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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
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'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
Solution:
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Verified Answer
The correct answer is:
$2 \mathrm{~N}$
$\mathrm{n}=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}=\frac{\mathrm{v}}{4 \ell}=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}+16}{\mathrm{n}}}=\frac{3 \mathrm{v}}{4 \ell}$
$\sqrt{\frac{T}{T+16}}=\frac{1}{3}$
$\frac{T}{T+16}=\frac{1}{9}$
$9 \mathrm{T}=\mathrm{T}+16 \quad \therefore 8 \mathrm{T}=16 \quad \therefore \mathrm{T}=2 \mathrm{~N}$
$\sqrt{\frac{T}{T+16}}=\frac{1}{3}$
$\frac{T}{T+16}=\frac{1}{9}$
$9 \mathrm{T}=\mathrm{T}+16 \quad \therefore 8 \mathrm{T}=16 \quad \therefore \mathrm{T}=2 \mathrm{~N}$
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