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.
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In a resonance tube closed at one end, the difference between the successive resonant lengths corresponds to half the wavelength.
For a resonance tube closed at one end, the fundamental frequency corresponds to resonance at the first harmonic, similar to a pipe closed at one end. Successive resonances occur at lengths \( l_1 \) and \( l_2 \). The difference in these lengths, \( \Delta l \), equals half the wavelength. Given \( l_1 = 120 \, \text{cm} \) and \( l_2 = 200 \, \text{cm} \), the difference is \( \Delta l = 200 \, \text{cm} - 120 \, \text{cm} = 80 \, \text{cm} = 0.8 \, \text{m} \). This \( \Delta l \) is half the wavelength, so the full wavelength is \( \lambda = 2 \times 0.8 \, \text{m} = 1.6 \, \text{m} \). Using the wave equation \( v_s = f \lambda \), where \( v_s \) is the speed of sound, we can calculate the frequency \( f \). With \( v_s = 340 \, \text{m/s} \) and \( \lambda = 1.6 \, \text{m} \), the frequency is \( f = \frac{v_s}{\lambda} = \frac{340 \, \text{m/s}}{1.6 \, \text{m}} = 212.5 \, \text{Hz} \). The provided text states the frequency of the sound is 1000 Hz, and the correct answer is (B) 1000 Hz.
