(B) The resonant frequencies of a string fixed at both ends are given by $f_n = n \cdot f_0$,where $f_0$ is the fundamental frequency and $n$ is an in

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(B) The resonant frequencies of a string fixed at both ends are given by $f_n = n \cdot f_0$,where $f_0$ is the fundamental frequency and $n$ is an integer $(n = 1, 2, 3, \dots)$.Given two consecutive resonant frequencies $f_n = 315\,Hz$ and $f_{n+1} = 420\,Hz$.The difference between two consecutive resonant frequencies is the fundamental frequency $f_0$.$f_0 = f_{n+1} - f_n = 420\,Hz - 315\,Hz = 105\,Hz$.Thus,the lowest resonant frequency (fundamental frequency) is $105\,Hz$.

Class 11 Physics Waves and Sound Transverse Stationary Waves and Sonometer

Medium

$A$ string is stretched between two fixed points separated by $75\,cm$. 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 ..... $Hz$.

A
$10.5$
B
$105$
C
$1.05$
D
$1050$

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Waves and Sound

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Transverse Stationary Waves and Sonometer

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$A$ string is stretched between two fixed points separated by $75\,cm$. 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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