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Awire having a linear mass density of `5.0xx10^(-3)kgm^(-1)` is stretched between two rigid supports with a tension of 45N. The wire resonates at a frequency of 420Hz. The next higher frequency at which the wire resonates is 490Hz. Find the length of the wire.

Text Solution

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Here, `m=5.0xx10^(-3)kg//m.` Let l be the length of the wire vibrating in p loops,
`T=450N`
`:. 420=(p)/(2l)sqrt((T)/(m))` …(i)
and `490=(p+1)/(2l)sqrt((T)/(m))`
Dividing, we get `(420)/(490)=(p)/(p+1)=(6)/(7)`
`7p=6p+6 or p=6.`
From (i) , `420=(6)/(2l)sqrt((450)/(5.0xx10^(-3)))=(3)/(l)xx300`
`l=(900)/(420)=2.14m`
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