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  4. The length of the wire between two ends of a sonometer is 100 cm. What should be the positions of two bridges below the wire so that the three segments of the wire have their fundamental frequencies in the ratio 1: 3: 5.

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Q. The length of the wire between two ends of a sonometer is $100\, cm$. What should be the positions of two bridges below the wire so that the three segments of the wire have their fundamental frequencies in the ratio $1 : 3 : 5$.

NEETNEET 2013Waves

Solution:

Let $L(= 100 \,cm)$ be the length of the wire and $L_1, L_2$ and $L_3$ are the lengths of the segments as shown in the figure.
image
Fundamental frequency , $\upsilon \propto \frac{1}{L}$
As the fundamental frequencies are in the ratio of $1 : 3 : 5$,
$\therefore L_1:L_2:L_3=\frac{1}{1}:\frac{1}{3}:\frac{1}{5}=15:5:3$
Let x be the common factor. Then
$15x + 5x + 3x = L = 100$
$23x = 100$ or $x=\frac{100}{23}$
$\therefore L_1=15\times\frac{100}{23}=\frac{1500}{23}\,cm$
$L_2=5\times\frac{100}{23}=\frac{500}{23}\,cm $
$L_2=3\times\frac{100}{23}=\frac{300}{23}\,cm $
$\therefore $ The bridges should be placed from $A$ at $ \frac{1500}{23}\,cm $ and $\frac{2000}{23}\,cm $respectively.