In an experiment with a sonometer,when a mass | vedclass

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(B) The fundamental frequency of a sonometer wire is given by $f = \frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the li

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$500$

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(B) The fundamental frequency of a sonometer wire is given by $f = \frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Since $T = Mg$ (where $M$ is the attached mass),we have $f \propto \sqrt{M}$.Therefore,$\frac{f_2}{f_1} = \sqrt{\frac{M_2}{M_1}}$.Given $f_1 = 30\,Hz$,$M_1 = 180\,g$,$f_2 = 50\,Hz$,and $M_2 = m$.Substituting the values: $\frac{50}{30} = \sqrt{\frac{m}{180}}$.Squaring both sides: $(\frac{5}{3})^2 = \frac{m}{180} \Rightarrow \frac{25}{9} = \frac{m}{180}$.Solving for $m$: $m = \frac{25}{9} 	imes 180 = 25 	imes 20 = 500\,g$.

In an experiment with a sonometer,when a mass of $180\,g$ is attached to the string,it vibrates with a fundamental frequency of $30\,Hz$. When a mass $m$ is attached,the string vibrates with a fundamental frequency of $50\,Hz$. The value of $m$ is $.........\,g$.

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