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Explainanalytically how thestationarywaves are formed.Hence show that the distance between node and adjacent antinode is (lambda)/(4). A set of 48 tuning forks is arrangedin a series ofdescending frequencies such thateach fork gives 4 beatsper secondwith preceding one. The frequenciesof first fork is 1.5 timesthe frequency of the last fork, fidn the frequency of the first and 42nd tuning fork. |
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Answer» SOLUTION :AMPLITUDE of antinodesis maximum, `A=+ -2a` `A = 2a cos '(2pix)/(lambda)` `:.+ -2a = 2a cos'(2pix)/(lambda)` `rArr cos'(2pix)/(lambda) = +-1` `:. (2pix)/(lambda) = 0,pi,2pi` or `(2pir)/(lambda) = Ppi` `:. x = (Plambda)/(2) = P(lambda/2)` (where `P = 0,1,2"....."`) For `x = 0, (lambda)/(2), lambda, (3lambda)/(2), "........."` antinodesare produced. Thus,distancebetween any two successive antinodesis `lambda/2`. AMPLITUDEOF nodes is zero , `A = 0` `:. A = 2a cos '(2pix)/(lambda)` `:. 0 = 2a cos'(2pix)/(lambda)` `:. cos '(2pix)/(lambda) = 0` `:. (2pix)/(lambda) = (pi)/(2), (3pi)/(2), (5pi)/(2) "...."` `:.x= (2p - 1) lambda/4` (where ` p = 1,2 "...."`) For `x= lambda/4, (3lambda)/(4), (5lambda)/(4) "...."`nodes areproduced. Thus, distance between any two successive nodes is `lambda/2`. The distance between node and adjacentantinodesis `lambda/4`. Numerical : Given : `n_(1) = 1.5 n_(48)` and 4 beat/second are produced with preceddingone. The set of tuningforks are arrangedin decreasingorderof frequencies . `:. n_(2) = n_(1) - 4` `n_(3)= n_(2) - 4 = n_(1) - 2 xx 4` `n_(48) =n_(47) - 4 = n_(1) - 47 xx 4` `rArr n_(48) =n_(1) - 188` `rArr n_(48) = 1.5n_(48) - 188 ( :' n_(1) = 1.5n_(48))` `rArr 0.5n_(48) = 188` `rArr n_(48) = 376` `rArr n_(1) = 1.5n_(48) = 1.5 xx 376= 564` `n_(42) = n_(1) - 164 = 564 - 164` `:. n_(42) = 400 Hz`. |
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