A wire having a linear density of `0.05gm // cm^(3)` . Is streched between two rigid supports with a tension `4.5 xx 10^(7)` dynes. It is observed that the wire resonates at a frequency of420 Hz. Thenext highest frequency at which thesame wire resonatesis 490Hz. Find the length of the wire.

Explanation

Let the frequency 420 Hz corresponds to j3th harmonic. The formula for pthharmonic isgiven by
`n_(p) = (p)/(2l) sqrt(((T)/(mu)))`
Hence, `420 = (p)/(2l) sqrt(((T)/(mu)))`
For thenext higher frequency P is (p+ 1), hence
`490 = (p+1)/(2l) sqrt(((T)/(mu)))`
From equations-(6.189) and(6.190), we have
`(490)/(420) = (p+1)/(p)`
Solving we get, p = 6
Substituting the valuep inequation-(6.189), we ge
`420 = (6)/(2l) ((4.5 xx10^(7))/(00.5))^(1//2)`
`l = 214.3 cm`