{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Problems 6 and 7 - A hat nun-ring at 3.2 infrar i5 Che-wing...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: A hat, nun-ring at 3.2 infrar i5 Che-wing a flyii'ig insert. The half (units a 45 ka (‘ilil‘p and remix-'05 hark an (who at kHz. The HIJIC‘ECd of sound in air is 338 111%. At what speed is the hat gaining; 011 its prey“? Cni‘i‘m-t answer: 1.234?4 infs (mien-111m i 1 are) . Explanation: Let : 'E_.-‘ : 338 infra. 'E-‘g, : 3.2 nifs. 12$ : 1.96526 infraT f = 45 ka. and f1, = 45.33 kHz. The 13m}; 1‘0(‘Cii-'er~a a frciqlicii'n‘fr’ (fl _ Fm) (t. _ “5) and The hat remix-'05 an (who with frequent}? “=f _ .F (1'1 + film) I fr — f b” Eigfl::&l fm _ f (-1.1 — vbfl-‘u —|— HI) T where =er ia the speed of the hat and 'E-‘I i5; T110. 513-00131 0f the prey. Solving for «aim. we obtain 1 fm (a: — 1.1.5) _ _ 1 _ f1. (-2.! — 1.1..) “[fw+m)+3‘*[l fw+mfl 1_nu~ww 1, _ E, f (1’ + vb) III — _ fr (1.? — "2.15) 1+fwwww Since fl, (-3: _ 1:3,) _ (45.33 kHz) (338 1115’s — 3.2 111,55) f (-2.1 —|— 11.5) _ (4.5 ka) 111335 —|— 3.2 111,55) : 0.988438. than 1 — 0.988438 1 -|— 0.988438. : 1.965261‘11ffi. 1.45, = (338 111;” A1: = 1:5, — 1:53 : 3.2 11135-1 — 1.96526111fo Two loudspeakers are plat-ad (111 a wall 4.11] 111 apart. A lia’ra1'1a1‘ atauda directly; in from of 0110 of the speakers. 81.1 111 110111 The wall. The apaakara are being driven by The 5111110, alar’rrit' signal genera-tad l1}: a l1a1‘111011i1' (mavillatar 0f 1110111101113? 4400 The speed ef the sennd in air is 343 infs. Tills-"That is the phase difference flail) hetweei'i the twe waves (generated h}! eaeh speaker) when they reaeh the listener"? Cerrert answer: 8.?1819 rad (teleranee i 1 s.) . Explanation: Each speaker prednees a spherical wave of the ferin (FF or: sin(.,:=). p : kr — wit — film}. where i' is the straight—line distai'iee from the speaker te the listener. The twe speakers have equal freqnei'ieies all = Leg 2 a: and equal initial phases (slim : sign} : dim]. but the}? are at different distant-es frein the listener: The distanee te the first speaker is :1 : 81.1 in but the distanee t0 the semnd speaker is r; : (81.1 in)2 —|— (4.11] in)2 : 81.2082 111. Censeqnently. there is a phase differenre he— tween the twe waves fisz—sl (k :2 — Let — shim) _ (t... _ _ am) it >< (r2 — r1). Tho Well-’(E—lllllflhfll‘ of oath sound Wan-'0 fol— lows-a from “Ia-urolougth whirh 111 Turn follows from tho froquout'y: k2??? 2$T1f A 1.. : 80.0000 I‘flCUlll. Consoquouth’. tho phi-15o diflorom'o is A; = k >< (r2 — F1) (80.0000 rodg’m) >< >< ((81.2082 111) — (81.1 111)] 8.?1810 rad. Question 11 part 2 of 2 10 points ...
View Full Document

{[ snackBarMessage ]}