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A metal wire of linear mass density of 9.8 g/m is stretched with a tension of 10 kg-wt 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

A Sreekhar , 15 Years ago
Grade 12
anser 3 Answers
anuj kumar sharma

Last Activity: 10 Years ago

the wire will vibrate with the same frequency as that of source .this can be considered as anexample of forced vibration.
T=10*9.8N=98N
M=9.8*10^-3Kg/m
Frequencyof wire,f=1/2Lroot(T/M)
                          f=1/2*1root(98/9.8*10^-3)
                           f=50Hz is the Answer.

Adeeba

Last Activity: 6 Years ago

f=1/2l√(T/m)As for1/2(1/2)/

Adeeba

Last Activity: 6 Years ago

The wire vibrate with the same frequency as that of source
f=1/2l√(T/m)
As the wire is at the middle point l=1/2
f=(2/2)/√(10*9.8)/9.8*10^-3
f=100

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