A wire of density 8 × 10 ^ 3 kg m ^ - 3 is stretched between two...

Using the relation of Young's modulus,

$\dfrac{T}{A} = Y\left(\dfrac{\DeltaL}{L}\right)$
$⇒ T = \left(\dfrac{Y\DeltaL}{L} \times A\right)$

The linear mass density is $\mu = \left(\dfrac{m}{L}\right)$.

So,
$\dfrac{T}{\mu} = \dfrac{Y\DeltaLA}{L\left(\dfrac{m}{L}\right)} = \dfrac{Y ( \DeltaL ) \times LA}{L ( m )} = \left(\dfrac{Y\DeltaL}{L}\right) \times \left(\dfrac{1}{ρ}\right)$

Substituting the values,
$\dfrac{T}{\mu} = \dfrac{8 \times {10}^{10} \times 3 . 2 \times {10}^{- 4}}{0 . 5} \times \left(\dfrac{1}{8 \times {10}^{3}}\right) = 6 . 4 \times {10}^{3}$
$⇒ \dfrac{T}{\mu} = 64 \times {10}^{2}$

The fundamental frequency is given by $f = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$.
$⇒ \sqrt{\dfrac{T}{\mu}} = 8 \times 10 = 80m{s}^{- 1}$
Therefore,

$f = \left(\dfrac{80}{1}\right) = 80Hz$