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Physics Helpline
L K Satapathy
Wave Motion Adv. Theory
Amplitudes of reflected & transmitted waves
Sound as a pressure wave
Speed of sound in a fluid
Intensity and loudness of sound
Physics Helpline
L K Satapathy
Wave Motion Adv. Theory
Amplitudes of reflected and transmitted waves
The fig. shows two strings AO and BO of equal area of cross-section joined at the
origin O. The composite string is stretched between rigid supports such that a
tension T exists in it.
A transverse wave pulse of amplitude Ai is sent along the string from A . The linear
densities of the parts AO and BO are different such that the speeds of the wave in
AO and BO are v1 and v2 respectively.
OA B
yi
yr
yt
As the pulse reaches O , it is partly reflected along OA and partly transmitted
along OB. Let Ai , Ar and At be the amplitudes of incident, reflected and
transmitted waves respectively.
Physics Helpline
L K Satapathy
Wave Motion Adv. Theory
Continuity of displacement at x = 0 gives
sin sin sini r tA t A t A t    
i r ty y y 
. . . (1)i r tA A A  
Continuity of slope at x = 0 gives i try yy
x x x
 
 
  
1 1 2
1 1 2
cos ( ) cos ( ) cos ( )i r tA t x v A t x v A t x v
v v v
  
  
     
            
     
For the incident wave (+ve x-direction)
For the reflected wave (-ve x-direction)
For the transmitted wave (+ve x-direction)
1sin ( )i iy A t x v 
1sin ( )r ry A t x v 
2sin ( )t ty A t x v 
Physics Helpline
L K Satapathy
Wave Motion Adv. Theory
1 1 2
2 2
(1) (2) 2 1i t t
v v v
A A A
v v
   
       
   
2
1 2
2
. . . (3)t i
v
A A
v v
 
   
 
2
1 2
2
(1) 1r t i i
v
Also A A A A
v v
 
     
 
2 1
1 2
. . . (4)r i
v v
A A
v v
 
   
 
1 1 2
0 , cos cos cosi r tAt x A t A t A t
v v v
  
  
     
          
     
1
1 1 2 2
. . . (2)
i r t
i r t
A A A v
A A A
v v v v
       
Physics Helpline
L K Satapathy
Wave Motion Adv. Theory
Sound as a pressure wave
Consider a sound wave , propagating through a section of the medium
with speed v as shown in the figure.
A BC D
y y+y
x
v A
In the undisturbed state ,
Length of the segment AB = x
Area of cross-section = A
 Volume of the segment V = A.x
When A is displaced to C such that AC = y
B is displaced to D such that BD = y + y
 Increase in length of the segment = y
 Increase in volume of the segment V = A.y
. . . (1)
V A y y
Strain
V A x x
  
  
 
Physics Helpline
L K Satapathy
Wave Motion Adv. Theory
Let the equation of the wave is given by sin ( )oy y t x v 
(1) cos ( )oyV y
t x v
V x v


 
     

Again bulk modulus
p V y
B p B B
V V V x
 
      
 
cos ( ) cos ( )o
o
B y
p t x v Or p p t x v
v

     
Where pressure amplitude o
o o
B y
p Bky k
v v
  
   
 
We observe that displacement is represented by a sine wave & pressure by a cosine
wave. Hence a displacement node corresponds to a pressure antinode i.e. where
pressure variation is maximum the displacement is minimum and vice versa.
Physics Helpline
L K Satapathy
Wave Motion Adv. Theory
Speed of sound in a fluid M N
p+pp
A
v
x+xx
Consider a sound wave propagating with speed v
through a segment of fluid of area of cross-section A
and bounded by M and N such that MN = x
Let the excess pressures at M and N be p and p + p
 Force towards right at point M = pA
and Force towards left at point N = ( p + p)A
 Net Force on the segment towards right = – A.p
From the previous section , we have sin . . . (1)o
x
y y t
v

 
  
 
cos . . . (2)oB y x
and p t
v v


 
  
 
Physics Helpline
L K Satapathy
Wave Motion Adv. Theory
2
2
sin . . sin .o oB y B yx x
p t x t x
v v v v v
 
 
      
               
      
2
2
. .sinoAB y x
F A p x t
v v


 
         
 
Mass of the segment is m = .A.x
 Acceleration
2
2
sin . . . (3)oB yF x
a t
m v v



  
      
2
2
2
(1) sin . . . (4)o
y x
Also a y t
t v
 
  
       
2
2
2 2
(3)&(4) 1o
o
B y B B
y v
v v


  
     
Physics Helpline
L K Satapathy
Wave Motion Adv. Theory
Intensity and loudness of sound
For a normal ear , the threshold intensity for audibility is 12 2
10oI W m

And the maximum intensity (threshold of pain) is 2
1I W m
 Loudness of sound of intensity Io is  10 1010log 10log 1 0o
o
o
I
L dB
I
 
   
 
For threshold of pain 12
10 10 1012
1
10log 10log 10log (10 ) 120
10o
I
L dB dB
I 
   
      
  
Loudness of sound of intensity I is defined as 1010log
o
I
L dB
I
 
  
 
Physics Helpline
L K Satapathy
Wave Motion Adv. Theory
(iii) Difference in loudness when I2 = 2 I1
1 2
1 10 2 1010log & 10log
o o
I I
L dB L dB
I I
   
    
   
2 1 2
2 1 10 10 10
1
10 log log 10log o
o o o
II I I
L L
I I I I
      
           
       
2
10 10
1
10log 10log 2 10 0.301
I
I
 
    
 
2 1 3.01 3L L dB   
(ii) For I = 100 Io , we have
10 1010log 10log (100) 20 2
o
I
L dB dB bel
I
 
    
 
(i) For I = 10 Io , we have 10 1010log 10log (10) 10 1
o
I
L dB dB bel
I
 
    
 
Physics Helpline
L K Satapathy
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Wave Motion Theory 6/ Advanced Theory

  • 1. Physics Helpline L K Satapathy Wave Motion Adv. Theory Amplitudes of reflected & transmitted waves Sound as a pressure wave Speed of sound in a fluid Intensity and loudness of sound
  • 2. Physics Helpline L K Satapathy Wave Motion Adv. Theory Amplitudes of reflected and transmitted waves The fig. shows two strings AO and BO of equal area of cross-section joined at the origin O. The composite string is stretched between rigid supports such that a tension T exists in it. A transverse wave pulse of amplitude Ai is sent along the string from A . The linear densities of the parts AO and BO are different such that the speeds of the wave in AO and BO are v1 and v2 respectively. OA B yi yr yt As the pulse reaches O , it is partly reflected along OA and partly transmitted along OB. Let Ai , Ar and At be the amplitudes of incident, reflected and transmitted waves respectively.
  • 3. Physics Helpline L K Satapathy Wave Motion Adv. Theory Continuity of displacement at x = 0 gives sin sin sini r tA t A t A t     i r ty y y  . . . (1)i r tA A A   Continuity of slope at x = 0 gives i try yy x x x        1 1 2 1 1 2 cos ( ) cos ( ) cos ( )i r tA t x v A t x v A t x v v v v                                For the incident wave (+ve x-direction) For the reflected wave (-ve x-direction) For the transmitted wave (+ve x-direction) 1sin ( )i iy A t x v  1sin ( )r ry A t x v  2sin ( )t ty A t x v 
  • 4. Physics Helpline L K Satapathy Wave Motion Adv. Theory 1 1 2 2 2 (1) (2) 2 1i t t v v v A A A v v                 2 1 2 2 . . . (3)t i v A A v v         2 1 2 2 (1) 1r t i i v Also A A A A v v           2 1 1 2 . . . (4)r i v v A A v v         1 1 2 0 , cos cos cosi r tAt x A t A t A t v v v                              1 1 1 2 2 . . . (2) i r t i r t A A A v A A A v v v v        
  • 5. Physics Helpline L K Satapathy Wave Motion Adv. Theory Sound as a pressure wave Consider a sound wave , propagating through a section of the medium with speed v as shown in the figure. A BC D y y+y x v A In the undisturbed state , Length of the segment AB = x Area of cross-section = A  Volume of the segment V = A.x When A is displaced to C such that AC = y B is displaced to D such that BD = y + y  Increase in length of the segment = y  Increase in volume of the segment V = A.y . . . (1) V A y y Strain V A x x        
  • 6. Physics Helpline L K Satapathy Wave Motion Adv. Theory Let the equation of the wave is given by sin ( )oy y t x v  (1) cos ( )oyV y t x v V x v            Again bulk modulus p V y B p B B V V V x            cos ( ) cos ( )o o B y p t x v Or p p t x v v        Where pressure amplitude o o o B y p Bky k v v          We observe that displacement is represented by a sine wave & pressure by a cosine wave. Hence a displacement node corresponds to a pressure antinode i.e. where pressure variation is maximum the displacement is minimum and vice versa.
  • 7. Physics Helpline L K Satapathy Wave Motion Adv. Theory Speed of sound in a fluid M N p+pp A v x+xx Consider a sound wave propagating with speed v through a segment of fluid of area of cross-section A and bounded by M and N such that MN = x Let the excess pressures at M and N be p and p + p  Force towards right at point M = pA and Force towards left at point N = ( p + p)A  Net Force on the segment towards right = – A.p From the previous section , we have sin . . . (1)o x y y t v         cos . . . (2)oB y x and p t v v         
  • 8. Physics Helpline L K Satapathy Wave Motion Adv. Theory 2 2 sin . . sin .o oB y B yx x p t x t x v v v v v                                   2 2 . .sinoAB y x F A p x t v v                 Mass of the segment is m = .A.x  Acceleration 2 2 sin . . . (3)oB yF x a t m v v              2 2 2 (1) sin . . . (4)o y x Also a y t t v              2 2 2 2 (3)&(4) 1o o B y B B y v v v           
  • 9. Physics Helpline L K Satapathy Wave Motion Adv. Theory Intensity and loudness of sound For a normal ear , the threshold intensity for audibility is 12 2 10oI W m  And the maximum intensity (threshold of pain) is 2 1I W m  Loudness of sound of intensity Io is  10 1010log 10log 1 0o o o I L dB I         For threshold of pain 12 10 10 1012 1 10log 10log 10log (10 ) 120 10o I L dB dB I                Loudness of sound of intensity I is defined as 1010log o I L dB I       
  • 10. Physics Helpline L K Satapathy Wave Motion Adv. Theory (iii) Difference in loudness when I2 = 2 I1 1 2 1 10 2 1010log & 10log o o I I L dB L dB I I              2 1 2 2 1 10 10 10 1 10 log log 10log o o o o II I I L L I I I I                            2 10 10 1 10log 10log 2 10 0.301 I I          2 1 3.01 3L L dB    (ii) For I = 100 Io , we have 10 1010log 10log (100) 20 2 o I L dB dB bel I          (i) For I = 10 Io , we have 10 1010log 10log (10) 10 1 o I L dB dB bel I         
  • 11. Physics Helpline L K Satapathy For More details: www.physics-helpline.com Subscribe our channel: youtube.com/physics-helpline Follow us on Facebook and Twitter: facebook.com/physics-helpline twitter.com/physics-helpline