A wire of density $9 \times 10^{-3}\ \text{kg cm}^{-3}$ is stretched between two clamps $1\ \text{m}$ apart. The resulting strain in the wire is $4.9 \times 10^{-4}$. The lowest frequency (in Hz) of the transverse vibrations in the wire (Young's modulus of wire $Y = 9 \times 10^{10}\ \text{N m}^{-2}$), to the nearest integer, is ______.
Numerical value type. Enter your answer.
Answer: 35
Density $\rho = 9\times10^{-3}\ \text{kg cm}^{-3} = 9000\ \text{kg m}^{-3}$. Tension per unit area $= Y\times\text{strain}$, and mass per unit length $= \rho A$, so
$$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{Y\,\text{strain}}{\rho}} = \sqrt{\frac{9\times10^{10}\times4.9\times10^{-4}}{9000}} = 70\ \text{m s}^{-1}$$
Fundamental: $f = \dfrac{v}{2L} = \dfrac{70}{2\times1} = 35$ Hz.
Solution by Sreeraj P, M.Sc Physics