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A wire having a linear density of 0.05 g...

A wire having a linear density of 0.05 gm/ cm is stretched between two rigid supports with a tension of `4.5 xx 10^(7)` dynes. It is observed that the wire resonates at a frequency of 420 cycles /sec. the next resonation frequency at which the same wire resonates is 490 cycles/ sec . the length of wire is approximately.

Text Solution

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Suppose the wire vibrates at 420 Hz in its nth harmonic and at 490 Hz in its (n +1)th harmonic.
`420s^(-1) = (n)/(2L) sqrt(F//mu)…(i)`
and `490 s^(-1) = ((n+1))/(2L)sqrt(F//mu) … (ii)`
This gives `(490)/(420) = (n+1)/(n)`
or n = 6.
Putting the value in (i).
`420 s^(-1) = (6)/(2L) sqrt(((450 N)/(5.0 xx 10^(-3) kg m^(-1))) = (900)/(L) ms^(-1)`
or `L = (900)/(420) m = 2.1 m`
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