Question

If $n_1,n_2$ and $n_3$ are the fundamental frequencies of three segments of a string of length $l$, then the original fundamental frequency $n$ of the string is given by

• A. $\frac{1}{n}=\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}$
• B. $\frac{1}{\sqrt{n}}=\frac{1}{\sqrt{n_1}}+\frac{1}{\sqrt{n_2}}+\frac{1}{\sqrt{n_3}}$
• C. $\sqrt{n}=\sqrt{n_1}+\sqrt{n_2}+\sqrt{n_3}$
• D. $n=n_1+n_2+n_3$

Answer 1

The correct option is A $\frac{1}{n}=\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}$


Total length of the string $l=l_1+l_2+l_3.......\left(1\right)$
We know that,
Fundamental frequency of a string,
$n=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$
$\Rightarrow l=\frac{1}{2n}\sqrt{\frac{T}{\mu }}$

Similarly,

$l_1=\frac{1}{2n_1}\sqrt{\frac{T}{\mu }};l_2=\frac{1}{2n_2}\sqrt{\frac{T}{\mu }};l_3=\frac{1}{2n_3}\sqrt{\frac{T}{\mu }}$

From $\left(1\right)$ we can write that,

$\frac{1}{2n}\sqrt{\frac{T}{\mu }}=\frac{1}{2n_1}\sqrt{\frac{T}{\mu }}+\frac{1}{2n_2}\sqrt{\frac{T}{\mu }}+\frac{1}{2n_3}\sqrt{\frac{T}{\mu }}$

$\Rightarrow \frac{1}{n}=\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}$

Thus, option (a) is the correct answer.