solutionshw5

# solutionshw5 - \$"&"""""...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ! # ! ! !! ! \$ ! % " ! % &'(" ) *" ! , ! ! " ! ! ! , "+ ! ! !. % % % ! ! " % / ! ! ! ! % ! !% 0 % ! "# " ! ! ! " % !! % ! ! !! % 1 2' % % 3 & 0 0 0 &3 &3" & 3 & &3 3 1 4 % !! " !! 5 % 67 ∆x = 9.00 + 4.00 − 3.00 = 13 − 3.00 = 0.606 m λ= v 3 43 m s = = 1.14 m f 300 Hz ∆x 0.606 = = 0.530 of a wave λ 1.14 ∆φ = 2π ( 0.530) = 3.33 rad # ! % 6%7 4 ∆x !! 5 % ! 8 " ! " ∆x ∆x = 0.500 = f λ v v 343 f= = = 283 Hz 2 ∆x 2 ( 0.606) 1 % 67 9 : ; :& 9 '"' 6 ;%7>& 9 0>& 9 '"'? * 60>&7 9 ' 0 9 6& ; 7 6%7 ! 60>&7 < ; %= 6 \$%7>& 9 \$ ' 6'7 , ! & 0>& 9 6& ; 7 ! 0 6 ' 79 : 0609'" 7 9 '"'? 0 % 6"&7 9 '"'&2 1 ∂y ∂x 2 2 % ) y = 2 A0 sin kx cos ω t ∂2 y ∂t 2 = − 2 A0 ω 2 sin kx cos ω t = −2 A0 k 2 sin kx cos ω t % 8 1 − 2 A0 k 2 sin kx cos ω t = 2 − 2 A0 ω 2 sin kx cos ω t v ( ) ω k ! v= 1 @ % &' ! 9 9 )" !! ! 8 &@ ' '"'' '"'' ) 0 2" > A& 9 ?2"& * # 4 ! B ' "B , 1 % &) '* '"' B'"B 4 ! 8 ! ! ! ! &@ B'"B "& "2 , &' '''' C 8 &'''' ??2 "2 ??2 0 "2 22)B , 1 % ? D @ 4 @ ! &@ ?? '")? & &) ") &) ") @ ?? &) ") ") &) ") 1% # ' \$8 3*3*3 ! ! L = 2dAA = λ " L= % v 5 100 m s = = 1.16 m 4 400 Hz f 1 67 % & % " 6%7 !8 % % 67 4 f2 f1 f= 521 Hz or 525 Hz ! " -! 8 526 Hz " v = λ Tµ 2L ! 8 " & & , , ! = 1T 2L µ 2 = T2 T1 2 f 523 Hz T2 = 2 T1 = T = 0.989T1 " 526 Hz 1 f1 ! % ! = T1 − T2 T1 = 1 − 0.989 = 0.011 4 = 1.14% " % reduced by 1.14% " ...
View Full Document

## This note was uploaded on 08/31/2010 for the course PHYSICS 7E taught by Professor Staff during the Fall '08 term at UC Irvine.

Ask a homework question - tutors are online