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 \(......\,Hz\).
- A \(80\)
- B \(60\)
- C \(40\)
- D \(20\)
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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