( 1.7 s ) ( 3.0 Hz) = 0.49 m

Transcription

1 Week 12 hmewrk IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAign verin thee prblem, variu detail have been changed, that the anwer will cme ut dierently. The methd t ind the lutin i the ame, but yu will need t repeat part the calculatin t ind ut what yur anwer huld have been. WebAign Prblem 1: A lngitudinal wave with a requency 3.0 Hz take 1.7 t travel the length a 2.5-m Slinky (ee Figure 16.3). Determine the wavelength the wave. REASONING AND SOLUTION Frm Equatin 16.1, we have λ = v/. But v = x/t, we ind λ = v = x t = 2.5 m ( 1.7 ) ( 3.0 Hz) = 0.49 m WebAign Prblem 2: Tunami are at-mving wave ten generated by underwater earthquake. In the deep cean their amplitude i barely nticeable, but upn reaching hre, they can rie up t the atnihing height a ix-try building. One tunami, generated the Aleutian iland in Alaka, had a wavelength 750 km and traveled a ditance 3700 km in 5.3 h. (a) What wa the peed (in m/) the wave? Fr reerence, the peed a 747 jetliner i abut 250 m/. Find the wave (b) requency and (c) perid. REASONING The peed a Tunami i equal t the ditance x it travel divided by the time t it take r the wave t travel that ditance. The requency the wave i equal t it peed divided by the wavelength λ, = v/λ (Equatin 16.1). The perid T the wave i related t it requency by Equatin 10.5, T = 1/. SOLUTION a. The peed the wave i (in m/) x m 1 h v = t 5.3 h m/ b. The requency the wave i v 190 m/ 4 = Hz λ m (16.1) c. The perid any wave i the reciprcal it requency:

2 T = Hz (10.5) WebAign Prblem 3: A teel cable cr-ectinal area i kept under a tenin. The denity teel i 7860 kg/m 3 (thi i nt the linear denity). At what peed de a tranvere wave mve alng the cable? REASONING AND SOLUTION The peed the wave n the cable i F F N v ( m / L) ρ A (7860 kg/m )( m ) m/ (16.2) WebAign Prblem 4: The drawing hw tw graph that repreent a tranvere wave n a tring. The wave i mving in the + x directin. Uing the inrmatin cntained in thee graph, write the mathematical exprein (imilar t Equatin 16.3 r 16.4) r the wave. REASONING AND SOLUTION We ind rm the graph n the let that λ = m m = m and Α = m. Frm the graph n the right we ind that Τ = = Then, = 1/(0.20 ) = 5.0 Hz. Subtituting thee int Equatin 16.3 we get 2π x y = Ain 2 π t and y = m in 10 π t 50π x λ ( ) ( ) WebAign Prblem 5: A mnatmic ideal ga (γ = 1.67) i cntained within a bx whe vlume i 2.5 m 3. The preure the ga i. The ttal ma the ga i 2.3 kg. Find the peed und in the ga. REASONING AND SOLUTION The peed und in an ideal ga i given by text Equatin 16.5: γ kt v = m

3 where γ = c p /c v, k i Bltzmann' cntant, T i the Kelvin temperature the ga, and m i the ma a ingle ga mlecule. I m TOTAL i the ma the ga ample and N i the ttal number mlecule in the ample, then the abve equatin can be written a γ kt γ N k T v (1) ( m / N) m TOTAL TOTAL Fr an ideal ga, PV = NkT, that Equatin (1) becme, TOTAL 5 3 γ PV (1.67)( Pa)(2.5 m ) 2 v = m/ m 2.3 kg WebAign Prblem 6: Suppe that und i emitted unirmly in all directin by a public addre ytem. The intenity at a lcatin 22 m away rm the und urce i. What i the intenity at a pt that i 78 m away? REASONING AND SOLUTION The intenity the und all accrding t the quare the ditance accrding t Equatin Therere, the intenity at the new lcatin will be I 78 2 ( ) 22 m = W/m = W/m 78 m WebAign Prblem 7: Yu are riding yur bicycle directly away rm a tatinary urce und and hear a requency that i 1.0% lwer than the emitted requency. The peed und i 343 m/. What i yur peed? REASONING Yu hear a requency that i 1.0% lwer than the requency emitted by the urce. Thi mean that the requency yu berve i 99.0% the emitted requency, that = Yu are an berver wh i mving away rm a tatinary urce und. Therere, the Dppler-hited requency that yu berve i peciied by Equatin 16.14, which can be lved r the bicycle peed v. SOLUTION Equatin 16.14, in which v dente the peed und, tate that 1 v = v Slving r v and uing the act that = reveal that

4 v = v 1 = ( 343 m/ ) 1 = 3.4 m/ WebAign Prblem 8: Interactive LearningWare 16.2 at prvide me pertinent backgrund r thi prblem. A cnvertible mve tward yu and then pae yu; all the while, it ludpeaker are prducing a und. The peed the car i a cntant 9.00 m/, and the peed und i 343 m/. What i the rati the requency yu hear while the car i appraching t the requency yu hear while the car i mving away? REASONING Thi prblem deal with the Dppler eect in a ituatin where the urce the und i mving and the berver i tatinary. Thu, the berved requency i given by Equatin when the car i appraching the berver and Equatin when the car i mving away rm the berver. Thee equatin relate the requency heard by the berver t the requency emitted by the urce, the peed v the urce, and the peed v und. They can be ued directly t calculate the deired rati the berved requencie. We nte that n inrmatin i given abut the requency emitted by the urce. We will ee, hwever, that nne i needed, ince will be eliminated algebraically rm the lutin. SOLUTION Equatin and are 1 1 (16.11) (16.12) Apprach Recede = = 1 v / v 1 + v / v The rati i 1 Apprach 9.00 m/ 1 v v + = = Recede 1 v / v 1 + / 343 m/ v 9.00 m/ / v v 343 m/ / v A mentined in the REASONING, the unknwn urce requency ha been eliminated algebraically rm thi calculatin.

5 WebAign Prblem 9: A 3.49-rad/ ( rpm) recrd ha a 5.00-kHz tne cut in the grve. I the grve i lcated m rm the center the recrd (ee drawing), what i the wavelength in the grve? REASONING AND SOLUTION Firt ind the peed the recrd at a ditance m rm the center: v = rω = (0.100 m)(3.49 rad/) = m/ (8.9) The wavelength i, then, λ = v/ = (0.349 m/)/( Hz) = m (16.1)

6 Practice cnceptual prblem: 6. A rpe ma m i hanging dwn rm the ceiling. Nthing i attached t the le end the rpe. A tranvere wave i traveling n the rpe. A the wave travel up the rpe, de the peed the wave increae, decreae, r remain the ame? Give a rean r yur chice. REASONING AND SOLUTION A rpe ma m i hanging dwn rm the ceiling. Nthing i attached t the le end the rpe. A tranvere wave i traveling up the rpe. The tenin in the rpe i nt cntant. The lwer prtin the rpe pull dwn n the higher prtin the rpe. I we imagine that the rpe i divided int mall egment, we ee that the egment near the tp the rpe are being pulled dwn by mre weight than the egment near the bttm. Therere, the tenin in the rpe increae a we mve up the rpe. The peed a tranvere wave n the rpe i given by Equatin 16.2: vwave = F /( m / L). Frm Equatin 16.2 we ee that, a the tenin F in the rpe increae, the peed the wave increae. Therere, a the tranvere wave travel up the rpe, the peed the wave increae. 13. Sme animal rely n an acute ene hearing r urvival, and the viible part the ear n uch animal i ten relatively large. Explain hw thi anatmical eature help t increae the enitivity the animal hearing r lw-intenity und. REASONING AND SOLUTION Animal that rely n an acute ene hearing r urvival ten have relatively large external ear part. Sund intenity i deined a the und pwer P that pae perpendicularly thrugh a urace divided by the area A that urace: I = P/ A. Fr lw intenity und, the pwer per unit area i mall. Relatively large uter ear have a greater area than maller uter ear. Hence, large uter ear intercept and direct mre und pwer int the auditry ytem than maller uter ear d. 17. A urce und prduce the ame requency underwater a it de in air. Thi urce ha the ame velcity in air a it de underwater. The berver the und i tatinary, bth in air and underwater. I the Dppler eect greater in air r underwater when the urce (a) apprache and (b) mve away rm the berver? Explain. REASONING AND SOLUTION Accrding t Table 16.1, the peed und in air at 20 C i v = 343 m/, while it value in water at the ame temperature i v = 1482 m/. Thee value inluence the Dppler eect, becaue which an berver hear a requency that i dierent rm the requency that i emitted by the urce und. Fr purpe thi quetin, we aume that = and that the peed at which the urce mve i v = 25 m/. Our cncluin, hwever, will be valid r any value and v.

7 a. The Dppler-hited requency when the urce apprache the berver i given by Equatin a = 1 ( v / v). Applying thi equatin r air and water, we ind Air Water 1 (25 m )/(343 m ) 1 (25 m )/(1482 m ) 1079 Hz 1017 Hz The change in requency due t the Dppler eect in air i greater. b. The Dppler-hited requency when the urce mve away rm the berver i given by Equatin a = 1 + ( v / v). Applying thi equatin r air and water, we ind Air 1 + (25 m )/(343 m ) 932 Hz Water 1 + (25 m )/(1482 m ) 983 Hz The change in requency due t the Dppler eect in air i greater.