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. $1:3$
• B. $2:1$
• C. $3:2$
• D. $3:4$

Answer 1

The correct option is A $1:3$

Given:
First overtone of $A=$ second overtone of $B$

(i.e) $2^{nd}$ harmonic of $A=3^{rd}$ harmonic of $B$
$2n_1=3n_2$

$\because n=\frac{1}{2l}\sqrt{\frac{T}{\mu }}[\because \mu =\frac{\text{mass}}{\text{length}}]$

So,
$2\times \frac{1}{2l_1}\sqrt{\frac{T}{{\mu }_1}}=3\times \frac{1}{2l_2}\sqrt{\frac{T}{{\mu }_2}}$

$\frac{l_1}{l_2}=\frac{2}{3}\sqrt{\frac{{\mu }_2}{{\mu }_1}}$

$=\frac{2}{3}\sqrt{\frac{a_2\rho }{a_1\rho }}$

$\frac{l_1}{l_2}=\frac{2}{3}\sqrt{\frac{a_2\rho }{a_1\rho }}$

$\frac{l_1}{l_2}=\frac{2}{3}\sqrt{\frac{a_2}{a_1}}$

$\frac{l_1}{l_2}=\frac{2}{3}\sqrt{\frac{\pi r_2^2}{\pi r_1^2}}=\frac{2}{3}\sqrt{\frac{r_2^2}{r_1^2}}$

$\because r_1=2r_2$

$\frac{l_1}{l_2}=\frac{2}{3}\sqrt{\frac{r_2^2}{(2r_2{)}^2}}=\frac{2}{3}\sqrt{\frac{1}{4}}=\frac{1}{3}$

Hence, option (d) is the correct answer.