Question

A $40cm$ long wire having a mass $3.2gm$ and area of c.s $1m{m}^2$ is stretched between the support $40.05cm$ apart. In its fundamental mode. It vibrate with a frequency $1000/64Hz$. Find the young's modulus of the wire.

• A. $1\times {10}^{9}N{m}^2$
• B. $2\times {10}^{9}N{m}^2$
• C. $1\times {10}^{8}N{m}^2$
• D. $4\times {10}^{9}N{m}^2$

Answer 1

The correct option is A $1\times {10}^{9}N{m}^2$

Due to stretching, tension develops in the string, as given by Hooke's Law,

$T=\frac{YA\Delta L}{L}$

Thus the fundamental frequency of wave=$\frac{1}{2L}\sqrt{\frac{T}{\mu }}=\frac{1}{2L}\sqrt{\frac{YA\Delta L/L}{m/L}}$

$⟹Y=1\times {10}^{9}N{m}^{-2}$