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Worked Examples · Example 6.2

Q.A sonometer wire of length 50 cm is stretched by keeping weights equivalent of 3.5 kg. The fundamental frequency of vibration is 125 Hz. Determine the linear density of the wire.

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✓ Free question

T = 3.5×9.8 = 34.3 N; m = T/(4n²l²) = 34.3/(4×125²×0.5²) = 2.195×10⁻³ kg/m.

Given l = 0.5 m, T = 3.5 × 9.8 = 34.3 N, n = 125 Hz. The fundamental frequency of a stretched wire is

n=12lTm⇒n2=14l2Tm⇒m=T4n2l2n = \frac{1}{2l}\sqrt{\frac{T}{m}} \Rightarrow n^2 = \frac{1}{4l^2}\frac{T}{m} \Rightarrow m = \frac{T}{4n^2 l^2}

m=34.34×(125)2×(0.5)2=2.195×10−3 kg m−1m = \frac{34.3}{4\times(125)^2\times(0.5)^2} = 2.195\times10^{-3}\ \mathrm{kg\,m^{-1}}

✓Final answer

m = 2.195 × 10⁻³ kg m⁻¹.

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