Question

Two uniform strings $A$ and $B$ made of steel are made to vibrate under the same tension. If the first overtone of $A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$, the ratio of the lengths of the strings is

• A. 1:02
• B. 1:03
• C. 1:04
• D. 1:06

Answer 1

Answer: (B)

1:03

Answer 2

First overtone of string $A$
= Second overtone of string $B$
$\Rightarrow$ 2nd harmonic of $A$ = 3rd harmonic of $B$
$\upsilon_A = \upsilon_B$
For a string $\upsilon = \frac{p}{l D} \sqrt{\frac{T}{\pi \rho}}$
For 2nd harmonic, $p = 2$, for 3rd harmonic,
$p = 3$
$\therefore \:\: \upsilon_A = \frac{2}{l_A D_A} \sqrt{\frac{T}{\pi \rho}}$ and $\upsilon_B = \frac{3}{l_B D_B} \sqrt{\frac{T}{\pi \rho}}$
$\frac{2}{l_A D_A} \sqrt{\frac{T}{\pi \rho}} = \frac{3}{l_B D_B} \sqrt{\frac{T}{\pi \rho}}$ (from eqn (i))
$\therefore \: \frac{2}{l_A D_A} = \frac{3}{l_B \frac{D_A}{2} } \Rightarrow \frac{l_A}{l_B} = \frac{2}{6} = \frac{1}{3} [ \because \, D_A = 2 D_B ]$