An air column in a tube $32\; cm$ long, closed at one end, is in resonance with a tuning fork. The air column in another tube, open at both ends, of length 66 cm is in resonance with another tuning fork. When these two tuning fork.The air column in another tube, open at both ends, of length 66 cm is in resonance with another tuning fork. When these two tuning forks are sounded together, they produce $8$ beats per seconds. Then the frequencies of the two tuning forks are , (Consider fundamental frequencies only)

Question

\[(a)\;250\;Hz, 258\; Hz \quad (b)\;240\;Hz,248\;Hz \quad (c)\;264\;Hz,256\;Hz \quad (d)\;280\;Hz,272\;Hz\]

Comments

Frequency of fork  A =  f1 frequency of fork  B = f2 beats =  | f1 - f2 | = 8  Hz f1 = resonance frequency of  closed Air column 32 cm     λ / 4 = 32 cm  => λ = 128 cm        =>          f1 = c / 128 Hz f2 = resonance frequency of open air column  66 cm      λ / 2  =  66 cm  =>  λ = 132 cm      =>    f2 = c / 132  Hz   f1 - f2  =  c * [1/128 - 1/132]=8Hz        => c = 33972 m/s =>  f1 = 264 Hz        and  f2 = 256Hz