The fundamental frequency of a sonometer wire | vedclass

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(B) The fundamental frequency $n$ of a sonometer wire is given by the formula $n = \frac{1}{2 \ell} \sqrt{\frac{T}{m}}$,where $\ell$ is the length,$T$

Vedclass · Accepted Answer

$30$

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(B) The fundamental frequency $n$ of a sonometer wire is given by the formula $n = \frac{1}{2 \ell} \sqrt{\frac{T}{m}}$,where $\ell$ is the length,$T$ is the tension,and $m$ is the mass per unit length.Since $\ell$ and $m$ are constant,we have $n \propto \sqrt{T}$.Let the initial frequency be $n$. When the tension $T$ is increased by $44\%$,the new tension $T' = T + 0.44T = 1.44T$.The new frequency $n'$ is given as $n + 6$.Using the proportionality $n \propto \sqrt{T}$,we get $\frac{n'}{n} = \sqrt{\frac{T'}{T}}$.Substituting the values: $\frac{n+6}{n} = \sqrt{\frac{1.44T}{T}} = \sqrt{1.44} = 1.2$.Thus,$n + 6 = 1.2n$.$0.2n = 6$.$n = \frac{6}{0.2} = 30\, Hz$.

The fundamental frequency of a sonometer wire increases by $6\, Hz$ if its tension is increased by $44\%$,keeping the length constant. The frequency of the wire is ...... $Hz$.

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