A stretched wire of length 260 cm is set into vibrations. It is divided into three segments whose frequencies are in the ratio 2 : 3 : 4 . Their lengths must be

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The frequency produced by a stretched wire is given by
$f = \frac{p}{2 l} \sqrt{\frac{T}{m}}$ … (i)
Where, p = number of loops formed in vibration,
$l$ = length of wire,
T = tension in wire,
and m = mass per unit length of wire
So, form Eq (i), we get
$f \propto \frac{1}{l}$
Here, the ratio of frequencies of three segments is
$f_{1} : f_{2} : f_{3} = 2 : 3 : 4$
$\Rightarrow l_{1} : l_{2} : l_{3} = \frac{1}{2} : \frac{1}{3} : \frac{1}{4}$
$= 6 : 4 : 3$
The length of wire, L = 260 cm, so
$l_{1} = \frac{6}{13} \times 260 = 120$ cm
$l_{2} = \frac{4}{13} \times 260 = 80$ cm
$a n d l_{3} = \frac{3}{13} \times 260 = 60$ cm

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