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A wire under tension vibrates with a fundamental frequency of 600 Hz. If the length of the wire is doubled, the radius is halved and the wire is made to vibrate under one ninth the tension . Then the fundamental frequency will become : |
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Answer» Solution :`v = (1)/(2l) sqrt((T)/(m))` `therefore 600 = (1)/(2l) sqrt((T)/(m)) ` v. = `(1)/(2l.) sqrt( (4 T.)/(rho PI D^(2).))` ` therefore v. = (1)/(2.2l) sqrt( (4 T//9)/(rho pi. (D^(2))/(4))) = (1)/(l)sqrt( ( T)/(9 rho pi D^(2))) = (1)/(3l) sqrt( (T)/(rho pi D^(2)))` ` therefore (v.)/(v) = (2l)/(3l)sqrt( ( T)/(rho pi. D^(2))) XX sqrt((rho pi D^(2))/(4T))= (1)/(3) ` so correct choice is a. |
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