A man can hear sounds in the frequency range | vedclass

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(B) The fundamental frequency of a vibrating string is given by $f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \frac{m}{L}$.Given: $T = 240 \,

Vedclass · Accepted Answer

$238$

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(B) The fundamental frequency of a vibrating string is given by $f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \frac{m}{L}$.Given: $T = 240 \, N$,$m = 3 	imes 10^{-3} \, kg$,$L = 8 \, m$.Linear mass density $\mu = \frac{3 	imes 10^{-3}}{8} \, kg/m$.Substituting the values: $f_n = \frac{n}{2 	imes 8} \sqrt{\frac{240}{3 	imes 10^{-3} / 8}} = \frac{n}{16} \sqrt{\frac{240 	imes 8}{3 	imes 10^{-3}}} = \frac{n}{16} \sqrt{640000} = \frac{n}{16} 	imes 800 = 50n \, Hz$.The audible range is $120 \, Hz \leq 50n \leq 12020 \, Hz$.Dividing by $50$: $2.4 \leq n \leq 240.4$.Since $n$ must be an integer,$n$ can take values from $3$ to $240$.The number of such frequencies is $240 - 3 + 1 = 238$.

$A$ man can hear sounds in the frequency range $120 \, Hz$ to $12020 \, Hz$ only. He is vibrating a piano string having a tension of $240 \, N$ and a mass of $3 \, g$. The string has a length of $8 \, m$. How many different frequencies can he hear?

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