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(A) For a string fixed at both ends,the resonance frequencies are given by $f_n = n f_1$,where $n = 1, 2, 3, \dots$ and $f_1$ is the fundamental frequ

Vedclass · Accepted Answer

$84$

Vedclass · Answer

(A) For a string fixed at both ends,the resonance frequencies are given by $f_n = n f_1$,where $n = 1, 2, 3, \dots$ and $f_1$ is the fundamental frequency.Given two consecutive resonance frequencies $f_n = 252 \ Hz$ and $f_{n+1} = 336 \ Hz$.Taking the ratio: $\frac{f_{n+1}}{f_n} = \frac{(n+1) f_1}{n f_1} = \frac{n+1}{n} = \frac{336}{252}$.Simplifying the fraction: $\frac{336}{252} = \frac{4}{3}$.Thus,$\frac{n+1}{n} = \frac{4}{3}$,which implies $n = 3$.Substituting $n = 3$ into $f_n = n f_1$,we get $252 = 3 f_1$.Therefore,$f_1 = \frac{252}{3} = 84 \ Hz$.

$A$ string of length $3 \ m$ and linear mass density $0.0025 \ kg/m$ is fixed at both ends. One of its resonance frequencies is $252 \ Hz$. The next higher resonance frequency is $336 \ Hz$. Then the fundamental frequency will be ..... $Hz$.

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