Question

Two identical wires, one made of iron and the other of aluminum are stretched along-side on a sonometer board by equal stretching forces. [Density of iron $=7.5gm/cc$, density of aluminum $=2.7g/cc]$. The frequency of lowest harmonic for which both wires vibrate in unison, given that the length of the wires is $1m$, their diameters $1mm$ and tension $75\pi$ is given as $\beta kHz$. Find the value of $12\beta .$

Answer 1

Answer: 6

(a) Given $\frac{P_{Fe}}{P_{Al}}=\frac{\sqrt{\frac{T}{{\rho }_{Al}}}}{\sqrt{\frac{T}{{\rho }_{Fe}}}}$
$=\sqrt{\frac{7.5}{2.7}}=\frac{5}{3}$
Here fifth harmonic of $F_e=$ third harmonic of $Al$ wire.
(b) Using $P_{Fe}=5;f=\frac{5}{2\times 1}\sqrt{\frac{75\pi \times 4}{3.14\times 10^{-6}\times 7.5\times 10^3}}$
$=500Hz$