A sonometer wire 'A' of diameter 'd' under te | vedclass

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(C) The fundamental frequency of a sonometer wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = 	ext{mass per unit length} = ho

Vedclass · Accepted Answer

$\rho_2=\frac{\rho_1}{2}$

Vedclass · Answer

(C) The fundamental frequency of a sonometer wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = 	ext{mass per unit length} = ho \cdot A = ho \cdot \pi \cdot (\frac{d}{2})^2 = \frac{ho \pi d^2}{4}$.Substituting $\mu$,we get $f = \frac{1}{2l} \sqrt{\frac{T}{\frac{ho \pi d^2}{4}}} = \frac{1}{ld} \sqrt{\frac{T}{\pi ho}}$.For wire $A$: $n = \frac{1}{ld} \sqrt{\frac{T}{\pi ho_1}}$.For wire $B$: $n = \frac{1}{l(2d)} \sqrt{\frac{2T}{\pi ho_2}}$.Equating the frequencies: $\frac{1}{ld} \sqrt{\frac{T}{\pi ho_1}} = \frac{1}{2ld} \sqrt{\frac{2T}{\pi ho_2}}$.Canceling common terms: $\sqrt{\frac{1}{ho_1}} = \frac{1}{2} \sqrt{\frac{2}{ho_2}}$.Squaring both sides: $\frac{1}{ho_1} = \frac{1}{4} \cdot \frac{2}{ho_2} = \frac{1}{2ho_2}$.Therefore,$2ho_2 = ho_1$,which gives $ho_2 = \frac{ho_1}{2}$.

$A$ sonometer wire '$A$' of diameter '$d$' under tension '$T$' having density '$\rho_1$' vibrates with fundamental frequency '$n$'. If we use another wire '$B$' which vibrates with the same frequency under tension '$2T$' and diameter '$2d$',then the density '$\rho_2$' of wire '$B$' will be:

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