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Two uniform strings A and B made of steel are made to vibrate under the same tension. If the first overtone of A is equal to the second overtone of B and if the radius of A is twice that of B, the ratio of the length of the strings is:
A. $1:2$
B. $1:3$
C. $1:4$
D. $1:5$

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Last updated date: 13th Jun 2024
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Answer
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Hint Using the expression of frequency for a string we can equate the first overtone of string A to the second overtone of string B, where overtone is essentially defined as the frequency produced which is greater than the fundamental frequency.
Formula used
$f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{\mu }} $ where $l$is the length of the string, $T$ is its tension and $\mu $is mass per unit length of the string.
$\mu = A \times \rho $ where $\mu $ is the mass per unit length, $\rho $ is the density and $A$ is the area.

Complete step by step answer
Let the tension of the two strings be $T$
Now, we know that the frequency of vibration of a string is given by the formula,
$f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{\mu }} $ where $l$is the length of the string, $T$is its tension and $\mu $ is mass per unit length of the string.
Now, an overtone is defined as any frequency produced by an instrument which is greater than the fundamental frequency.
Therefore the frequency for string A be ${f_A} = \dfrac{1}{{2{l_A}}}\sqrt {\dfrac{T}{{A\rho }}} $ where ${l_A}$ is the length of string A, $T$ is the tension in the string, $A$ is its area and $\rho $is its density.
The mass per unit length can be written as the product of area and density by matching their dimensions as given below,
$
  \mu = \dfrac{{A \times l \times \rho }}{l} \\
   \Rightarrow \mu = A \times \rho \\
$
The frequency for string B is ${f_B} = \dfrac{1}{{2{l_B}}}\sqrt {\dfrac{T}{{A\rho }}} = \dfrac{1}{{2{l_B}}}\sqrt {\dfrac{T}{{\pi r_B^2}}} $
Now it’s given that the first overtone of A is equal to the second overtone of B,
So, $2{f_A} = 3{f_B}$
So comparing the above two equations we get,
$
  \dfrac{2}{{2{l_A}}}\sqrt {\dfrac{T}{{\pi r_A^2}}} = \dfrac{3}{{2{l_B}}}\sqrt {\dfrac{T}{{\pi r_B^2}}} \\
    \\
\ $
It’s given in the question that ${r_A} = 2{r_B}$
Substituting this value we get,
$
  \dfrac{2}{{{l_A}}}\sqrt {\dfrac{1}{{4r_B^2}}} = \dfrac{3}{{{l_B}}}\sqrt {\dfrac{1}{{r_B^2}}} \\
   \Rightarrow \dfrac{{{l_A}}}{{{l_B}}} = \dfrac{1}{3} \\
$

Therefore, the correct option is option B.

Note The term harmonic means the integer multiple (whole number) of the fundamental frequency and the term overtone is used to refer to any resonant frequency above the fundamental frequency. So, an overtone may or may not be a harmonic.