A wire of density 9 times 10^3 text{ kg/m}^3 | vedclass

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(B) The fundamental (lowest) frequency of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $T$ is the tension and $m$ is the m

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(B) The fundamental (lowest) frequency of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $T$ is the tension and $m$ is the mass per unit length.Given: Density $ho = 9 	imes 10^3 	ext{ kg/m}^3$,length $l = 1 	ext{ m}$,extension $\Delta l = 4.9 	imes 10^{-4} 	ext{ m}$,and Young's modulus $Y = 9 	imes 10^{10} 	ext{ N/m}^2$.Mass per unit length $m = \frac{M}{l} = \frac{A \cdot l \cdot ho}{l} = Aho$.From Young's modulus,$Y = \frac{T/A}{\Delta l/l} \implies T = \frac{Y \cdot A \cdot \Delta l}{l}$.Substituting $T$ and $m$ into the frequency formula:$n = \frac{1}{2l} \sqrt{\frac{Y \cdot A \cdot \Delta l / l}{A \cdot ho}} = \frac{1}{2l} \sqrt{\frac{Y \cdot \Delta l}{l \cdot ho}}$.Substituting the values:$n = \frac{1}{2 	imes 1} \sqrt{\frac{9 	imes 10^{10} 	imes 4.9 	imes 10^{-4}}{1 	imes 9 	imes 10^3}} = \frac{1}{2} \sqrt{\frac{9 	imes 4.9 	imes 10^6}{9 	imes 10^3}} = \frac{1}{2} \sqrt{4.9 	imes 10^3} = \frac{1}{2} \sqrt{4900} = \frac{70}{2} = 35 	ext{ Hz}$.

$A$ wire of density $9 \times 10^3 \text{ kg/m}^3$ is stretched between two clamps $1 \text{ m}$ apart and is subjected to an extension of $4.9 \times 10^{-4} \text{ m}$. The lowest frequency of transverse vibration in the wire is ..... $\text{Hz}$ $(Y = 9 \times 10^{10} \text{ N/m}^2)$.

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