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A string under a tension of 129.6 N produces 10 beats persecond when it is vibrated along with a tuning fork. When the tension in the string is increased to 160 N, it sounds in unison with the same tuning fork. Calculate the fundamental frequency oftuning fork. . |
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Answer» Solution :Let n be the frequency of TURNING FORK. The frequency of string will be either (n+10) or (n - 10) As the tension in the string increase, its frequency INCREASES `(n infty sqrt(T))` and becomes n. This shows that the initial frequency (at = T = 129.6N) of string will be (n - 10). Hence `n - 10 = (1)/(2L) sqrt(((129.6)/(mu)))` `n = (1)/(2l) sqrt(((160)/(mu)))` Dividingequation-(6.246)by equation-(6.247),weg `(n - 10)/(n) = sqrt(((129.6)/(160)))` Solving we get n = 100 Hz. |
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