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(A) For a string fixed at both ends,the resonant frequencies are given by $v_n = n \cdot v_0$,where $v_0$ is the fundamental frequency (lowest resonan

Vedclass · Accepted Answer

$105$

Vedclass · Answer

(A) For a string fixed at both ends,the resonant frequencies are given by $v_n = n \cdot v_0$,where $v_0$ is the fundamental frequency (lowest resonant frequency) and $n = 1, 2, 3, \dots$.The difference between any two consecutive resonant frequencies is $\Delta v = v_{n+1} - v_n = (n+1)v_0 - nv_0 = v_0$.Given that $420\, Hz$ and $315\, Hz$ are resonant frequencies with no other resonant frequencies between them,they must be consecutive harmonics.Therefore,the fundamental frequency $v_0$ is the difference between these two frequencies:$v_0 = 420\, Hz - 315\, Hz = 105\, Hz$.

$A$ string is stretched between fixed points separated by $75.0\, cm$. It is observed to have resonant frequencies of $420\, Hz$ and $315\, Hz$. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is .... $Hz$.

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