A wire having linear mass density $9 \times 10^3\text{ kg/m}^3$ is stretched between two clamps $1\text{m}$…
A wire having linear mass density $9 \times 10^3\text{ kg/m}^3$ is stretched between two clamps $1\text{m}$ apart and is subjected to an extension of $4.9 \times 10^{-4}\text{ m}$. The lowest frequency of wave produced in the wire is
$(\text{Take}, Y = 9 \times 10^{10}\text{ N/m}^2)$ [UP CPMT 2015]
$47\text{ Hz}$
$42\text{ Hz}$
$35\text{ Hz}$
$37\text{ Hz}$
Solution
The fundamental frequency of vibrations,
$f = \frac{1}{2L}\sqrt{\frac{F}{\alpha}} = \frac{1}{2L}\sqrt{\frac{T}{\rho A}}$
(where, $T = \text{tension}$ and $\alpha = \rho A$)
Tension due to elasticity, $T = YA \left(\frac{\Delta L}{L}\right)$ $\left(\because Y = \frac{\text{Stress}}{\text{Strain}}\right)$
$\Rightarrow f = \frac{1}{2L}\sqrt{\frac{YA}{\rho A} \times \frac{\Delta L}{L}} = \frac{1}{2L}\sqrt{\frac{Y\Delta L}{\rho L}}$
Substituting the given values in above equation, we get
$f = \frac{1}{2 \times 1}\sqrt{\frac{9 \times 10^{10} \times 4.9 \times 10^{-4}}{9 \times 10^3 \times 1}} = 35\text{ Hz}$