Question

The fundamental frequency of transverse wave of a stretched string subjected to a tension \( T_1 \) is 300 Hz. If the length of the string is doubled and subjected to a tension of \( T_2 \), the fundamental frequency of the transverse wave in the string becomes 100 Hz, then \( T_2 : T_1 = \) (Linear density of the string is constant)

• A. 1:2
• B. 3:4
• C. 2:3
• D. 4:9

Hint:

Write out the proportionality relation \( f \propto \frac{\sqrt{T}}{L} \) to simplify ratio problems quickly.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:

The fundamental frequency of a stretched string depends on its length, tension, and linear mass density.
Step 2: Key Formula or Approach:

Formula for fundamental frequency: \( f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} \) where \( L \) is length, \( T \) is tension, and \( \mu \) is linear mass density.
Step 3: Detailed Explanation:

Given: \( f_1 = 300 \) Hz, \( L_1 = L \), \( T_1 \) \( f_2 = 100 \) Hz, \( L_2 = 2L \), \( T_2 \) \( \mu \) is constant. Write the ratio \( \frac{f_2}{f_1} \): \[ \frac{f_2}{f_1} = \frac{\frac{1}{2(2L)}\sqrt{\frac{T_2}{\mu}}}{\frac{1}{2L}\sqrt{\frac{T_1}{\mu}}} \] \[ \frac{100}{300} = \frac{1}{2} \sqrt{\frac{T_2}{T_1}} \] \[ \frac{1}{3} = \frac{1}{2} \sqrt{\frac{T_2}{T_1}} \] \[ \frac{2}{3} = \sqrt{\frac{T_2}{T_1}} \] Squaring both sides: \[ \frac{T_2}{T_1} = \left( \frac{2}{3} \right)^2 = \frac{4}{9} \]
Step 4: Final Answer:

The ratio \( T_2 : T_1 \) is 4:9.