{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Sound_0212_handout

# Sound_0212_handout - Sound and Human Hearing This...

This preview shows pages 1–3. Sign up to view the full content.

PHYS 0212 Sound and Human Hearing 1 Sound and Human Hearing This experiment has five parts: 3. Study the relationship between frequency and musical scales 2. Determine the velocity of sound in air with an open-open tube 4. Measure the beat frequency between two sources that are close in pitch 1. Determine the velocity of sound in air with an open-closed tube 5. Measure the harmonics of your voice The first two parts of this lab will use standing waves to determine the velocity of sound. First, let us consider standing waves on a string. Waves on a string are transverse – the displacement is perpendicular to the direction of travel. Travel y x Displacement PHYS 0212 Sound and Human Hearing 2 y x ( )() ,s i n yxt A t k x ω =+ 2 2 f T π ωπ == 2 k λ = The equation for a wave is: is the angular frequency: k is the wave number: is the wavelength This gives you the displacement y at a position x and at a time t . The frequency f is the number of crests that pass a given point per second. The units are Hertz (Hz) where 1 Hz = 1/s. The period T is the time for one complete oscillation. PHYS 0212 Sound and Human Hearing 3 When a wave encounters a barrier, it reflects and travels back the way it came. If the boundary is fixed, the reflected wave will be inverted. y x The incident and reflected waves will overlap and their displacements will add together. () , Incident wave Reflected wave We can use the trigonometric identity: ( ) sin sin cos cos sin A BA B A B ±= ± To rewrite this as: () () ,2 s i n c o s A k x t = PHYS 0212 Sound and Human Hearing 4 ( ) ( i n s i n A x A x ωω −− ( ) ( ) ( ) , sin cos cos sin sin cos cos sin y x t A t kx t kx A t kx t kx ( ) ( ) , sin cos sin cos cos sin cos sin y x t A t kx t kx t kx t kx ωω ωω =− + + ( ) ( ) ( ) c o s s i n A t k x = Note that y = 0 , when kx is a multiple of π : , 0,1,2, kx n n If the boundary is placed at x = L , so that L is the length of the string, then 22 nn Ln k πλ  =   This says that if L is equal to an integer multiple of a half-wavelength, both ends of the string ( x = 0 and x = L ) will always have zero displacement. ( ) 0, 0 (,) 0 yt yLt = = These two points are called nodes and the resulting wave is called a standing wave . 1,2,3, kL n n = =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
PHYS 0212 Sound and Human Hearing 5 X = 0 X = L Node Node Antinode n = 1 2 L λ = The nodes are half a wavelength apart. The antinode is the x position where the maximum amplitude occurs. It is half way between two nodes, so it is one quarter wavelength from either node. 2 4 PHYS 0212 Sound and Human Hearing 6 vf = 2 L n = v f = Relationship between wave velocity, frequency and wavelength. n = 2 2 2 L = Antinode Node Node Antinode 2 2 Node 2 L n = 22 vv v fn L nL  == =   2 n v L = PHYS 0212 Sound and Human Hearing 7 2 n v L = These frequencies are called harmonics. Note that if n = 1 , then 1 2 v f L = Now we can rewrite the harmonics as 1 n f nf = This frequency is called the fundamental.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

Sound_0212_handout - Sound and Human Hearing This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online