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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. |
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Answer» 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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