Transcript lecture10

Speed of Waves

Speed of waves in strings (review):

v

F

  

m

/

L F

 - tension - linear density m - mass

L

– length of the string

Speed of longitudinal waves in a solid rod:

v

E

  

m

/

V E

 Young’s modulus - density m - mass V - volume

Young’s modulus (optional):

E

  

Stress: Strain:

   

F

 

L A L

0 The Young’s modulus is the stress divided by the strain.

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Speed of Waves (continued)

Speed of sound waves in a fluid:

v

B

B

  

P

V V

0

B

 - bulk modulus density

V

- volume

P

- pressure

Speed of sound waves in an ideal gas:

v

 

RT M T

t

 273 .

15  - ratio of heat capacities T - temperature in Kelvin t - temperature in Celsius M - molar mass R - gas constant R=8.31451 J/mol K

For air:

  1 .

4

v M

   28 .

9645 20 .

047

T g

/ 

m mol

/

s

  2 .

8 9645  10  2  331 .

3 

kg

0 .

606

t

m

/ /

mol s

2

Speed of Sound in Some Common Substances

Substance Speed (m/s) 1. Air (20 o C) 344 2. Helium (20 o C) 999 3. Water (0 o C) 1,402 4. Lead 1,200 5. Human tissue 1,540 6. Aluminum 6,420 7. Iron and steel 5,941

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a) Mirror reflection

 1   2  1  2

Reflection

in normal to surface out  1  2 S S ’

source image wall (mirror) b) Diffuse reflection c) Reflection from curved surfaces

acoustical mirrors (curved mirrors)

whispering galleries

4

 1

In optics:

n

1 sin  1

Example:

 2 

n

2 sin  2  1 n 1  2  2 n 2

Refraction: Snell’s Law

sin sin   2 1 

v

2

v

1    2 1

n

1

n

2 

v

2

v

1    2 1 n 3  3

f

2 

f

1 index of refraction:

n

c v n

1 sin  1 

n

2 sin  2 

n

3 sin  3  3 is independen If

n

3 

n

1   3 t from

n

2   1 •

Sound waves in atmosphere when temperature varies with height

Sound traveling against the wind

Mirage

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Doppler Effect

What do you hear when a sound-emitting object (train, car) passes you?

The sound changes in pitch (frequency) as the object goes.

Why does this happen?

v

f

 •As the object travels towards you, the distance between wavefronts is compressed; this makes it seem like  is smaller and frequency is higher.

•As the object travels away from you, the distance between wavefronts is extended; this makes it seem like  is larger and frequency is lower.

Because the speed of a wave in a medium is constant, change in  will affect change in

f

.

Velocity of listener (L):

v L

Velocity of source (S):

v S

Velocity of sound:

v f L

f S



v v

v L v S

 

v L

or

v S

is “+” if in the same direction as from listener to source and is “-” otherwise 6

Diffraction

1) What it is?

The bending of waves behind obstacles or apertures into the ”shadow region”, that can be considered as interference of many waves.

2) Haw to observe?

Diffraction is most pronounced when the wavelength of the wave is similar to the size of the obstacle or aperture.

For example, the diffraction of sound waves is commonly observed because the wavelength of sound is similar to the size of doors, etc.

The waves spread out from the opening!

•Waves will diffract around a single slit or obstacle.

•The resulting pattern of light and dark stripes on a screen is called a diffraction pattern (fringes).

•This pattern arises because different points along a slit create wavelets that interfere with each other just as a double slit would.

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3a) Diffraction from a single slit (intensity)

Minima (dark fringes):

D

sin

m

 

m

m

  1 ,  2 ,  3 ,...

Example:

In order to obtain a good single slit diffraction pattern, the slit width could be: A.  /100 ; B.  /10; C.  ; D. 10  ; E.100

 8

Example:

Light of wavelength 610 nm is incident on a slit 0.20 mm wide and the diffraction pattern is produced on a screen that is 1.5 m from the slit. What is the width of the central maximum? 

D

  610

nm

0 .

20

mm x

 1 .

5

m

2

y

 ?

y

2

y

m

 

x D

2  1 .

5

m

610  10  9 0 .

20  10  3  9 .

1510  3

m

 9 .

2

mm

Example:

Light of wavelength 687 nm is incident on a single slit 0.75 mm wide. At what distance from the slit should a screen be placed if the second dark fringe in the diffraction pattern is to be 1.7 mm from the center of the screen? 

D

  687

nm

0 .

75

mm y

 1 .

7

mm x

 ?

y

x m

D x

x

yD m

  1 .

7   3 10 2

m

 687   0 .

10 75  9 

m

10   3

m

    2 .

 7 0 .

75 0 .

687  

m

 0 .

93

m

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