Higher 1A T2 2012 Solutions

75 s1 substitute into eqn2 eqn2 1275 s1

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Unformatted text preview: p;!:(!&8+!81B(0(23&'.!0+23!#0(1,$23! &+!/+2.&*I/&$B(!$2&(*-(*(2/(%! S2P " S1P = 13 " 12 = 1m thus # = 1/2 = 0.5 m !!!!!!!!!!!!!! v = #f !! v 350 f= = = 700 Hz # 0.5 !!!!!!!!!!#$$%!K+*!">E!)(*$+,.!,$--(*(2/(J!&'(!2(8!)1&'!,$--(*(2/(!5I.&!:(!">E! 81B(0(23&'.!0+23!!!!#8'$/'!,+(.!0(1,!&+!,(.&*I/&$B(!$2&(*-(*(2/(%>!L+&'!!,$.&12/(.! ! C"M!12,!CGM!8$00!$2/*(1.(!:4!."!12,!.G>! (S2P + s2) " (S1P + s1) = 0.75 from Pythagoras : eqn(1) (S2P + s2) 2 = 25 + (S1P + s1) 2 (S2P + s2) = 12.75 + s1 !!!!!!! Substitute into eqn(2) eqn(2) ! (12.75 + s1) 2 = (12 + s1) 2 + 25 which can be solved for s1 The answer is then 12 + s1 First find s1: "3.2 s1 = = 4.29 # 4.3 m "0.75 !!!!!!!!!!!!!! new distance from S2 : 13 + s2 = 12.75 + 4.3! ! new distance from S2 = 17 m (c) Have to do (ii) before (i) to find the train speed. ! (ii) If the frequency appears lower to the observer, the source and the observer must be moving away from each other: their velocities will be negative # v...
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