Step 1: Formula for Fundamental Frequency:
The fundamental frequency of a string is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( L \) is length, \( T \) is tension, and \( \mu \) is linear mass density.
Step 2: Initial Condition:
Let initial length be \( L_1 \) and tension be \( T_1 \).
\[ f_1 = \frac{1}{2L_1} \sqrt{\frac{T_1}{\mu}} \]
Step 3: Final Condition:
Length is decreased by 25%, so \( L_2 = L_1 - 0.25 L_1 = 0.75 L_1 = \frac{3}{4} L_1 \).
Frequency increases by 100%, so \( f_2 = f_1 + 1.00 f_1 = 2f_1 \).
New tension is \( T_2 \).
\[ f_2 = \frac{1}{2L_2} \sqrt{\frac{T_2}{\mu}} \]
Step 4: Ratio Calculation:
Substitute \( f_2 = 2f_1 \) and \( L_2 = \frac{3}{4} L_1 \):
\[ 2f_1 = \frac{1}{2(\frac{3}{4}L_1)} \sqrt{\frac{T_2}{\mu}} \]
\[ 2f_1 = \frac{4}{3} \cdot \frac{1}{2L_1} \sqrt{\frac{T_2}{\mu}} \]
We know \( \frac{1}{2L_1\sqrt{\mu}} = \frac{f_1}{\sqrt{T_1}} \).
So,
\[ 2f_1 = \frac{4}{3} f_1 \sqrt{\frac{T_2}{T_1}} \]
Cancel \( f_1 \):
\[ 2 = \frac{4}{3} \sqrt{\frac{T_2}{T_1}} \]
\[ \sqrt{\frac{T_2}{T_1}} = \frac{6}{4} = \frac{3}{2} \]
Squaring both sides:
\[ \frac{T_2}{T_1} = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \]