  gaseous  medium  because  of  the   difference  in  the  compressibility  and  density  between  the  two  phases.       But  wait,  what  about  the  concomitant  increase  in  density  from  gases   to  liquids?  Would  this  not  compensate  for  the  difference  in  bulk   modulus  between  the  two  phases?  It  turns  out  that  liquids  are  indeed   so  resistant  to  compression  that  this  is  not  the  case.  Lets  look  at   another  quesLon  to  drive  this  point  home.     (Hawkes  et  al.,  2014,  p.  425-­‐426)    
  • 14. QuesLon  2   Mercury  is  a  very  dense  liquid     (ρ  =  13  534  kg/m3).  Its  bulk  modulus  is  2.85  x   1010  Pa.  Determine  the  speed  of  sound  in  liquid   mercury.    
  • 15. SoluLon   v  =  √(B/ρ)   v  =  √(2.85  x  1010  Pa    /  13  534  kg/m3)   v  =  1  450  m/s     The  speed  of  sound  in  liquid  mercury  is     1  450  m/s.    
  • 16. References   Dr.’s  Bates  and  Ropler.  2015.  PHYS  101,  202   Lecture  Slides.       Hawkes  et  al.  2014.  Physics  for  ScienLsts  and   Engineers:  An  interacLve  Approach.  Revised   Custom  Volume  1:  PHYS  101.  Custom  Ed  UBC.