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A metal wire of diameter1 mm is held on two knife by a distance50 cm . The tension in the wire is 100 N. The wire vibrating with its fundamental frequency and a vibrationg tuning fork together produce5 beats//s. The tension in the wire is then reduced to 81 N. When the two excited , beats are heard at the same rate . Calculate i. the frequency of the fork and ii. the density of material of wire |
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Answer» Solution :Let the FREQUENCY of the TUNING fork be `n` . When tension in the string is decreased , the number of beats remains unchanged. This means initially the frequency of metal wire is higher than that of fork . As number of beats ` = 5 per second` . Therefore initial frequency of wire ` n_(1) = n + 5 per second`. Therefore initial frequency of wire ` n_(1) = n + 5` So , `n_(1) = n + 5 (1)/(2 l) SQRT(((T_(1))/(m))) = (1)/( 2 xx 0.5 ) sqrt(((100)/(m)))`(i) [ Since `T = 100 N , l = 50 cm = 0.50 m`] When tension is reduced to `T_(2) = 81 N`, the frequency of wire becomes `n_(2) = n - 5` . Therefore , `n_(2) = n - 5 = (1)/(2 l) sqrt (((T_(2))/(m))) = (1)/( 2 xx 0.5 ) sqrt(((81)/(m)))`(ii) i. Dividing Eq. (i) by Eq. (ii), we GET `( n + 5)/( n - 5) = sqrt (((100)/(81))) = (10)/(9)` Solving frequency of fork , `n = 95 cycles//s.` ii. We have ` m = "mass per unit LENGTH" = Ad = pi r^(2) d`. Substituting this in Eq. (i) , we get ` n + 5 = (1)/( 2 xx 0.5) sqrt (((100)/( pi r^(2) d)))` here `n = 95 , r = ( 1mm)/( 2) = (1)/(2) xx 10^(-3) m = 5 xx 10^(-4) m` Putting all value s` 95 + 5 = (1)/( 1.0) xx(100)/([ sqrt (3.14 xx(5 xx 10^(-4))^(2) d)])` ` d = 12.7 xx 10^(3) kg// m^(3)` |
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