Physics, asked by LakyaMuni, 1 year ago

Two tuning forks Aand B when sounded together give 4 beats per second. The fork A is in resonance with a closed column of air of length 15 cm, while the second is in resonance with an open column of length 30.5 cm.Calculate their frequencies.

Answers

Answered by kvnmurty
35
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


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