A resonance tube has an unknown end correction. A 500 Hz tuning fork gives first resonance at air-column length 12 cm, while a 250 Hz tuning fork gives first resonance at 29 cm in the same tube. What is the speed of sound?
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Question

A resonance tube has an unknown end correction. A 500 Hz tuning fork gives first resonance at air-column length 12 cm, while a 250 Hz tuning fork gives first resonance at 29 cm in the same tube. What is the speed of sound?

Options

G1 Physics · Accepted Answer

For first resonance, l+e = v/(4f). For the 500 Hz tuning fork, 0.12+e = v/(2000). Therefore, v = 2000(0.12+e). For the 250 Hz tuning fork, 0.29+e = v/(1000). Therefore, v = 1000(0.29+e). Equate the two expressions: 2000(0.12+e) = 1000(0.29+e). Divide by 1000: 2(0.12+e) = 0.29+e. Thus, 0.24+2e = 0.29+e, so e = 0.05 m. Hence, v = 2000(0.12+0.05) = 2000(0.17) = 340 m/s. Correct option: C.