Question

If $n_1,n_2$ and $n_3$ are the fundamental frequencies of three segments into which a string is divided, 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

Answer: (A)

$n_1=\frac{1}{2l_1}\sqrt{\frac{T}{\mu }}...(i)$
$n_2=\frac{1}{2l_2}\sqrt{\frac{T}{\mu }}...(ii)$
$n_3=\frac{1}{2l_3}\sqrt{\frac{T}{\mu }}...(iii)$
$n=\frac{1}{2l}\sqrt{\frac{T}{\mu }}...(iv)$
From eqn. (iv), we get
$\frac{1}{n}=\frac{2l}{\sqrt{\frac{T}{\mu }}}$
As $l=l_1+l_2+l_3$
$\therefore \frac{1}{n}=\frac{2(l_1+l_2+l_3)}{\sqrt{\frac{T}{\mu }}}$
$=\frac{2l_1}{\sqrt{\frac{T}{\mu }}}+\frac{2l_2}{\sqrt{\frac{T}{\mu }}}+\frac{2l_3}{\sqrt{\frac{T}{\mu }}}$
$=\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}$ [Using (i), (ii) and (iii)]