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14.Waves and Sound
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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 $1$ metre 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 ..... $Hz$
A
$25$
B
$50$
C
$100$
D
$200$
(AIEEE-2003)
Solution
(b) 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$$Hz$
Standard 11
Physics
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