Physics 102 Homework Solutions: Ch 16

Transcription

1 Physics 0 Hoework Solutions: Ch 6. SSM REASONING Since light behaes as a wae, its speed, requency, and waelength λ are related to according to = λ (Equation 6.). We can sole this equation or the requency in ters o the speed and the waelength. SOUTION Soling Equation 6. or the requency, we ind that /s Hz REASONING a. The period is the tie required or one coplete cycle o the wae to pass. The period is also the tie or two successie crests to pass the person. b. The requency is the reciprocal o the period, according to Equation 0.5. c. The waelength is the horizontal length o one cycle o the wae, or the horizontal distance between two successie crests. d. The speed o the wae is equal to its requency ties its waelength (see Equation 6.). e. The aplitude A o a wae is the axiu excursion o a water particle ro the particle s undisturbed position. SOUTION a. Ater the initial crest passes, 5 additional crests pass in a tie o 50.0 s. The period T o the wae is 50.0 s T 0.0 s 5 b. Since the requency and period T are related by = /T (Equation 0.5), we hae T 0.0 s 0.00 Hz c. The horizontal distance between two successie crests is gien as 3. This is also the waelength o the wae, so 3 d. According to Equation 6., the speed o the wae is 0.00 Hz3 3. /s e. There is no inoration gien, either directly or indirectly, about the aplitude o the wae. Thereore, it is not possible to deterine the aplitude.

2 6. REASONING AND SOUTION The period o the wae is the sae as the period o the person, so T = 5.00 s. a. T 5.00 s 0.00 Hz (0.5) b. = = (0.0 )(0.00 Hz) = 4.00 /s (6.). REASONING The speed o a transerse wae on a wire is gien by / / (Equation 6.), where is the tension and / is the ass per unit length (or linear density) o the wire. We are gien that and are the sae or the two wires, and that one is twice as long as the other. This inoration, along with knowledge o the wae speed on the wire, will allow us to deterine the speed o the wae on the wire. SOUTION The speeds on the and wires are: / / Diiding the expression or by that or gies / / Noting that = 40 /s and that =, the speed o the wae on the wire is 40 /s 40 /s 340 /s 4. REASONING The length o the string is one o the actors that aects the speed o a wae traeling on it, in so ar as the speed depends on the ass per unit length / according to (Equation 6.). The other actor / aecting the speed is the tension. The speed is not directly gien here. Howeer, the requency and the waelength λ are gien, and the speed is related to the according to = λ (Equation 6.). Substituting Equation 6. into Equation 6. will gie us an equation that can be soled or the length. SOUTION Substituting Equation 6. into Equation 6. gies / Soling or the length, we ind that 3 60 Hz kg N

3 3. REASONING The waelength o a sound wae is equal to its speed diided by its requency (Equation 6.): The speed o sound in a gas is gien by Equation 6.5 as kt /, where T is the Kelin teperature and is the ass o a single air olecule. The Kelin teperature is related to the Celsius teperature T c by T = T c (Equation.), so the speed o sound can be expressed as k T c 73.5 () We know that the speed o sound is 343 /s at 0.0 C, so that k 0.0 C /s (3) Diiding Equation () by (3) gies k Tc 73.5 Tc /s k 0.0 C C 73.5 (4) SOUTION Soling Equation (4) or and substituting the result into Equation () gies T c C /s 343 /s 0.0 C C Hz 36. REASONING The speed o sound in an ideal gas depends on its teperature T through Equation 6.5 as kt, where =.40 or air, k is Boltzann s constant, and is the ass o a olecule in the air. Thus, the speed o sound depends on the air teperature, with a lower teperature giing rise to a saller speed o sound. The speed o sound is greater in the aternoon at the warer teperature and is the speed that the car ust exceed. kt SOUTION The speed o sound in air is, where the teperature T ust be expressed on the Kelin scale (T = T c + 73, Equation.). Taking the ratio o the speed o sound at 43 C to that at 0 C, we hae 43 C 0C k k K 73 K

4 The speed o sound at 43 C is 36 K 36 K 33 /s 356 /s 73 K 73 K 43 C 0 C Thereore, the speed o your car ust exceed car = 356 /s. 38. REASONING Under the assuptions gien in the proble, both the bullet and the sound o the rile discharge d trael at constant speeds. Thereore, we can use (Equation., where is speed, d is distance, and t is t elapsed tie) to deterine the distance that the bullet traels, as well as the tie it takes the rile report to reach the obserer. The speed o the bullet is gien, and we know that the speed o sound in air is 343 /s, because the air teperature is 0 C (See Table 6.). SOUTION Applying Equation., we can express the distance d b traeled by the bullet as d b t () where b = 840 /s and t is the tie needed or the report o the rile to reach the obserer. The obserer is a distance d = 5 ro the arksan, so we can use Equation. with the speed o sound = 343 /s to deterine t: b d t () Substituting Equation () into Equation (), we ind the distance traeled by the bullet beore the obserer hears the report: 840 /s5 d d b db bt b 343 /s 6 4. REASONING AND SOUTION a. In order to deterine the order o arrial o the three waes, we need to know the speeds o each wae. The speeds or air, water and the etal are: a = 343 /s, w = 48 /s, = 5040 /s The order o arrial is etal wae irst, water wae second, air wae third. b. Calculate the length o tie each wae takes to trael 5. t = (5 )/(5040 /s) = 0.05 s t w = (5 )/(48 /s) = s t a = (5 )/(343 /s)= s Thereore, the delay ties are: t = t w t = s 0.05 s = s t 3 = t a t = s 0.05 s = s

5 53. SSM WWW REASONING AND SOUTION Since the sound radiates uniorly in all directions, at a distance r ro the source, the energy o the sound wae is distributed oer the area o a sphere o radius r. Thereore, according P to I (Equation 6.9) with r = 3.8, the power radiated ro the source is 4 r P 4Ir 4 (3.60 W/ )(3.8 ) 6.5 W 55. REASONING AND SOUTION Since the sound spreads out uniorly in all directions, the intensity is unior oer any sphere centered on the source. ro text Equation 6.9, Then, I P 4 r Soling or r, we obtain /(4 ) /(4 ) I P r r I P r r I 6.00 W/ r I W/ r SSM REASONING AND SOUTION The intensity leel in decibels (db) is related to the sound intensity I according to Equation 6.0: I 0 db log I 0 where the quantity I 0 is the reerence intensity. Thereore, we hae I I I/ I 0 I 0 db log 0 db log 0 db log 0 db log I 0 I 0 I / I 0 I Soling or the ratio I / I, we ind I I db 0 db log or I I Thus, we conclude that the sound intensity increases by a actor o 000.

6 76. REASONING The obserer o the sound (the bird-watcher) is stationary, while the source (the bird) is oing toward the obserer. Thereore, the Doppler-shited obsered requency is gien by Equation 6.. This expression can be soled to gie the ratio o the bird s speed to the speed o sound, ro which the desired percentage ollows directly. SOUTION According to Equation 6., the obsered requency o is related to the requency s o the source, and the ratio o the speed o the source s to the speed o sound by Soling or s / gies o o s s s or or s / s s / o s s 50 Hz Hz o This ratio corresponds to 3.%. 78. REASONING The dolphin is the source o the clicks, and eits the at a requency s. The arine biologist easures a lower, Doppler-shited click requency o, because the dolphin is swiing directly away. The dierence between the requencies is the source requency inus the obsered requency: s o. We will use o s (Equation 6.), where s is the speed o the dolphin and is the speed o sound in seawater, to s deterine the dierence between the requencies. SOUTION Soling o s s (Equation 6.) or the unknown source requency s, we obtain s s o Thereore, the dierence between the source and obsered requencies is s s s o o o o s 8.0 /s o 500 Hz 3 Hz 5 /s