Question

A wire of density '$\rho$' and Youngs modulus '$Y$' is stretched between two rigid supports separated by a distance '$L$' under tension '$T$'. Derive an expression for its frequency in fundamental mode. Hence show that $\eta =\frac{1}{2L}\sqrt{\frac{Yl}{\rho L}}$, where symbol have their usual meanings.
When the length of a simple pendulum is decreased by $20cm$, the period changes by $10$%. Find the original length of the pendulum.

Answer 1

Consider a wire stretched between two rigid supports a distance $L$ apart. Let
$T=$ the tension in the wire
$r=$ the radius of cross section of the wire
$Y,p=$ young's modulus and mass density of the material of the wire
$M,m=$ mass and linear density of the wire
Then,
$M=\left[\pi r^{2}L\right]\rho$ and $m=\frac{M}{L}=\pi r^2\rho ....\left(i\right)$
$\because$ the stress in the wire $=\frac{T}{\pi r^2}$
$\frac{T}{m}=\frac{T}{\pi r^2\rho }=\frac{stress}{\rho }....\left(ii\right)$
The fundamental frequency of vibration of the wire,
$n=\frac{1}{2L}\sqrt{\frac{T}{m}}=\frac{1}{2L}\sqrt{\frac{stress}{\rho }}....\left(iii\right)$
If $△L=l$ is the elastic extension of the wire under tension $T$, strain=$l/L$
Since $Y=\frac{Stress}{Strain}$
$\Rightarrow Stress=Y\times strain=Y\frac{l}{L}...\left(iv\right)$
$\eta =\frac{1}{2L}\sqrt{\frac{Yl}{\rho L}}...\left(v\right)$
Which is the required expression
Let the original length $=l$
Length decreased by $20cm$
and period changes by $10$%
$T=2\pi \sqrt{\frac{l}{g}}....\left(1\right)$
$=T-\frac{10}{100}\times T=T-\frac{T}{10}=\frac{9T}{10}$
$\Rightarrow \frac{9T}{10}=2\pi \sqrt{\frac{l-20}{g}}....\left(2\right)$
$\Rightarrow \frac{T}{\frac{9T}{10}}=\frac{2\pi \sqrt{\frac{l}{g}}}{2\pi \sqrt{\frac{l-20}{g}}}$
$\Rightarrow \frac{10}{9}=\sqrt{\frac{l}{l-20}}$
$\Rightarrow {\left(\frac{10}{9}\right)}^2=\frac{l}{l-20}$
$\Rightarrow \frac{100}{81}=\frac{l}{l-20}$
$\Rightarrow 100(l-20)=81l$
$\Rightarrow$ $100l-81l=2000$
$\Rightarrow$ $19l=2000$
$\Rightarrow$ $l=\frac{2000}{19}cm=\frac{20}{19}m$
$\therefore$ $l=1.05m$