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Q. 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 $=7.5 \,gm / cc$, density of aluminum $=2.7 \,g / cc ]$. The frequency of lowest harmonic for which both wires vibrate in unison, given that the length of the wires is $1\, m$, their diameters $1 \,mm$ and tension $75 \pi$ is given as $\beta \,kHz$. Find the value of $12\, \beta .$

Waves

Solution:

(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 $P_{ Fe }=5 ; f=\frac{5}{2 \times 1} \sqrt{\frac{75 \pi \times 4}{3.14 \times 10^{-6} \times 7.5 \times 10^{3}}}$
$=500\, Hz$