The length of the wire between two ends of a sonometer is 100 cm… | NEET 2013

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Stationary Waves & Vibrating Strings

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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Answer: (d)

$f \propto \dfrac{1}{l}$ : $l_1 : l_2 : l_3 = 1 : \dfrac{1}{3} : \dfrac{1}{5} = 15 : 5 : 3$ .
Total parts $= 23$ : $l_1 = \dfrac{1500}{23}$ , $l_2 = \dfrac{500}{23}$ , $l_3 = \dfrac{300}{23}\text{ cm}$ .
Bridges at $\dfrac{1500}{23}\text{ cm}$ and $\dfrac{1500 + 500}{23} = \dfrac{2000}{23}\text{ cm}$ from one end.
So the answer is (d) $\frac{1500}{23}\text{ cm}, \frac{2000}{23}\text{ cm}$ .

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