Question

If the fundamental frequency of the stretched string of length 1 m under a given tension is 3 Hz, then the fundamental frequency of the stretched string of length 0.75 m under the same tension is:

• A. 1 Hz
• B. 2 Hz
• C. 6 Hz
• D. 4 Hz
• E. 5 Hz

Hint:

The frequency of a stretched string is inversely proportional to its length. Reducing the length increases the frequency.

Answer 1

The Correct Option is D

The fundamental frequency \( f \) of a stretched string is given by the formula: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \] where: - \( L \) is the length of the string,
- \( T \) is the tension in the string,
- \( \mu \) is the mass per unit length of the string.
Since the tension \( T \) and mass per unit length \( \mu \) remain constant, the frequency is inversely proportional to the length of the string: \[ f \propto \frac{1}{L} \] Let \( f_1 \) and \( f_2 \) be the fundamental frequencies for strings of lengths \( L_1 = 1 \, {m} \) and \( L_2 = 0.75 \, {m} \), respectively. From the proportionality, we have: \[ \frac{f_1}{f_2} = \frac{L_2}{L_1} \] Substitute the given values: \[ \frac{3}{f_2} = \frac{0.75}{1} \quad \Rightarrow \quad f_2 = \frac{3}{0.75} = 4 \, {Hz} \] Thus, the fundamental frequency of the stretched string of length 0.75 m is 4 Hz.
Therefore, the correct answer is option (D), 4 Hz.