Q119 / 200 · NEET 2012 (Screening) · Physics
When a string is divided into three segments of lengths $l_1, l_2$ and $l_3$, the fundamental frequencies of these three segments are $v_1$, $v_2$ and $v_3$ respectively. The original fundamental frequency $(v)$ of the string is
- A$\sqrt{v}=\sqrt{v_1}+\sqrt{v_2}+\sqrt{v_3}$
- B$v=v_1+v_2+v_3$
- Correct answerC$\frac{1}{v}=\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}$
- D$\frac{1}{\sqrt{v}}=\frac{1}{\sqrt{v_1}}+\frac{1}{\sqrt{v_2}}+\frac{1}{\sqrt{v_3}}$
Solution
The fundamental frequency of string
$\begin{aligned}
v & =\frac{1}{2 l} \sqrt{\frac{T}{m}} \\
\therefore v_1 l_1 & =v_2 l_2=v_2 l_3=k
\end{aligned}$
From Eq. (i)
$l_1=\frac{k}{v_1}, l_2=\frac{k}{v_2}, l_3=\frac{k}{v_3}$
Original length
$\begin{aligned}
l & =\frac{k}{v} \\
l & =l_1+l_2+l_3 \\
\frac{k}{v} & =\frac{k}{v_1}+\frac{k}{v_2}+\frac{k}{v_3} \\
\frac{1}{v} & =\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}
\end{aligned}$
Here,
$\begin{aligned}
& l=l_1+l_2+l_3 \\
& \frac{k}{v}=\frac{k}{v_1}+\frac{k}{v_2}+\frac{k}{v_3} \\
& \frac{1}{v}=\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}
\end{aligned}$
Hence, the correct answer is option: 'C'.