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(B) 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

(B) 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 $n$ is an integer $(n = 1, 2, 3, \dots)$.Given two consecutive resonant frequencies are $f_n = 315 \ Hz$ and $f_{n+1} = 420 \ Hz$.We know that $f_{n+1} - f_n = (n+1)f_1 - nf_1 = f_1$.Therefore,the difference between any two consecutive resonant frequencies is equal to the fundamental frequency $f_1$.$f_1 = 420 \ Hz - 315 \ Hz = 105 \ Hz$.Since $105 \ Hz$ is the fundamental frequency,it is also the lowest resonant frequency for the string.

$A$ 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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