Question

Two wires are fixed on a sonometer. Their tensions are in the ratio $8:1$, the lengths in the ratio $36:35$, the diameters in the ratio $4:1$ and densities in the ratio $1:2$. If the string of higher pitch has a frequency of $360Hz$, the frequency of the other string is

• A. $350Hz$
• B. $450Hz$
• C. $550Hz$
• D. $650Hz$

Answer 1

The correct option is A $350Hz$

Given, $\frac{T_1}{T_2}=\frac{8}{1},\frac{r_1}{r_2}=\frac{4}{1}$
$\frac{l_1}{l_2}=\frac{36}{35},\frac{{\rho }_1}{{\rho }_2}=\frac{1}{2}$
$f_2=360Hz.$
frequency $=\frac{\nu }{2L}$
$=\frac{\sqrt{T/\rho \pi r^2}}{2l}$
So, $\frac{f_1}{f_2}=\frac{\sqrt{T_1/{\rho }_1\pi {r_1}^2}}{2l_1}\times \frac{2l_2}{\sqrt{T_2/{\rho }_2\pi {r_2}^2}}$
$=\sqrt{\frac{T_1}{T_2}\times \frac{{\rho }_2}{{\rho }_1}\times \frac{{r_2}^2}{{r_1}^2}}\times \frac{l_2}{l_1}$
$=\sqrt{\frac{8}{1}\times \frac{2}{1}\times {\left(\frac{1}{4}\right)}^2}\times \frac{35}{36}$
$=\frac{35}{36}$
So, $f_1=\frac{35}{36}\times f_2$
$=\frac{35}{36}\times 360$
$=350Hz.$