Question
A sonometer wire when vibrated in full length has frequency $$n.$$ Now, it is divided by the help of bridges into a number of segments of lengths $${l_1},{l_2},{l_3},\,....$$ When vibrated these segments have frequencies $${n_1},{n_2},{n_3},\,....$$ Then, the correct relation is
A.
$$n = {n_1} + {n_2} + {n_3} + ....$$
B.
$${n^2} = n_1^2 + n_2^2 + n_3^2 + \,....$$
C.
$$\frac{1}{n} = \frac{1}{{{n_1}}} + \frac{1}{{{n_2}}} + \frac{1}{{{n_3}}} + \,.....$$
D.
$$\frac{1}{{\sqrt n }} = \frac{1}{{\sqrt {{n_1}} }} + \frac{1}{{\sqrt {{n_2}} }} + \frac{1}{{\sqrt {{n_3}} }} + \,.....$$
Answer :
$$\frac{1}{n} = \frac{1}{{{n_1}}} + \frac{1}{{{n_2}}} + \frac{1}{{{n_3}}} + \,.....$$
Solution :
From law of length, the frequency of vibrating string is inversely proportional to its length,
\[{\text{i}}{\text{.e}}{\text{.}}\,\,n \propto \frac{1}{l}\,\,\left[ {\begin{array}{*{20}{c}}
{n = {\text{frequency of string}}} \\
{l = {\text{length of string}}}
\end{array}} \right]\]
$$\eqalign{
& {\text{or}}\,\,nl = {\text{constant}}\left( {{\text{say}}\,k} \right) \cr
& {\text{or}}\,\,l = \frac{k}{n} \cr} $$
The segments of string of length $${l_1},{l_2},{l_3},\,....$$ have frequencies $${n_1},{n_2},{n_3},\,....$$
Total length of string is $$l.$$
$$\eqalign{
& {\text{So,}}\,\,l = {l_1} + {l_2} + {l_3} + \,.... \cr
& \therefore \frac{k}{n} = \frac{k}{{{n_1}}} + \frac{k}{{{n_2}}} + \frac{k}{{{n_3}}} + \,.... \cr
& {\text{or}}\,\,\frac{1}{n} = \frac{1}{{{n_1}}} + \frac{1}{{{n_2}}} + \frac{1}{{{n_3}}} + \,.... \cr} $$