Question

A string vibrates in its fundamental mode when a tension \(T_1\) is applied to it. If the length of the string is decreased by 25% and the tension applied is changed to \(T_2\), the fundamental frequency of the string increases by 100%, then \(\frac{T_2}{T_1}=\)

• A. \(\frac{3}{8}\)
• B. \(\frac{2}{3}\)
• C. \(\frac{8}{9}\)
• D. \(\frac{9}{4}\)

Hint:

For problems involving changes in parameters, setting up a ratio is the most efficient method. It allows you to cancel out constants (\(\mu\) in this case) and work directly with the proportionalities. Remember \(f \propto \frac{1}{L}\) and \(f \propto \sqrt{T}\).

Answer 1

The Correct Option is D

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: \[ \frac{T_2}{T_1} = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \]