Question

The length of a sonometer wire has a total length of $1m$ between the fixed ends. Where should the two bridges be placed below the wire so the three segments of the wire have their fundamental frequencies in the ratio. $1:2:3$ ?

Answer 1

Let the lengths of the three segments be $x,y,1-(x+y)$.

Fundamental frequencies of each segment is inversely proportional to their lengths.

Let the frequencies be $k^{\prime },2k^{\prime },3k^{\prime }$ since they are in ratio $1:2:3$

Thus $\frac{1}{2x}\sqrt{\frac{T}{\mu }}=k^{\prime }$

$\frac{1}{2y}\sqrt{\frac{T}{\mu }}=2k^{\prime }$

$\frac{1}{2(1-(x+y\left)\right)}\sqrt{\frac{T}{\mu }}=3k^{\prime }$

$⟹x=\frac{1}{k}$

$y=\frac{1}{2k}$

$1-(x+y)=\frac{1}{3k}$

Solving the above three equations gives $k=\frac{11}{6}$

Thus $x=\frac{6}{11}m$

$y=\frac{3}{11}$m

Distance from the other end=$\frac{2}{11}m$