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MHT CET · Physics · Waves and Sound

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. A \(1 \mathrm{~N}\)
  2. B \(2 \mathrm{~N}\)
  3. C \(1 \cdot 5 \mathrm{~N}\)
  4. D \(0 \cdot 5 \mathrm{~N}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(2 \mathrm{~N}\)

Step-by-step Solution

Detailed explanation

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