Solve this following
Question:

A wire of density $9 \times 10^{-3} \mathrm{~kg} \mathrm{~cm}^{-3}$ is stretched between two clamps $1 \mathrm{~m}$ apart. The resulting strain in the wire is $4.9 \times 10^{-4}$. The lowest frequency of the transverse vibrations in the wire is (Young’s modulus of wire $\mathrm{Y}=9 \times 10^{10} \mathrm{Nm}^{-2}$, (to the nearest integer),

Solution:

$\rho_{\text {wire }}=9 \times 10^{-3} \frac{\mathrm{kg}}{\mathrm{cm}^{3}}=\frac{9 \times 10^{-3}}{10^{-6}} \mathrm{~kg} / \mathrm{m}^{3}$

$=9000 \mathrm{~kg} / \mathrm{m}^{2}$

$(\mathrm{A}=\mathrm{CSA}$ of wire $)$

$\left(\mathrm{Y}=9 \times 10^{10} \mathrm{Nm}^{2}\right)$

$\left(\right.$ Strain $\left.=4.9 \times 10^{-4}\right)$

$\Rightarrow \mathrm{L}=1 \mathrm{~m}=\frac{\lambda}{2} \Rightarrow \lambda=2 \mathrm{~m}$

$\Rightarrow \mathrm{v}=\mathrm{f} \lambda \Rightarrow \sqrt{\frac{\mathrm{T}}{\mu}}=\mathrm{f} \lambda$

Where $\mathrm{Y}=\frac{\mathrm{T} / \mathrm{A}}{\text { strain }} \Rightarrow \mathrm{T}=\mathrm{Y} . \mathrm{A} .$ strain

$\Rightarrow \sqrt{\frac{\text { Y.A. strain }}{\mathrm{m} / \mathrm{L}}}=\mathrm{f} \times 2 \Rightarrow \sqrt{\frac{\text { Y.A.L. strain }}{\mathrm{M}}}=\mathrm{f} \times 2$

$\Rightarrow \sqrt{\frac{Y \times V \times \text { strain }}{M}}=f \times 2 \Rightarrow \sqrt{\frac{Y \times \text { strain }}{\rho}}=f \times 2$

$\mathrm{f}=\frac{1}{2} \cdot \sqrt{\frac{\mathrm{Y} \times \mathrm{strain}}{\rho}}=\frac{1}{2} \sqrt{\frac{9 \times 10^{10} \times 4.9 \times 10^{-4}}{9000}}$

$\mathrm{f}=\frac{1}{2} \cdot \sqrt{\frac{9 \times 10^{3}}{9} \times 4.9}=\frac{1}{2} \sqrt{4900}=\frac{70}{2}=35 \mathrm{~Hz}$

 

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