Question 462: Waves

Question

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

Accepted Answer

The frequency produced by a stretched wire is determined by the equation:
$ f = \frac{p}{2l} \sqrt{\frac{T}{m}} \quad	ext{.... (i)} $
where:
$ p $ is the number of loops formed in vibration,
$ l $ is the length of the wire,
$ T $ is the tension in the wire,
$ m $ is the mass per unit length of the wire.
From equation (i), it follows that:
$ f \propto \frac{1}{l} $
Given that the ratio of frequencies of the three segments is $ 2:3:4 $, we can deduce:
$ \begin{aligned} & 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 \end{aligned} $
The total length of the wire is $ 260 \, 	ext{cm} $. Therefore, the lengths $ l_1 $, $ l_2 $, and $ l_3 $ are calculated as follows:
$ \begin{aligned} l_1 & = \frac{6}{13} 	imes 260 = 120 \, 	ext{cm} \ l_2 & = \frac{4}{13} 	imes 260 = 80 \, 	ext{cm} \ l_3 & = \frac{3}{13} 	imes 260 = 60 \, 	ext{cm} \end{aligned} $

Answer

$80 \mathrm{~cm}, 60 \mathrm{~cm}, 120 \mathrm{~cm}$

Answer

$120 \mathrm{~cm}, 80 \mathrm{~cm}, 60 \mathrm{~cm}$

Answer

$60 \mathrm{~cm}, 80 \mathrm{~cm}, 120 \mathrm{~cm}$

Answer

$120 \mathrm{~cm}, 60 \mathrm{~cm}, 80 \mathrm{~cm}$

Waves

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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