Question

A 40 cm wire having a mass of 3.2 g is stretched between two fixed supports 40.05 cm apart. In its fundamental mode, the wire vibrates at 220 Hz. If the area of cross section of the wire is $1.0m{m}^2$, find its Young modulus.

Answer 1

$L=40cm=0,4m,$

mass $=3.2kg=3.2\times {10}^{-3}kg$

$\therefore$ Mass per unit length,

$m=\frac{3.2}{0.4}=8\times {10}^{-3}kg/m$

Change in length,

$\Delta L=40.05-40=0.05\times {10}^{-2}m$

Strain $=\frac{\Delta L}{L}=\frac{(0.05\times {10}^{-2})}{0.4}=0.125\times {10}^{-2}$

$f=220Hz$

$f=\frac{1}{2L}\sqrt{\frac{T}{m}}$

$=\frac{1}{2\times \left(0.4005\right)}\sqrt{\frac{T}{8\times {10}^{-3}}}$

$\Rightarrow 220\times 220=\left[\frac{1}{(0.801{)}^2}\right]\times T\times \left(\frac{{10}^3}{8}\right)$

$\Rightarrow T\times 1000=220\times 220\times 0641\times 0.8$

$\Rightarrow T=248.19N$

Strain $=\frac{248.19}{1m{m}^2}$

$=\frac{248.19}{{10}^{-6}}=248.19\times {10}^6$

$Y=\frac{\text{stress}}{strain}$

$=\frac{(248.19\times {10}^6)}{(0.125\times {10}^{-2})}$

$=1985.52\times {10}^8$

$=1.985\times {10}^{11}N/m^2$