A metal wire of linear mass density of 9.8 g/m is stretched with a tension of 10 kg weight between two rigid supports 1 meter apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency n. The frequency n of the alternating source is
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In condition of resonance, frequency of a.c. will be equal to natural frequency of wire
$n = \frac{1}{2l} \sqrt{\frac{T}{m}} = \frac{1}{2 \times 1} \sqrt{\frac{10 \times 9.8}{9.8 \times 10^{-3}}} = \frac{100}{2} = 50 \mathrm{Hz}$
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