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 forks are sounded together, they produce 8 beats per second. Then the frequencies of the two tuning forks are, (Consider fundamental frequencies only)

Step-by-step solution

We knows frequency of a closed end an column n₁= v 4 / 1 We knows frequency of a open end an column n₂= v 2 l₂ Given, l₁=32 ~cm , I₂=66 ~cm and n₁-n₂=8 heat / s So, n₁= v 4 32 = v 128 and n₂= v 2 66 = v 132 In given condition, aligned & v 128 - v 132 =8 & v=8448 4 & v=33792 & Hence, n₁= 33792 128 & n₁=264 ~Hz & and & n₂= 33792 132 & n₂=256 ~Hz & aligned