Physics 107 TUTORIAL ASSIGNMENT #7

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1 Physics 07 TUTORIL SSIGNMENT #7 Cutnell & Johnson, 7 th edition Chapter 6: Problems 5, 65, 79, 93 Chapter 7: Problems 7,, 9, 37, 48 Chapter 6 5 Suppose that sound is emitted uniormly in all directions by a public address system. The intensity at a location m away rom the sound source is 3.0 x 0-4 W/m. What is the intensity at a spot that is 78 m away? *65 Two identical riles are shot at the same time, and the sound intensity leel is 80.0 d. What would be the sound intensity leel i only one rile were shot? (Hint: The answer is not 40.0 d.) *79 ssm Two submarines are underwater and approaching each other head-on. Sub has a speed o m/s and sub has a speed o 8 m/s. Sub sends out a 550-Hz sonar wae that traels at a speed o 5 m/s. (a) What is the requency detected by sub? (b) Part o the sonar wae is relected rom and returns to. What requency does detect or this relected wae? 93 ssm listener doubles his distance rom a source that emits sound uniormly in all directions. y how many decibels does the sound intensity leel change? Chapter 7 7 ssm The drawing shows a loudspeaker and point C, where a listener is positioned. second loud-speaker is located somewhere to the right o. oth speakers ibrate in phase and are playing a 68.6-Hz tone. The speed o sound is 343 m/s. What is the closest to speaker that speaker can be located, so that the listener hears no sound? ssm Consult Multiple-Concept Example 3 or background pertinent to this problem. speaker has a diameter o 0.30 m. (a) ssuming that the speed o sound is 343 m/s, ind the diraction angle θ or a.0-khz tone. (b) What speaker diameter D should be used to generate a 6.0-kHz tone whose diraction angle is as wide as that or the.0-khz tone in part (a)?

2 9 string has a linear density o kg/m and is under a tension o 80 N. The string is.8 m long, is ixed at both ends, and is ibrating in the standing wae pattern shown in the drawing. Determine the (a) speed, (b) waelength, and (c) requency o the traeling waes that make up the standing wae. 37 Reer to Interactie Solution 7.37 to reiew a method by which this problem can be soled. The undamental requencies o two air columns are the same. Column is open at both ends, while column is open at only one end. The length o column is 0.70 m. What is the length o column? 48 Two ultrasonic sound waes combine and orm a beat requency that is in the range o human hearing. The requency o one o the ultrasonic waes is 70 khz. What is (a) the smallest possible and (b) the largest possible alue or the requency o the other ultrasonic wae?

3 SOLUTIONS Chapter 6 5. RESONING ND SOLUTION The intensity o the sound alls o according to the square o the distance according to Equation 6.9. Thereore, the intensity at the new location will be I 78 ( ) m W/m.4 0 W/m 78 m RESONING ND SOLUTION Let β and I denote the intensity leel and the intensity, respectiely, when two riles are shot. Let β and I denote the intensity leel and intensity, respectiely, when one rile is shot. Then I I β ( 0 d) log ( 0 d) log I 0 I 0 where I is the intensity o a single rile. Thereore, I I 0 β /( 0 d) 0 I I 0 β ( ).00 0 W/m W/m / 0 d I W/m β ( 0 d) log ( 0 d) log 77.0 d I W/m 79. SSM RESONING a. Since the two submarines are approaching each other head on, the requency o detected by the obserer (sub ) is related to the requency s emitted by the source (sub ) by o o + s s (6.5) where o and s are the speed o the obserer and source, respectiely, and is the speed o the underwater sound

4 b. The sound relected rom submarine has the same requency that it detects, namely, o. Now sub becomes the source o sound and sub is the obserer. We can still use Equation 6.5 to ind the requency detected by sub. SOLUTION a. The requency o detected by sub is o o 8 m/s m/s s ( 550 Hz) 570 Hz s m/s 5 m/s b. The sound relected rom submarine has the same requency that it detects, namely, 570 Hz. Now sub is the source o sound whose requency is s 570 Hz. The speed o sub is s 8 m/s. The requency detected by sub (whose speed is o m/s) is o o m/s m/s s ( 570 Hz) 590 Hz s 8 m/s 5 m/s 93. SSM RESONING ccording to Equation 6.0, the sound intensity leel β in decibels β 0 d log ( I / I ), where the (d) is related to the sound intensity I according to ( ) 0 quantity I 0 is the reerence intensity. Since the sound is emitted uniormly in all directions, the intensity, or power per unit area, is gien by position can be written as written as π π I P /(4 π r ). Thus, the sound intensity at I P /(4 r ), while the sound intensity at position can be I P /(4 r ). We can obtain the sound intensity leels rom Equation 6.0 or these two positions by using these expressions or the intensities. SOLUTION Using Equation 6.0 and the expressions or the intensities at the two positions, we can write the dierence in the sound intensity leels β between the two positions as ollows: I I β β β ( 0 d) log ( 0 d) log I 0 I 0 I / I 0 I ( 0 d) log ( 0 d ) log I / I 0 I

5 β P /(4 π r ) r r ( 0 d) log ( 0 d) log ( 0 d) log P /(4 π r ) r r r r ( 0 d) log ( 0 d) log ( 0 d) log ( / ) 6.0 d r r The negatie sign indicates that the sound intensity leel decreases. Chapter 7 7. SSM WWW RESONING The geometry o the positions o the loudspeakers and the listener is shown in the ollowing drawing. C d.00 m d y 60.0 x x The listener at C will hear either a loud sound or no sound, depending upon whether the intererence occurring at C is constructie or destructie. I the listener hears no sound, destructie intererence occurs, so nλ d d n, 3, 5, () SOLUTION Since λ, according to Equation 6., the waelength o the tone is 343 m/s λ 5.00 m 68.6 Hz Speaker will be closest to Speaker when n in Equation () aboe, so d nλ 5.00 m + d +.00 m 3.50 m From the igure aboe we hae that,

6 x (.00 m) cos m Then y (.00 m) sin m x + y d (3.50 m) or x (3.50 m) (0.866 m) 3.39 m Thereore, the closest that speaker can be to speaker so that the listener hears no sound is x + x m m 3.89 m.. SSM RESONING The diraction angle or the irst minimum or a circular opening is gien by Equation 7.: sinθ.λ / D, where D is the diameter o the opening. SOLUTION a. Using Equation 6., we must irst ind the waelength o the.0-khz tone: λ 343 m/s Hz 0.7 m The diraction angle or a.0-khz tone is, thereore, b. The waelength o a 6.0-kHz tone is θ sin. 0.7 m m λ 343 m/s Hz m Thereore, i we wish to generate a 6.0-kHz tone whose diraction angle is as wide as that or the.0-khz tone in part (a), we will need a speaker o diameter D, where D. λ (.)(0.057 m) 0.0 m sinθ sin 44

7 9. RESONING standing wae is composed o two oppositely traeling waes. The speed F o these waes is gien by (Equation 6.), where F is the tension in the string m / L and m/l is its linear density (mass per unit length). oth F and m/l are gien in the statement o the problem. The waelength λ o the waes can be obtained by isually inspecting the standing wae pattern. The requency o the waes is related to the speed o the waes and their waelength by /λ (Equation 6.). SOLUTION a. The speed o the waes is F 80 N m / L kg/m 80 m/s b. Two loops o any standing wae comprise one waelength. Since the string is.8 m long and consists o three loops (see the drawing), the waelength is.8 m ( ) λ.8 m. m 3 c. The requency o the waes is 80 m/s λ. m 50 Hz λ 37. RESONING The undamental requency o air column, which is open at both ends, is gien by Equation 7.4 with n : ( ), where is the speed o sound L in air and L is the length o the air column. Similarly, the undamental requency o air column, which is open at only one end, can be expressed using Equation 7.5 with n : ( ). These two relations will allow us to determine the length o air column. 4L SOLUTION Since the undamental requencies o the two air columns are the same, so that or 0.70 m 0.35 m L 4L ( ) ( ) L L ( )

8 48. RESONING ND SOLUTION Two ultrasonic sound waes combine and orm a beat requency that is in the range o human hearing. We know that the requency range o human hearing is rom 0 Hz to 0 khz. The requency o one o the ultrasonic waes is 70 khz. The beat requency is the dierence between the two sound requencies. The smallest possible alue or the ultrasonic requency can be ound by subtracting the upper limit o human hearing rom the alue o 70 khz. The largest possible alue or the ultrasonic requency can be determined by adding the upper limit o human hearing to the alue o 70 khz. a. The smallest possible requency o the other ultrasonic wae is 70 khz 0 khz 50 khz which results in a beat requency o 70 khz 50 khz 0 khz. b. The largest possible requency or the other wae is 70 khz + 0 khz 90 khz which results in a beat requency o 90 khz 70 khz 0 khz.