Question

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

Answer 1

Step 1: Given data

Length of the wire, $L=40cm$

Length of the stretched wire, $l=40.05cm$

Mass of the wire, $m=3.2g$

Frequency of the vibration, $f=220Hz$

Cross section area of the wire, $A=1.0m{m}^2$

Step 2: To find out the difference between the original length and stretched wire, $\mathbf{∆}\mathit{L}$

The difference between the original length and stretched wire, $∆L=40.05cm-40cm$

$=0.5cm$

Step 3: To find out the value of strain

Value of strain, $=\left|\frac{∆L}{L}\right|$

$=\left|\frac{0.05}{40}\right|$

$=0.125x{10}^{-2}$

Step 4: To find out the tension, $\mathit{T}$

Frequency of the vibration, $f=\left(1/2L\right)\sqrt{\left(T/m\right)}$

$$\begin{gathered} 220=\left[1/\left(2\times 40m\right)\right]\sqrt{T/3.2\times {10}^{-3}kg} \\ T=248.19N \end{gathered}$$

Step 5: To find out the stress, $\mathit{S}$

The value of the stress, $S=T/A$

$$\begin{gathered} =248.19N/1\times {10}^{-6}{m}^2 \\ =248.19\times {10}^{6}N/m^2 \end{gathered}$$

Step 6: To find out the young's modulus, $Y$

The value of young's modulus, $Y=Stress/Strain$

$$\begin{gathered} =248.19\times {10}^{6}N/0.125\times {10}^{-2} \\ =1985.52\times {10}^{8}N/m^2 \\ =1.985\times {10}^{11}N/m^2 \end{gathered}$$

Hence, young's modulus, $Y=1.985\times {10}^{11}N/m^2$