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A string is cut into three parts, having fundamental frequencies $n_1$, $n_2$ and $n_3$ respectively. Then original fundamental frequency 'n' is related by the expression as :
A
$\dfrac{1}{n} = \dfrac{1}{n_1} + \dfrac{1}{n_2} + \dfrac{1}{n_3}$
B
$n = n_1 \times n_2 \times n_3$
C
$n = n_1 + n_2 + n_3$
D
$n = \dfrac{n_1 + n_2 + n_3}{3}$
Detailed Solution
For the same tension and mass per unit length, the fundamental frequency $n = \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}$, so $l = \dfrac{k}{n}$ with k constant.
The lengths of the pieces add up to the original length: $l = l_1 + l_2 + l_3$
$\dfrac{k}{n} = \dfrac{k}{n_1} + \dfrac{k}{n_2} + \dfrac{k}{n_3}$
$\Rightarrow \dfrac{1}{n} = \dfrac{1}{n_1} + \dfrac{1}{n_2} + \dfrac{1}{n_3}$
The lengths of the pieces add up to the original length: $l = l_1 + l_2 + l_3$
$\dfrac{k}{n} = \dfrac{k}{n_1} + \dfrac{k}{n_2} + \dfrac{k}{n_3}$
$\Rightarrow \dfrac{1}{n} = \dfrac{1}{n_1} + \dfrac{1}{n_2} + \dfrac{1}{n_3}$