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JEE Mains · Physics · STD 11 - 14. waves and sound

A wire of density \(8 \times 10^3\,kg / m ^3\) is stretched between two clamps \(0.5\,m\) apart. The extension developed in the wire is \(3.2 \times 10^{-4}\,m\). If \(Y =8 \times 10^{10}\,N / m ^2\), the fundamental frequency of vibration in the wire will be \(......\,Hz\).

  1. A \(80\)
  2. B \(60\)
  3. C \(40\)
  4. D \(20\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(80\)

Step-by-step Solution

Detailed explanation

\(f =\frac{1}{2 L } \sqrt{\frac{ T }{\mu}}=\frac{1}{2 L } \sqrt{\frac{ YA \Delta L }{\rho A L}}\) \(f =80\,Hz\)
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