CHAPTER 9 SOUND WAVES. - As a tuning fork vibrates a series of condensations and rarefactions moves

Transcription

1 CHAPTER 9 SOUND WAVES 9.1 Producing a sound waes - As a tuning fork ibrates a series of condensations and rarefactions moes away outward, away from the fork. - Sound is a wae that propagate in solids, liquid and gas. Figure 9.1-Pressure fluctuations in air. - Figure 6.1 showing, as the tuning fork ibrates, a series of condensations and rarefactions moes outward, away from the fork. 9.2 Characteristic of sound waes -Sound waes are longitudinal waes traeling through a medium such as air. -Longitudinal waes means that the motion of the particles is back and forth along the direction in which the wae traels. 1

2 -Sound waes, while traeling through the air, making an alternating series of regions of molecular density and air pressure. 2

3 - Regions of high density and air pressure are called compression or condensation. - Regions of lower than normal density are called rarefaction. - Sound frequency spectrum consists of three regions, f < 20Hz infrasonic region 20 Hz < f < 20 khz audible region 20 khz < f ultrasonic region 9.3 The speed of sound - The speed of sound in a medium depends on : 1. elasticity 2. mass/density 3. temperature of the medium - The speed of sound in solid = ρ where Y = Young s modulus = density of the medium. - The speed of sound in fluid = B ρ 3

4 where B = Bulk modulus = density of the medium. - The speed also depends on the temperature of the medium. And it is written as: = (331 m/s ) 1 T 273 where 331 m/s is the speed of sound in air at 0 0 C and T is the temperature in degrees Celsius. Example 9.1 A sound wae has a frequency of 700Hz in air and a waelength of 0.50m. What is the temperature of the air? Solution: The speed of the sound wae is = f = (0.500 m) x (700 Hz) = 350 m/s. Therefore, = 350 m/s = (331 m/s) 1 + T 273 from which T = 32.2 C Example 9.2 The density of a certain metal solid is kg/m 3 and its Young s modulus is N/m 2. What is the elocity of sound in this metal? 4

5 Answer: 3700 m/s Example 9.3 The range of human hearing extends from approximately 20 Hz to Hz. Find the waelengths of these extremes at a temperature of 27 0 C. Solution: The speed of sound at 27 C is: = (331 m/s) = 347 m s. Thus, the waelengths are: 20 Hz = f = 347 m/s 20.0 Hz = 17.3 m, And 20,000 Hz = f = 347 m/s 2.0 x 10 4 Hz = 1.7 x 10-2 m. 9.4 The Doppler Effect - In general, a Doppler effect is experienced wheneer there is relatie motion between a source of waes and an obserer. - When the source and obserer are moing toward each other, the obserer hears a frequency higher than the frequency of the source in the absence of relatie motion. - When the source and obserer are moing away from each other, the obserer hears a frequency lower than the source frequency. 5

6 - This effect is study base on three main cases: A. Obserer moing (towards or away) while the source stationary. B. Source moing (towards or away) while the obserer stationary. C. Source (towards or away) and obserer (towards or away) in motion. - First case: A. Obserer moing while the source stationary. Hence the frequency will be heard when the obserer moing toward the source: f 0 = f s 0 The frequency heard by the obserer traeling away from the source: f o = f s 0 these two equation can be incorporate into one: f o = f s 0 The positie sign is used when the obserer is moing toward the source, and the negatie sign is used when the obserer is moing away from the source. 6

7 - Second case: B. Source moing while the obserer stationary. The frequency heard by obserer A as the source moing toward the obserer: f o = f s ( s ) When the source moing away, the stationary obserer will hears a decreased frequency of: f o = f s ( s ) - So for the source moing (towards and away) while the obserer stationary, the equation can be written as: f o = f s s - Third case: C. Source and obserer in motion. If both source and the obserer are in motion, one finds the following general relationship for the obsered frequency: 7

8 f o = f s 0 s the upper signs refer to motion of one toward the other, and the lower signs refer to motion of one away from the other. Examples 9.7 (Obserer moing while the source stationary) A train at rest emits a sound at a frequency of 1000Hz. An Obserer in a car traels away from the sound source at a speed of 30 m/s. What is the frequency heard by the obserer? (Taking = 345 m/s as the sound of air). Solutions: Using, f o = f s 0 f o = (1000 Hz) 345 m/s m/s 345 m/s = 913 Hz. 8

9 Examples 9.8 (Source moing while the obserer stationary) An airplane traeling with half the speed of sound (=172 m/s) emits a sound of frequency 5.00 khz. At what frequency does a stationary listener hear the sound as the plane approaches? Solutions: If the source is approaching at half the speed of sound (s = - 2 ), We hae : f o =f s ( ) = 2f = 10.0 khz. - /2 Examples 9.9 (Source and obserer in motion) Two trains on separate tracks moe toward one another. Train 1 has a speed of 130km/h and train 2 has a speed 90.0 km/h. Train 2 blows its horn, emitting a frequency of 500 Hz. What is the frequency heard by the engineer on train 1? Solution: ± o Both obserer and source are moing, so we use f o = f( ± s )with V s = 90.0 km/h = 25.0 m/s ( source moes toward obserer) and V 0 = 130 km/h = 36.1 m/s ( obserer moes toward source) 345 m/s m/s Thus, f o = (500 Hz) 345 m/s m/s = 595 Hz. 9

10 9.5 Interference of sound waes - Since sound is a wae, it has all the wae characteristics can be reflected, refracted and diffracted. - Sound wae interfere when they meet. - Constructie interference - when two waes meet in a region where they are in phase. - Destructie interference when two waes meet in a region where they are not in phase. - Interference can be constructie or destructie depending on the path-length difference ( L ) of the waes. L n, n=0,1,2, constructie interference L m / Standing waes, m=1,3,5 destructie interference - A combination of incident and reflected waes according to superposition principle standing waes. - Interfering waes of the same frequency, waelength and amplitude in opposite directions in a rope can produce a standing wae. - It was called standing waes because it appears as it is not moing. Node amplitude is zero Antinode the wae ibrating at the highest amplitude. 10

11 Figure (a) : no ibration Figure (b) : L First harmonic Figure (c) : 2 L 2 2 Second harmonic Figure (d) : 3 L 3 2 Third harmonic General equation : n L n 2 where n = 1,2,3, or n n - In general, f n n n nf n 1 n F T 11

12 where n = 1,2,3,4. f 2 = L = 2 = 2 f 1 - The next highest frequency, f 3 = 3 = 3 f 1 - The frequency of = f 1, 2f 1, 3 f 1 form a harmonic series where f 1 the lowest frequency of the ibrating string and is called as fundamental frequency / first harmonic. Example 9.10 Find the first four harmonic of a 1.0m long string if the string has a mass perunit length of kg/m and is under a tension of 80N. Solution: The speed of the wae on the string is = F = 200 m/s then, f 1 = 1 F = 100 Hz. The frequencies of the next three modes are f 2 = 2f 1 f 3 =3f 1 f 4 = 4f 1 that is, 12

13 f 2 = 200Hz f 3 = 300Hz f 4 = 400Hz. 9.7 Forced ibrations and resonance. - Forced ibration This is a phenomenon where a system is forced to ibrate by giing it a driing force. - Resonance A phenomenon where a system of spring ibrate at a maximum amplitude because the frequency of the driing force equal to the system natural frequency, f 0 ( or called the resonant frequency) 9.8 Standing waes in air columns. - Musical instrument proide good examples of standing waes. - Stringed musical instrument produce notes by setting up transerse standing waes in strings with different fundamental frequencies. : eg: guitar / iolin - Standing waes are also set up in wind instrument. For example, consider a pipe organ with fixed pipe lengths, which may be open (Open at both ends) or closed (Closed at one end and open at the other end). 13

14 1. Open Tube a pipe open at both ends; - Natural frequency for a pipe open at both ends: f n n n = 1,2,3 2. Closed Tube a pipe closed at one end and open at the other end Natural frequency for a pipe open at one ends and closed the other : f m m m = 1,3,5 4L 14

15 Example 9.11 A pipe has a length of 2.46m a) Determine the frequencies of the first three harmonics if the pipe is open at each end. Take 345 m/s as the speed of sound in air. b) What are the lowest possible frequencies if the pipe is closed at one end and open at the other? Solutions a) n = 1, using f 1 = 345m/s = 2(2.46m) = 70 Hz. So, f 2 = 2f 1 = 140Hz ; f 3 = 3f 1 = 210 Hz. Using the same method, b) n =1, f 1 = 4 L = 345m/s 4(2.46m) = 35Hz, odd harmonics, f 3 = 105 Hz ; f 5 = 175Hz Example 9.12 A 3.00 m long pipe is in a room where the temperature is 20 0 C. Case 1 : Open pipe a) What is the fundamental frequency? b) What is the frequency of the second harmonic? Case 2 : Closed pipe Solution: a) What is the fundamental frequency? b) What is the frequency of the second harmonic? The speed of sound at 20 C is: =(331 m/s ).a) For an open pipe f n n nf1 n= 1,2,3 T 1 =343 m 273 s. 15

16 f 343m / s Hz 2(3) 2 1 and f f 2(57.2) 114Hz b) For a closed pipe f m m m = 1,3,5 4L f 343m / s Hz 4L 4(3) 6 1 Een harmonic cannot exist in a closed pipe. There is no second harmonic. 9.9 Beats. - Another interesting interference effect occurs when two tones of the nearly same frequency ( f 1 f 2 ) are sounded simultaneously. - The ear sense pulsations in loudness known as beats. - The beat frequency is equal to the difference between the two frequencies. 16

17 Beat frequency ( f b ) gien by, f b = f 1 - f 2 Example 9.24 How many beats will be heard in 5 second, when two forks with frequencies of 516 Hz and 513 Hz are sounded simultaneously. Solutions f b = f 1 - f 2 = 516 Hz Hz = 3 Hz ( 3 beats in each second) so, 3 5 = 15 beats will be heard. 17