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Unformatted text preview: 2 + 2 Drops to half of P max at ( 2 2 ) 2 / 2 + 2 = 2 2 This is true when 2 2 = This denes two frequencies: 1 , the halfpower point above the resonance 2 , the halfpower point below the resonance 2 1 1 2 = 0 [1] 2 2 + 2 2 = 0 [1] Solve the to nd the width of the resonance at halfpower [1][2] : 2 1 2 2 ( 1 + 2 ) = 0 Factorise the difference of squares to get a term like ( 1 + 2 ) : ( 1 2 )( 1 + 2 ) = 0 Since we dont have a physical interpretation for negative frequencies the second root is a false solution. The other root gives us 1 2 = The difference in frequencies between the two midpoints is, again, This is true even at low Q ....
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This note was uploaded on 07/10/2011 for the course PHY 293 taught by Professor Pierresavaria during the Fall '07 term at University of Toronto Toronto.
 Fall '07
 PierreSavaria
 mechanics, Energy, Power, Statistical Mechanics

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