The linear density of mass of a metal wire is ` 9 . 8 g * m ^(-1)` . It is kept between two rigid supports, 1 m apart, with a tension of 98 N . The mid-point of the wire is kept between the poles of an electromagnet driven by an alternating current . Find out the frequency of this alternating current that will produce resonance in the wire .

Explanation

Mass per unit length of the wire,
`m = 9 . 8 g * m^(-1) = 9 . 8 xx 10 ^(-3) kg * m ^(-1) ` ,
length of the wire, ` l = 1 m ` ,
tension in the wire, ` T = 98 N `
` :.` Fundamental frequency,
` n = (1) / (2l) sqrt((T)/(m)) = (1)/(2 xx 1) sqrt((98)/(9 . 8 xx 10 ^(-3))) = 50 ` Hz
At resonance, the frequency of the alternating current is also 50 Hz .