MHT CET · Physics · Waves and Sound
A uniform wire of length \(L\), diameter \(D\) and density \(\rho\) is stretched by a tension \(T\) length \(L\) and the diameter \(D\) is
- A \(f \propto \frac{L}{D}\)
- B \(f \propto \frac{1}{L D}\)
- C \(f \propto \frac{1}{L \sqrt{D}}\)
- D \(f \propto \frac{1}{L D^2}\)
Answer & Solution
Correct Answer
(B) \(f \propto \frac{1}{L D}\)
Step-by-step Solution
Detailed explanation
Given,
Frequency \(=f\), Diameter \(=D\), Length \(=L\), Density \(=\rho\) and Tension \(=T\)
Now, we know that \(f=\frac{c}{\lambda}\),
\(\therefore f=\frac{1}{2 L} \times \sqrt{\frac{T}{\mu}}\)
where \(\mu\) is mass per unit length
And mass per unit length is related to density via,
\(\mu=\left(\pi \times \frac{D^2}{4}\right) \frac{1}{\rho}\)
So, the frequency is \(f=\frac{1}{2 L} \times \sqrt{\frac{T}{\left(\frac{\pi D^2}{4 \rho}\right)}}\)
\(\therefore f \propto \frac{1}{L D}\)
Frequency \(=f\), Diameter \(=D\), Length \(=L\), Density \(=\rho\) and Tension \(=T\)
Now, we know that \(f=\frac{c}{\lambda}\),
\(\therefore f=\frac{1}{2 L} \times \sqrt{\frac{T}{\mu}}\)
where \(\mu\) is mass per unit length
And mass per unit length is related to density via,
\(\mu=\left(\pi \times \frac{D^2}{4}\right) \frac{1}{\rho}\)
So, the frequency is \(f=\frac{1}{2 L} \times \sqrt{\frac{T}{\left(\frac{\pi D^2}{4 \rho}\right)}}\)
\(\therefore f \propto \frac{1}{L D}\)
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