Two tuning forks and when sounded together give 4 beats per second. The fork is in resonance with a closed column of air of length , while the second is in resonance with an open column of length .Calculate their frequencies.
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I assume that the air column in resonance with fork A is closed at one end and open at one end. Then the second column that is in resonance with fork B is open at both ends. Further we take that the resonance frequencies in question are the fundamental frequencies and not hormonics. Let velocity of sound be = c.
Frequency of fork A = f1
frequency of fork B = f2
beats = | f1 - f2 | = 4 Hz
f1 = resonance frequency of closed Air column 15 cm
λ / 4 = 15 cm => λ = 60 cm => f1 = c / 60 Hz
f2 = resonance frequency of open air column 30.5 cm
λ / 2 = 30.5 cm => λ = 61 cm => f2 = c / 61 Hz
f1 - f2 = c * [1/60 - 1/61] = c / 3660 = 4 Hz
=> c = 14, 640 m/s
=> f1 = 244 Hz and f2 = 240 Hz
Frequency of fork A = f1
frequency of fork B = f2
beats = | f1 - f2 | = 4 Hz
f1 = resonance frequency of closed Air column 15 cm
λ / 4 = 15 cm => λ = 60 cm => f1 = c / 60 Hz
f2 = resonance frequency of open air column 30.5 cm
λ / 2 = 30.5 cm => λ = 61 cm => f2 = c / 61 Hz
f1 - f2 = c * [1/60 - 1/61] = c / 3660 = 4 Hz
=> c = 14, 640 m/s
=> f1 = 244 Hz and f2 = 240 Hz
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