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Question
Two wires of the same material and same cross-section are stretched on a sonometer. One wire is loaded with 1 kg and another is loaded with 9 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.
Solution
Given:
M1 = 1 kg, M2 = 9 kg,
A1 = A2,
m1 = m2 (where m is linear density of wire)
L1 = 60 cm = 0.6 m
(n0)1 = (n0)2
To find: Vibrating length of wire (L2)
Formula:
n = `1/(2"L")sqrt("T"/"m")`
Calculation:
From formula,
`("n")_1 = 1/(2"L"_1)sqrt("T"_1/"m"_1)`
`("n")_2 = 1/(2"L"_2)sqrt("T"_2/"m"_2)`
∵ (n)1 = (n)2
∴ `1/(2"L"_1)sqrt("T"_1/"m"_1) = 1/(2"L"_2)sqrt("T"_2/"m"_2)`
∴ `1/(4"L"_1^2) "T"_1/"m"_1 = 1/(4"L"_2^2) "T"_2/"m"_2`
∴ L2 = `sqrt("L"_1^2 xx "T"_2/"m"_2 xx "m"_1/"T"_1)`
= `sqrt((0.6)^2 xx (9 "g")/(1 "g"))`
= 0.6 × 3
= 1.8 m
The vibrating length of the wire is 1.8 m.
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