Phys 2214 hw5 Solutions - Phys 2214 Assignment #5 Solutions...

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Phys 2214 Assignment #5 Solutions 1. Optical standing waves in a laser (a) standing wave frequencies: L nc c n f n f L n 2 so , 2 1 1 1 , n = 1,2,3, … (b) (i) f c L f f f f n n n 4 )] ( [ 0 0 min max and Δ n = 6 (ii) For Δ n = 1, cm m f c L 5 . 7 075 . 0 4 3. Superposition and Energy ) ( y and ) ( 2 2 1 1 vt x f vt x f y . Define vt x u . (a) For an individual wave: 2 1 2 1 2 1 1 , 1 , 1 , 1 , 2 1 2 2 u f x u u f x y u u u u P P K Tot , where the pulse equation was used. This is allowed since we’re dealing with a single wave. For the wave described by ) ( 2 1 vt x f y , 1 , 2 1 2 1 2 1 , 4 4 ) 2 ( 2 1 2 2 Tot P Tot u u f x u u f x y u u . Here the pulse equation could be used again since the two waves are moving in the same direction (see lecture notes). (b) The two pulses are shown in the figure. (i) For an individual pulse, the energy density is: 2 2 2 2 4 2 / 2 W A W A x y u u P T , and the total energy is W A W u E T T 2 4 . (ii) Since there are two separated pulses, the total energy of the string is just the sum of the energies of the two pulses: W A E T 2 8 ) string ( . (iii) When the two pulses are completely overlapping, the velocity of the string is zero everywhere. For example, the right edge of the pulse is made up of the leading edge of the original left pulse and the trailing edge of the original right pulse. The leading edge of the original left pulse has a velocity given by (using the pulse equation): W Av W A v t y 2 2 / . A W A W v v W 2 A
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This note was uploaded on 01/02/2012 for the course PHYSICS 2214 at Cornell University (Engineering School).

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Phys 2214 hw5 Solutions - Phys 2214 Assignment #5 Solutions...

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