Two uniform stretched steel strings $A$ and $B$ are vibrating under the same tension. The first overtone of…
Two uniform stretched steel strings $A$ and $B$ are vibrating under the same tension. The first overtone of $A$ is equal to the second overtone of $B$. If the radius of $A$ is twice that of $B$, then the ratio of the lengths of the strings is
2 : 3
1 : 2
1 : 3
1 : 4
Solution
Frequency in stretched string,
$
f=\frac{n}{2 l} \cdot v=\frac{n}{2 l} \sqrt{\frac{T}{m}}
$
where, $l=$ length, $T=$ tension and
$m=$ mass per unit length
So, $\quad m=\frac{\pi r^2 l \cdot d}{l}$, where, $d=$ density and
$
m=\pi r^2 d \text {. }
$
Given, $f_A$ (first overtone, $n=2$ )
$
\begin{aligned}
& =f_B(\text { second overtone, } n=3) \\
\Rightarrow \frac{2}{2 l_A} \sqrt{\frac{T}{m_A}} & =\frac{3}{2 l_B} \sqrt{\frac{T}{m_B}} \Rightarrow \frac{1}{l_A \sqrt{\pi r_A^2 d}}=\frac{3}{2 l_B \sqrt{\pi r_B^2 d}} \\
\Rightarrow \quad \frac{l_A}{l_B} & =\frac{2}{3} \times \frac{r_B}{r_A}=\frac{2}{3} \times \frac{1}{2} \Rightarrow \frac{l_A}{l_B}=\frac{1}{3}
\end{aligned}
$