If the length of a stretched string is shortened by 40% and the tension is increased by 44%, then the ratio of the final and initial fundamental frequencies is:
A. 2:1
B. 3:2
C. 3:4
D. 1:3

Hint: To find the fundamental frequency of the transverse wave on a string, we use the formula of transverse wave speed on string. Then comparing the string with the air column closed at both the ends, we get the fundamental frequency.

Formula used:
\[v = \sqrt {\dfrac{T}{\mu }} \],
where v is the speed of the transverse wave on a string with \[\mu \] as the mass per unit length of the string, i.e. \[\mu = \dfrac{m}{l}\] and T is the applied tension.
\[{\nu _0} = \dfrac{v}{{2l}}\],
where \[{\nu _0}\]is the fundamental frequency of a wave on a string of length l fixed at both the ends.

Complete step by step solution:
If the length of the string is l , the mass of the string is m and the applied tension is T. Then the speed of the transverse wave on the string is,
\[v = \sqrt {\dfrac{T}{\mu }} \]
here \[\mu \] is the mass per unit length of the string, i.e. \[\mu = \dfrac{m}{l}\].

Both the ends of the string are fixed, so the fundamental frequency is equivalent to the air column problem with two ends closed.
\[{f_0} = \dfrac{v}{{2l}}\]
Using the expression of the speed of wave, we get the fundamental frequency of the transverse wave on the string as,
\[{f_0} = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{\mu }} \]

Let the initial length is \[{l_1}\] and the applied tension is \[{T_1}\]. Then using formula for the velocity of the transverse wave on the string, the initial velocity is,
\[{f_{01}} = \dfrac{1}{{2{l_1}}}\sqrt {\dfrac{{{T_1}}}{\mu }} \]
Let the initial length is \[{l_2}\] and the applied tension is \[{T_2}\]. Then using formula for the velocity of the transverse wave on the string, the initial velocity is,
\[{f_{02}} = \dfrac{1}{{2{l_2}}}\sqrt {\dfrac{{{T_2}}}{\mu }} \]

Dividing first fundamental frequency by the second, we get
\[\dfrac{{{f_{01}}}}{{{f_{02}}}} = \dfrac{{\dfrac{1}{{2{l_1}}}\sqrt {\dfrac{{{T_1}}}{\mu }} }}{{\dfrac{1}{{2{l_2}}}\sqrt {\dfrac{{{T_2}}}{\mu }} }} \\ \]
\[\Rightarrow \dfrac{{{f_{01}}}}{{{f_{02}}}} = \dfrac{{{l_2}}}{{{l_1}}}\sqrt {\dfrac{{{T_1}}}{{{T_2}}}} \\ \]
\[\Rightarrow \dfrac{{{f_{01}}}}{{{f_{02}}}} = \dfrac{{{l_1}\left( {1 - \dfrac{{40}}{{100}}} \right)}}{{{l_1}}} \times \sqrt {\dfrac{{{T_1}}}{{{T_1}\left( {1 + \dfrac{{44}}{{100}}} \right)}}} \\ \]
\[\Rightarrow \dfrac{{{f_{01}}}}{{{f_{02}}}} = \dfrac{{60}}{{100}} \times \dfrac{{10}}{{12}} = \dfrac{1}{2} \\ \]
\[\therefore \dfrac{{{f_{02}}}}{{{f_{01}}}} = \dfrac{2}{1}\]
Hence, the ratio of the final fundamental frequency to the initial fundamental frequency is 2:1

Therefore, the correct option is A.

Note: We should be careful about whether the string is compressed or shortened. In case of shortening of length, mass per unit length remains same. In case of compression, the mass per unit length changes.