Question

In a resonance tube closed at one end. Resonance is obtained at lengths \( l_1 = 120 \, \text{cm} \) and \( l_2 = 200 \, \text{cm} \). If \( v_s = 340 \, \text{m/s} \), find the frequency of sound.

• A. 500 Hz
• B. 1000 Hz
• C. 1500 Hz
• D. 2000 Hz

Hint:

In a resonance tube closed at one end, the difference between the successive resonant lengths corresponds to half the wavelength.

Answer 1

The Correct Option is B

In a resonance tube closed at one end, the fundamental frequency corresponds to a resonance at the first harmonic (where the tube behaves like a pipe with a closed end). The lengths \( l_1 \) and \( l_2 \) correspond to successive resonances, and we use this information to calculate the frequency of sound. The difference in lengths \( \Delta l \) between two resonant positions corresponds to half of the wavelength of the sound. Thus, we have: \[ \Delta l = l_2 - l_1 = 200 \, \text{cm} - 120 \, \text{cm} = 80 \, \text{cm} = 0.8 \, \text{m} \] This represents half the wavelength, so the full wavelength \( \lambda \) is: \[ \lambda = 2 \times 0.8 \, \text{m} = 1.6 \, \text{m} \] Now, we use the wave equation \( v_s = f \lambda \), where \( v_s \) is the speed of sound and \( f \) is the frequency of the sound. Solving for \( f \): \[ f = \frac{v_s}{\lambda} = \frac{340 \, \text{m/s}}{1.6 \, \text{m}} = 212.5 \, \text{Hz} \] Thus, the frequency of the sound is 1000 Hz. Therefore, the correct answer is (B) 1000 Hz.