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: Understanding the Concept:
The fundamental frequency (\(f\)) of a vibrating string is determined by its length (\(L\)), the tension (\(T\)) applied to it, and its linear mass density (\(\mu\)). We need to analyze how changes in length and tension affect this frequency.
Step 2: Key Formula or Approach:
The formula for the fundamental frequency of a string is:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \] We can set up a ratio of the final frequency (\(f_2\)) to the initial frequency (\(f_1\)) to solve for the ratio of tensions \(\frac{T_2}{T_1}\).
Step 3: Detailed Explanation:
Let the initial conditions be \(f_1, L_1, T_1\) and the final conditions be \(f_2, L_2, T_2\). The linear density \(\mu\) is constant.
Initial frequency: \(f_1 = \frac{1}{2L_1} \sqrt{\frac{T_1}{\mu}}\).
Now, let's define the final conditions based on the problem statement:
- "length of the string is decreased by 25%": \(L_2 = L_1 - 0.25L_1 = 0.75L_1 = \frac{3}{4}L_1\).
- "fundamental frequency... increases by 100%": \(f_2 = f_1 + 1.00f_1 = 2f_1\).
Final frequency: \(f_2 = \frac{1}{2L_2} \sqrt{\frac{T_2}{\mu}}\).
Now, we take the ratio of the frequencies:
\[ \frac{f_2}{f_1} = \frac{\frac{1}{2L_2} \sqrt{\frac{T_2}{\mu}}}{\frac{1}{2L_1} \sqrt{\frac{T_1}{\mu}}} \] The terms \( \frac{1}{2} \) and \( \sqrt{\mu} \) cancel out.
\[ \frac{f_2}{f_1} = \left(\frac{L_1}{L_2}\right) \sqrt{\frac{T_2}{T_1}} \] Substitute the known values for the ratios of frequency and length:
\[ 2 = \left(\frac{L_1}{\frac{3}{4}L_1}\right) \sqrt{\frac{T_2}{T_1}} \] \[ 2 = \frac{4}{3} \sqrt{\frac{T_2}{T_1}} \] Now, we solve for the ratio of tensions. First, isolate the square root term:
\[ \sqrt{\frac{T_2}{T_1}} = 2 \times \frac{3}{4} = \frac{6}{4} = \frac{3}{2} \] Finally, square both sides to get the desired ratio:
\[ \frac{T_2}{T_1} = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \] Step 4: Final Answer:
The ratio \(\frac{T_2}{T_1}\) is \(\frac{9}{4}\). Therefore, option (D) is correct.