The length of a sonometer wire is $1.21 m$ . It is divided into three segments. What should be the lengths of the three segments for their fundamental frequencies to be in the ratio $1 : 2 : 3$ ?

0.66 m, 0.33 m, 0.22 m

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0.72 m, 0.30 m, 0.18 m

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0.56 m, 0.44 m, 0.21 m

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0.82 m, 0.18 m, 0.21 m

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Solution

The correct option is A $0.66 m, 0.33 m, 0.22 m$
Given that,
Length of the sonometer wire, $L = 1.21 m$
$\Rightarrow f = \frac{n}{2 L} \sqrt{\frac{T}{μ}}$

Suppose the lengths of the three segments are $L_{1}, L_{2}$ and $L_{3}$ respectively. The fundamental frequencies are,
$f_{1} = \frac{1}{2 L_{1}} \sqrt{\frac{T}{μ}} . . . (A)$
$f_{2} = \frac{1}{2 L_{2}} \sqrt{\frac{T}{μ}} . . . (B)$
$f_{3} = \frac{1}{2 L_{3}} \sqrt{\frac{T}{μ}} . . . (C)$

Given, $f_{1} : f_{2} : f_{3} = 1 : 2 : 3 . . . (i)$
and $f_{1} L_{1} = f_{2} L_{2} = f_{3} L_{3} . . (i i)$
[From $A, B$ and $C$ ]

From the equation $(i)$ , we can write.
$f_{2} = 2 f_{1}$ and $f_{3} = 3 f_{1}$
So, by equation $(i i)$ ,
$L_{2} = \frac{f_{1}}{f_{2}} L_{1} = \frac{L_{1}}{2}$
& $L_{3} = \frac{f_{1}}{f_{3}} L_{1} = \frac{L_{1}}{3}$

$L_{1} + L_{2} + L_{3} = 1.21 m$
$\Rightarrow L_{1} + \frac{L_{1}}{2} + \frac{L_{1}}{3} = 1.21 m \Rightarrow L_{1} = 0.66 m$
$∴ L_{2} = \frac{L_{1}}{2} = \frac{0.66}{2} = 0.33 m$
& $L_{3} = \frac{L_{1}}{3} = \frac{0.66}{3} = 0.22 m$
Thus, option (a) is correct.

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The length of a sonometer wire is 1.21 m. It is divided into three segments. What should be the lengths of the three segments for their fundamental frequencies to be in the ratio 1:2:3?

The length of a sonometer wire is $1.21 m$ . It is divided into three segments. What should be the lengths of the three segments for their fundamental frequencies to be in the ratio $1 : 2 : 3$ ?