Producing sound waves Displacement

1.4 Sound Density Pressure Producing sound waves Speed of sound •Produced by compression and rarefaction of media (air) Energy and Intensity •Sound waves are longitudinal Spherical and Plane waves. resulting in displacement in the direction of propagation. Interference of sound waves • The displacements result in oscillations in density and pressure.

Frequencies of sound wave Speed of sound

Speed of sound in a fluid B infra-sonic Audible Sound ultra- sonic v = ρ ∆P B =− Bulk modulus 10 20,000 ∆V/V m Frequency (Hz) ρ= Density V Similarity to speed of a transverse wave on a string 30 0.015 elastic _property Wavelength (m) in air v = int ertial _property

Speed of sound in air

B γP v = v = ρ ρ γ is a constant that depends on the nature of the gas γ =7/5 for air. Why is the speed of P - Pressure sound higher in Helium ρ -Density than in air? Why is the speed of Since P is proportional to the absolute temperature T by the sound higher in water ideal gas law. PV=nRT than in air? v is dependent on T T v= 331 (m/s) 273

1 Energy and Intensity of sound waves o energy Find the speed of sound in air at 20 C. P = time T area A v331= 273

273+ 20 v== 331 343m/ s 273

For calculations use v=340 m/s power P 2 Intensity I = = (units W/m ) area A

Sound intensity level The ear is capable of distinguishing a wide range of sound intensities. The is a measure of the level

⎛⎞I β=10log⎜⎟ (dB) ⎝⎠Io

-12 2 Io = 10 W/m the threshold of hearing

note - decibel is a logarithmic unit. It covers a wide range of intensities.

Question Spherical and plane waves

2 What is the intensity A4r= π area of sphere of sound at a rock 2 concert? (W/m ) As sound spreads out uniformly from a point source ⎛⎞I β=10log⎜⎟ = 120 The intensity decreases as 1/r2 ⎝⎠Io ⎛⎞I 120 P log⎜⎟== 12 I 10 I = 2 ⎝⎠0 4rπ I = 1012 I0

12 12− 12 2 I = 10 I0 =⋅10 10 = 1 W/m

2 Question 1 Suppose you are standing near a that can The sound intensity of an ipod earphone can be as is blasting away with 100 W of . How far much as 120 dB. How is this possible? away from the speaker should you stand if you want to hear a sound level of 120 dB. ( assume that the sound is emitted uniformly in all directions.) A) The ipod is very powerful PP B) The area of the earphone is very small I == C) The ipod is a digital device A4rπ 2 D) Rock music can be very loud P 100W r = ==2.8m 4Iπ 4(1W/m)π 2

Question 2 The sound intensity of an ipod earphone can be as much as 120 dB. How is this possible? The sound level in a truck is 20 dB greater than the sound level in a Strarbucks cafe. If the intensity in the cafe is 10-7 W/m2 the intensity in the truck is______The earphone is placed directly in the ear. The intensity W/m2. at the earphone is the power divided by a small area.

-7 Say the area is about 1cm2. A) 20 X10 B) 10-9 -5 242−− 4 C)10 P== IA 1w /m (10 m ) = 10 W D) 20

A small amount of power produces a high intensity.

Interference of sound waves Noise canceling headphones Two sound waves superimposed

Constructive Interference Noise Wave 1

Wave 2 Destructive Interference Anti-noise

3 Interference due to path Interference difference δ=0 Constructive Interference path difference =δ =r2 –r1 r1 Wave 1 A Superposition of waves at A shows interference r2 Wave 2 due to path differences Sum In phase at x=0

Condition for constructive interference δ = mλ 1 Condition for destructive interference δ =+()m λ λ 2 where m is any integer m = 0 + 1, + 2,….

Interference Interference λ Destructive Interference δ= 2 δ=λ Constructive Interference

Interference Interference

3λ Destructive Interference Constructive interference δ= 2 δ=2λ

4 Interference of sound waves Example

Phase shift due to path differences An experiment is performed to measure the speed of sound using by separating the sound from a single source along two separate paths with different path lengths and combining them at the detector. For a frequency of 3.0

kHz (assume vsound =340 m/s);

x A) What would the smallest path difference be to observe a minimum in intensity λ v 340m / s r –r = = = = 5.7cm 2 1 2 2f 2(3x103 s−1) B) What would the smallest (non-zero) path difference be When r2 –r1 =mλ Constructive Interference to observe a maximum in intensity. r –r = λ = 11cm When r2 –r1 = (m+½) λ Destructive Interference 2 1 m is any integer

Example 14.6 Path difference for two sources.

r2-r1 =0.13 m

At position P the listener hears the first minimum in sound intensity. Find the frequency of the oscillation.

vsound =340 m/s At position P the path difference is equal to λ/2. (first minimum) destructive interference. λ =−=r r 0.13m 2 21 λ=2(0.13) = 0.26m v 340m/ s f1.31x10Hz== = 3 λ 0.26m

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