Physics 11 HW #9 Solutions

Transcription

1 Phyic HW #9 Solution Chapter 6: ocu On Concept: 3, 8 Problem: 3,, 5, 86, 9 Chapter 7: ocu On Concept: 8, Problem:,, 33, 53, 6 ocu On Concept 6-3 (d) The amplitude peciie the maximum excurion o the pot rom the pot unditurbed poition, and the pot moe through thi ditance our time during each cycle. or intance, in one cycle tarting rom it unditurbed poition, the pot moe upward a ditance A, downward a ditance A (returning to it unditurbed poition), downward again a ditance A, and inally upward again a ditance A (returning to it unditurbed poition). ocu On Concept 6-8 (c) The Doppler eect caue the obered requency to be hited toward higher alue when the ource moe toward a tationary oberer and when the oberer moe toward a tationary ource. The hit to higher obered requencie i een greater when the ource and the oberer moe toward each other, a they do here. In Equation 6.5 the plu ign applie in the numerator and the minu ign in the denominator. ocu On Concept 7-8 (d) The trombone produce 6 beat eery econd, o the beat requency i 3 Hz. The econd trombone can be producing a ound whoe requency i either 438 Hz 3 Hz = 435 Hz or 438 Hz + 3 Hz = 44 Hz. ocu On Concept 7- (b) The requency o a tanding wae i directly proportional to the peed o the traeling wae that orm it (ee Equation 7.3). The peed o the wae, on the other hand, depend on the ma m o the tring through the relation = / ( m/ ), o the maller the ma, the greater i the peed and, hence, the greater the requency o the tanding wae. Problem 6-3 REASONING a. The period i the time required or one complete cycle o the wae to pa. The period i alo the time or two ucceie cret to pa the peron. b. The requency i the reciprocal o the period, according to Equation 0.5. c. The waelength i the horizontal length o one cycle o the wae, or the horizontal ditance between two ucceie cret.

2 d. The peed o the wae i equal to it requency time it waelength (ee Equation 6.). e. The amplitude A o a wae i the maximum excurion o a water particle rom the particle unditurbed poition. SOUTION a. Ater the initial cret pae, 5 additional cret pa in a time o The period T o the wae i 50.0 T = = b. Since the requency and period T are related by = /T (Equation 0.5), we hae = = = T Hz c. The horizontal ditance between two ucceie cret i gien a 3 m. Thi i alo the waelength λ o the wae, o λ = 3 m d. According to Equation 6., the peed o the wae i = λ = ( 0.00 Hz)( 3 m) = 3. m/ e. There i no inormation gien, either directly or indirectly, about the amplitude o the wae. Thereore, it i not poible to determine the amplitude. Problem 6- REASONING The length o the tring i one o the actor that aect the peed o a wae traeling on it, in o ar a the peed depend on the ma per unit length m/ according to = (Equation 6.). The other actor aecting the peed i the m/ tenion. The peed i not directly gien here. Howeer, the requency and the waelength λ are gien, and the peed i related to them according to = λ (Equation 6.). Subtituting Equation 6. into Equation 6. will gie u an equation that can be oled or the length. SOUTION Subtituting Equation 6. into Equation 6. gie = λ = Soling or the length, we ind that m/

3 3 ( ) ( ) ( ) λ m 60 Hz 0.60 m kg = = = 80 N 0.68 m Problem 6-5 REASONING The peed o the tranere pule on the wire i determined by the tenion in the wire and the ma per unit length m/ o the wire, according to = (Equation 6.). The ball ha a ma M. Since the wire upport the weight m/ Mg o the ball and ince the weight o the wire i negligible, it i only the ball weight that determine the tenion in the wire, = Mg. Thereore, we can ue Equation 6. with thi alue o the tenion and ole it or the acceleration g due to graity. The peed o the tranere pule i not gien, but we know that the pule trael the length o the wire in a time t and that the peed i = /t. SOUTION Subtituting the tenion = Mg and the peed = /t into Equation 6. or the peed o the pule on the tring gie = or = Mg m/ t m/ Soling or the acceleration g due to graity, we obtain 0.95 m 4 ( m/ ) (. 0 kg/m) t 0.06 g = = = M kg 7.7 m/ Problem 6-86 REASONING a. Since the ound ource and the oberer are tationary, there i no Doppler eect. The waelength remain the ame and the requency o the ound heard by the oberer remain the ame a that emitted by the ound ource. b. When the ound ource moe toward a tationary oberer, the waelength decreae (ee igure 6.8b). Thi decreae arie becaue the condenation bunch-up a the ource moe toward the oberer. The requency heard by the oberer increae, becaue, according to Equation 6., the requency i inerely proportional to the waelength; a maller waelength gie rie to a greater requency. c. The waelength decreae or the ame reaon gien in part (b). The increae in requency i due to two eect; the decreae in waelength, and the act that the oberer intercept more wae cycle per econd a he moe toward the ound ource.

4 SOUTION a. The requency o the ound i the ame a that emitted by the iren; = = 450 Hz The waelength i gien by Equation 6. a o 343 m / λ = = = 450 Hz 0.40 m b. According to the dicuion in Section 6.9 (ee the ubection Moing ource ) the waelength λ o the ound i gien by λ = λ T, where i the peed o the ource and T i the period o the ound. Howeer, T = / o that 6.8 m / 0.40 m 0.9 m = T= = = 450 Hz λ λ λ The requency o heard by the oberer i equal to the peed o ound diided by the hortened waelength λ : 343 m / = = = 660 Hz o λ 0.9 m c. The waelength i the ame a that in part (b), o λ = 0.9 m. The requency heard by the oberer can be obtained rom Equation 6.5, where we ue the act that the oberer i moing toward the ound ource: o o 4.0 m/ m/ = = ( 450 Hz) = 770 Hz 6.8 m/ 343 m/ Problem 6-9 REASONING Both the oberer (you) and the ource (the eagle) are moing toward each other. According to Equation 6.5, the requency o heard by the oberer i related to the requency o the ound emitted by the ource by o o + = where o and are, repectiely, the peed o the oberer and ource.

5 SOUTION Subtituting in the gien data, we ind that the requency heard by the oberer i o o 39 m/ + + ( ) 330 m/ 3 = = 3400 Hz = Hz 8 m/ 330 m/ Problem 7- REASONING When the dierence in path length traeled by the two ound wae i a,,, o waelength, the wae are out o phae and hal-integer number ( ) detructie intererence occur at the litener. The mallet eparation d between the peaker i when the dierence in path length i o a waelength, o d = λ. The waelength i, according to Equation 6., i equal to the peed o ound diided by the requency ; λ = /. SOUTION Subtituting λ = / into d = λ gie 343 m/ d = λ = = = 45 Hz m Problem 7- REASONING The beat requency i the dierence between two ound requencie. Thereore, the original requency o the guitar tring (beore it wa tightened) wa either 3 Hz lower than that o the tuning ork (440.0 Hz 3 Hz = 337 Hz) or 3 Hz higher (440.0 Hz + 3 Hz = 443 Hz): Hz 443 Hz 437 Hz } } 3-Hz beat requency 3-Hz beat requency To determine which o thee requencie i the correct one (437 or 443 Hz), we will ue the inormation that the beat requency decreae when the guitar tring i tightened SOUTION When the guitar tring i tightened, it requency o ibration (either 437 or 443 Hz) increae. A the drawing below how, when the 437-Hz requency increae, it become cloer to Hz, o the beat requency decreae. When the 443-Hz requency increae, it become arther rom Hz, o the beat requency increae. Since the problem tate that the beat requency decreae, the original requency o the guitar tring wa 437 Hz.

6 440.0 Hz 443 Hz 437 Hz } Beat requency increae } Beat requency decreae Tuning ork Original tring Tightened tring Problem 7-33 REASONING A tanding wae i compoed o two oppoitely traeling wae. The peed o thee wae i gien by = (Equation 6.), where i the tenion in the tring m/ and m/ i it linear denity (ma per unit length). Both and m/ are gien in the tatement o the problem. The waelength λ o the wae can be obtained by iually inpecting the tanding wae pattern. The requency o the wae i related to the peed o the wae and their waelength by = /λ (Equation 6.). SOUTION a. The peed o the wae i 80 N = = = 80 m/ m/ kg/m b. Two loop o any tanding wae comprie one waelength. Since the tring i.8 m long and conit o three loop (ee the drawing), the waelength i 3 ( ) λ =.8 m =. m λ.8 m c. The requency o the wae i 80 m/ = = = λ. m 50 Hz Problem 7-53 REASONING According to Equation 7.3, the length o the tring i related to it third harmonic (n = 3) requency 3 and the peed o the wae on the tring by 3 3 = 3 or = () 3 The peed o the wae i ound rom Equation 6.:

7 = (6.) m Here, i the tenion in the tring and the ratio m/ i it linear denity. SOUTION Subtituting Equation 6. into Equation (), then, gie the length o the tring a: 3 3 = = () 3 3 m Although appear on both ide o Equation (), no urther algebra i required. Thi i becaue appear in the ratio m/ on the right ide. Thi ratio i the linear denity o the tring, which ha a known alue o kg/m. Thereore, the length o the tring i N = = = m 30 Hz kg/m 3 3 ( ) 0.8 m Problem 7-6 REASONING When the dierence in path length traeled by the two ound,,, o waelength, detructie intererence wae i a hal-integer number ( ) occur at the litener. When the dierence in path length i zero or an integer number (,, 3, ) o waelength, contructie intererence occur. Thereore, we will diide the ditance by the waelength o the ound to determine i contructie or detructie intererence occur. The waelength i, according to Equation 6., the peed o ound diided by the requency ; λ = /. SOUTION a. The ditance and can be determined by applying the Pythagorean theorem to the two right triangle in the drawing: ( ) ( ) =.00 m +.83 m =.85 m.83 m.87 m.00 m ( ) ( ) =.00 m +.87 m =.500 m P Thereore, = 0.35 m. The 343 m/ waelength o the ound i λ = = = 0.34 m. Diiding the ditance 466 Hz by the waelength λ gie the number o waelength in thi ditance: 0.35 m Number o waelength = = =.5 l 0.33 m

8 Since the number o waelength i a hal-integer number ( ) occur at the litener., detructie intererence 343 m/ b. The waelength o the ound i now λ = = = 0.35 m. Diiding the ditance 977 Hz by the waelength λ gie the number o waelength in that ditance: 0.35 m Number o waelength = = = l 0.35 m Since the number o waelength i an integer number ( ), contructie intererence occur at the litener.