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Two wires of the same material and the same cross-section are stretched on a sonometer. One wire is loaded with 1.5 kg and another is loaded with 6 kg. - Physics

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Sum

Two wires of the same material and the same cross-section are stretched on a sonometer. One wire is loaded with 1.5 kg and another is loaded with 6 kg. The vibrating length of the first wire is 60 cm and its fundamental frequency of vibration is the same as that of the second wire. Calculate the vibrating length of the other wire.

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Solution

Data: m1 = m2 = m,
Length of first wire L1 = 60 cm = 0.6 m,
One Wire is loaded with T1 = 1.5 kg = 14.7 N,
Other wire is loaded with T2 = 6 kg = 58.8 N

n1 = `1/(2"L"_1)sqrt("T"_1/"m")` and n2 = `1/(2"L"_2)sqrt("T"_2/"m")`

But, n1 = n2

`∴1/(2"L"_1)sqrt("T"_1/"m")=1/(2"L"_2)sqrt("T"_2/"m")`

`("L"_2)/("L"_1) = sqrt("T"_2/"m") xx sqrt("m"/"T"_1)`

L2 = `sqrt("T"_2/"T"_1)xx"L"_1`

= `sqrt(58.8/14.7)xx0.6` = `sqrt4xx0.6` = 1.2 m

The second wire has a vibrating length of 1.2 metres.

Concept: Sonometer
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APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 6 Superposition of Waves
Exercises | Q 14 | Page 157
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