KCET · Physics · Waves and Sound
A uniform wire of length \(\mathrm{L}\), diameter \(\mathrm{D}\) and density \(\rho\) is stretched under a tension T. The correct relation between its fundamental frequency \(f\), the length \(L\) and the diameter \(D\) is
- A \(f \propto \frac{1}{L D}\)
- B \(f \propto \frac{1}{L \sqrt{D}}\)
- C \(\mathrm{f} \propto \frac{1}{\mathrm{D}^{2}}\)
- D \(f \propto \frac{1}{L D^{2}}\)
Answer & Solution
Correct Answer
(A) \(f \propto \frac{1}{L D}\)
Step-by-step Solution
Detailed explanation
The fundamental frequency is \(f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}\)
\(f=\frac{1}{2 L} \sqrt{\frac{T}{\rho \pi \frac{D^{2}}{4}}}=\frac{1}{L D} \sqrt{\frac{T}{\pi \rho}}\)
\(\therefore\) \(f \propto \frac{1}{L D}\)
\(f=\frac{1}{2 L} \sqrt{\frac{T}{\rho \pi \frac{D^{2}}{4}}}=\frac{1}{L D} \sqrt{\frac{T}{\pi \rho}}\)
\(\therefore\) \(f \propto \frac{1}{L D}\)
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