Question

Two uniform wires are vibrating simultaneously in their fundamental modes. The tensions, lengths, diameters and the densities of the two wires are in the ratio 8:1, 36:35, 4:1 and 1:2 respectively. If the note of the higher pitch has a frequency 360 hertz, the number of beats produced per second is:

• A. 5
• B. 10
• C. 15
• D. 20

Answer 1

The correct option is B 10

Fundamental frequency $\nu =\frac{1}{2L}\sqrt{\frac{T}{\mu }}$
where, mass per unit length $\mu =\left(\frac{\pi d^2}{4}\right)\rho$
$⟹$ $\frac{{\nu }_1}{{\nu }_2}=\frac{L_2}{L_1}\sqrt{\frac{T_1}{T_2}.\frac{{\rho }_2{d}_2^2}{{\rho }_1{d}_1^2}}$
Given : $\frac{T_1}{T_2}=\frac{8}{1}$, $\frac{L_1}{L_2}=\frac{36}{35}$, $\frac{d_1}{d_2}=\frac{4}{1}$ and $\frac{{\rho }_1}{{\rho }_2}=\frac{1}{2}$
$\therefore$ $\frac{{\nu }_1}{{\nu }_2}=\frac{35}{36}\sqrt{\frac{8}{1}.\frac{2\times 1^2}{1\times 4^2}}$
We get $\frac{{\nu }_1}{{\nu }_2}=\frac{35}{36}$
Higher frequency ${\nu }_2=360Hz$
Or $\frac{{\nu }_1}{360}=\frac{35}{36}$
$⟹{\nu }_1=350Hz$
Beat frequency $b={\nu }_2-{\nu }_1=360-350=10Hz$