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Question

(A) The resonant frequencies of a string fixed at both ends are given by $f_n = n 	imes f_1$,where $f_1 = \frac{v}{2L}$ is the fundamental frequency

Vedclass · Accepted Answer

$105$

Vedclass · Answer

(A) The resonant frequencies of a string fixed at both ends are given by $f_n = n 	imes f_1$,where $f_1 = \frac{v}{2L}$ is the fundamental frequency and $n = 1, 2, 3, \dots$ is an integer.Given two consecutive resonant frequencies $f_n = 315\, Hz$ and $f_{n+1} = 420\, Hz$.We know that the difference between two consecutive resonant frequencies is equal to the fundamental frequency $f_1$.$f_1 = f_{n+1} - f_n = 420\, Hz - 315\, Hz = 105\, Hz$.Alternatively,$\frac{f_{n+1}}{f_n} = \frac{n+1}{n} = \frac{420}{315} = \frac{4}{3}$.This implies $n = 3$,so $f_3 = 315\, Hz$ and $f_4 = 420\, Hz$.Since $f_3 = 3 	imes f_1 = 315\, Hz$,we get $f_1 = \frac{315}{3} = 105\, Hz$.The lowest resonant frequency is the fundamental frequency $f_1 = 105\, Hz$.

$A$ 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. Then,the lowest resonant frequency for this string is .... $Hz$

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