Waves

JEE Main level, numerical answer, medium

67 of 142

Question 67 of 142

A wire of length 30cm30cm, stretched between rigid supports, has it's nth{n}^{th} and (n+1)th{\left(n+1\right)}^{th} harmonics at 400Hz400Hz and 450Hz450Hz, respectively. If tension in the string is 2700N2700N, it's linear mass density is _____ kgm−1kg{m}^{-1}.

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Answer: 3

Frequency of nth{n}^{th} harmonic is nv2l=400⇒nv0.6=400...(1)\dfrac{nv}{2l}=400\Rightarrow \dfrac{nv}{0.6}=400...\left(1\right)
And frequency of (n+1)th{\left(n+1\right)}^{th} harmonic is (n+1)v2l=450⇒(n+1)v0.6=450...(2)\dfrac{\left(n+1\right)v}{2l}=450\Rightarrow \dfrac{\left(n+1\right)v}{0.6}=450...\left(2\right), here, vv is speed of wave on string.
From equation (1)\left(1\right), we have
n=400×0.6vn=\dfrac{400\times 0.6}{v}
Putting in equation (2)\left(2\right),
⇒[0.6×400v+1]v0.6=450\Rightarrow \left[\dfrac{0.6\times 400}{v}+1\right]\dfrac{v}{0.6}=450
⇒[400+v0.6]=450\Rightarrow \left[400+\dfrac{v}{0.6}\right]=450
⇒v=30ms−1\Rightarrow v=30m{s}^{-1}
Speed of wave on string is stated as v=Tμ=30v=\sqrt{\dfrac{T}{\mu }}=30, here, μ\mu is linear mass density.
So, μ=Tv2=2700900⇒μ=3kgm−1\mu =\dfrac{T}{{v}^{2}}=\dfrac{2700}{900}\Rightarrow \mu =3kg{m}^{-1}

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