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A uniform wire of length $L$, diameter $D$ and density $\rho$ is stretched by a tension $T$ length $L$ and the diameter $D$ is
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Verified Answer
The correct answer is:
$f \propto \frac{1}{L D}$
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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