First slide

Sonometer

Question

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.

Moderate

A

$\frac{1500}{23} cm, \frac{500}{23} cm$
B

$\frac{1500}{23} cm, \frac{300}{23} cm$
C

$\frac{300}{23} cm, \frac{1500}{23} cm$
D

$\frac{1500}{23} cm, \frac{2000}{23} cm$

Solution

$Let L (= 100 cm) be the length of the wire and L_{1}, L_{2} and L_{3} are the lengths of the segments as shown in the figure.$
$Fundamental frequency, v \propto \frac{1}{L} As the fundamental frequencies are in the ratio of 1 : 3 : 5, ∴ L_{1} : L_{2} : L_{3} = \frac{1}{1} : \frac{1}{3} : \frac{1}{5} = 15 : 5 : 3 Let x be the common factor. Then 15 x+5 x+3 x=L=100 23 x = 100 or x = \frac{100}{23} ∴ L_{1} = 15 \times \frac{100}{23} = \frac{1500}{23} cm L_{2} = 5 \times \frac{100}{23} = \frac{500}{23} cm L_{3} = 3 \times \frac{100}{23} = \frac{300}{23} cm ∴ The bridges should be placed from A at \frac{1500}{23} cm and \frac{2000}{23} cm respectively.$

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