Question

If we are given \[{{n}_{1}}\], \[{{n}_{2}}\] 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). \[\dfrac{1}{n}=\dfrac{1}{{{n}_{1}}}+\dfrac{1}{{{n}_{2}}}+\dfrac{1}{{{n}_{3}}}\]
B).\[\dfrac{1}{\sqrt{n}}=\dfrac{1}{\sqrt{{{n}_{1}}}}+\dfrac{1}{\sqrt{{{n}_{2}}}}+\dfrac{1}{\sqrt{{{n}_{3}}}}\]
C). \[\sqrt{n}=\sqrt{{{n}_{1}}}+\sqrt{{{n}_{2}}}+\sqrt{{{n}_{3}}}\]
D). \[n={{n}_{1}}+{{n}_{2}}+{{n}_{3}}\]​

Answer 1

Hint: Here, frequency of all three strings has been given. Analyzing the equation for the fundamental frequency of oscillations, we can see that the total length of a string is related to its frequency. Hence, take the length of strings as the factor leading to the equation and also change the frequency equation in a suitable form to derive the solution.

Formula used:
\[f=\left( \dfrac{1}{2L} \right)\sqrt{\dfrac{T}{\mu }}\]

Complete step-by-step solution:
Given,
\[\text{n }\!\!~\!\!\text{ - fundamental frequency of the total string}\]
\[{{\text{n}}_{\text{1}}}\text{ }\!\!~\!\!\text{ - fundamental frequency of string 1}\]
\[{{\text{n}}_{\text{2}}}\text{ }\!\!~\!\!\text{ - fundamental frequency of string 2}\]
\[{{\text{n}}_{\text{3}}}\text{ }\!\!~\!\!\text{ - fundamental frequency of string 3}\]
We have,
Fundamental frequency of oscillation, \[f=\left( \dfrac{1}{2L}
\right)\sqrt{\dfrac{T}{\mu }}\]
Where,
\[\text{L - length of the string in centimeters}\]
\[\text{T - String tension in gcm/}{{\text{s}}^{2}}\]
\[\text{ }\!\!\mu\!\!\text{ - Linear density or mass per unit length of the string in g/cm}\]
Here,
\[\text{f = n}\]
Then,
\[n=\left( \dfrac{1}{2L} \right)\sqrt{\dfrac{T}{\mu }}\]
Or, we can write it as,
\[\dfrac{1}{n}=\dfrac{2L}{\sqrt{\dfrac{T}{\mu }}}\] -------- (1)
Take,
\[{{\text{L}}_{\text{1}}}\text{-length of string1}\]
\[{{L}_{2}}\text{- length of string 2}\]
\[{{L}_{3}}\text{- length of string 3}\].
Then,
\[\Rightarrow \dfrac{1}{{{n}_{1}}}=\dfrac{2{{L}_{1}}}{\sqrt{\dfrac{T}{\mu }}}\]
\[\Rightarrow \dfrac{1}{{{n}_{2}}}=\dfrac{2{{L}_{2}}}{\sqrt{\dfrac{T}{\mu }}}\]
\[\Rightarrow \dfrac{1}{{{n}_{3}}}=\dfrac{2{{L}_{3}}}{\sqrt{\dfrac{T}{\mu }}}\]
Total length of the string, \[L={{L}_{1}}+{{L}_{2}}+{{L}_{3}}\] ------(2)
Substitute 2 in equation 1.
Then,
\[\dfrac{1}{n}=\dfrac{2\left({{L}_{1}}+{{L}_{2}}+{{L}_{3}}\right)}{\sqrt{\dfrac{T}{\mu}}}=\dfrac{2{{L}_{1}}}{\sqrt{\dfrac{T}{\mu}}}+\dfrac{2{{L}_{2}}}{\sqrt{\dfrac{T}{\mu}}}+\dfrac{2{{L}_{3}}}{\sqrt{\dfrac{T}{\mu}}}=\dfrac{1}{{{n}_{1}}}+\dfrac{1}{{{n}_{2}}}+\dfrac{1}{{{n}_{3}}}\]
Therefore,
\[\dfrac{1}{n}=\dfrac{1}{{{n}_{1}}}+\dfrac{1}{{{n}_{2}}}+\dfrac{1}{{{n}_{3}}}\]
Hence, the answer is option A.

Additional information:
Oscillation is considered as a type of periodic motion. A motion is considered to be periodic, only if it repeats itself after regular intervals of time. The motion of a particle is said to be oscillatory or vibratory if it moves back and forth along the same path, and the frequency of this motion is one of its most important physical characteristics.

Note: Typically, the S.I unit of tension, length, and linear density are Newton, meter, and kg/m respectively. But, for the calculations with strings, these units are inconvenient. Thus, the smaller units of these quantities are used. But the frequency of oscillation of a string is measured in Hertz \[\left( Hz \right)\].