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  • An object of mass 'm' is suspended at the end of a mass less wire of length L and area of cross-section A. Young's modules of the maternal of the wire is Y. if the mass is pulled down slightly its freq. of oscillation along the vertical dir

An object of mass 'm' is suspended at the end of a mass less wire of length L and area of cross-section A. Young's modules of the maternal of the wire is Y. if the mass is pulled down slightly its freq. of oscillation along the vertical direction is :
Option: 1 f=\frac{1}{2\pi }\sqrt{\frac{mA}{YL}}
Option: 2 f=\frac{1}{2\pi }\sqrt{\frac{YL}{mA}}
Option: 3 f=\frac{1}{2\pi }\sqrt{\frac{YA}{mL}}
Option: 4 f=\frac{1}{2\pi }\sqrt{\frac{mL}{YA}}

Answers (1)

best_answer

At equilibrium

T=mg ......(i)

Y=\frac{T / A}{l / L}$

\Rightarrow T=\frac{Y A l}{L} \ldots$(ii)

From
(i) and (ii)

m g=\frac{Y A l}{L} \ldots$ (iii)

Now for a small displacement x

\begin{aligned} &\text { Restoring force } &\left.=-\left[T^{\prime}-m g\right]=-\left[\frac{Y A(l+x)}{L}-\frac{Y A l}{L}\right] \right] &=\frac{-Y A x}{L} \end{aligned}Comparing with  F=-m \omega^{2} x

\begin{array}{l} m \omega^{2}=\frac{Y A}{L}\\ \Rightarrow \omega=\sqrt{\frac{Y A}{m L}} \Rightarrow \frac{2 \pi}{T}=\sqrt{\frac{Y A}{m L}} \\ \text { Frequency } =f=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{Y A}{m L}} \end{array}

 

 

 

 

 

Posted by

Deependra Verma

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