A stretched uniform wire of length L under tension $\mathrm{T}$ is vibrating with frequency 'n'. A closed…

A stretched uniform wire of length L under tension $\mathrm{T}$ is vibrating with frequency 'n'. A closed pipe of same length is also vibrating with same fundamental frequency 'n'. If $\mathrm{T}$ is increased by $16 \mathrm{~N}$, it is in resonance with $2^{\text {nd }}$ harmonic of same closed pipe. The initial tension in the wire is
  1. $1 \mathrm{~N}$
  2. $2 \mathrm{~N}$
  3. $1 \cdot 5 \mathrm{~N}$
  4. $0 \cdot 5 \mathrm{~N}$

Solution

$\mathrm{n}=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}=\frac{\mathrm{v}}{4 \ell}=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}+16}{\mathrm{n}}}=\frac{3 \mathrm{v}}{4 \ell}$ $\sqrt{\frac{T}{T+16}}=\frac{1}{3}$ $\frac{T}{T+16}=\frac{1}{9}$ $9 \mathrm{T}=\mathrm{T}+16 \quad \therefore 8 \mathrm{T}=16 \quad \therefore \mathrm{T}=2 \mathrm{~N}$

Asked in: MHT CET 2020 (20 Oct Shift 1)

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