Physics, asked by devangrai9676, 1 year ago

a stretched wire of length 110 cm is divided into three segments whose frequency are in ratio 1:2:3 their length must be

Answers

Answered by lidaralbany
22

Answer:

The length of the wire is divided into three segments 60 cm, 30 cm and 20 cm.

Explanation:

Given that,

Length l = 110 cm

A stretched wire is divided into three segments whose frequency are in 1:2:3.

We know that,

The frequency of the wave is inversely proportional to the wavelength.

f\propto\dfrac{1}{l}

f_{1}:f_{2}:f_{3}=1:2:3

l_{1}:l_{2}:l_{3}=\dfrac{1}{1}:\dfrac{1}{2}:\dfrac{1}{3}

l_{1}:l_{2}:l_{3}=6:3:2

The length of first frequency is

l_{1}=\dfrac{6}{11}\times110

l_{1}=60

The length of first frequency is

l_{2}=\dfrac{3}{11}\times110

l_{2}=30

The length of first frequency is

l_{3}=\dfrac{2}{11}\times110

l_{3}=20

Hence, The length of the wire is divided into three segments 60 cm, 30 cm and 20 cm.

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