An open organ pipe and closed organ pipe of same length produce 2 beats per second, when they are set into vibrations together, in fundamental mode. The length of open pipe is made half and that of closed pipe is doubled. The number of beats produced per second will be (neglect end correction)

Step-by-step solution

Let L be the initial length of both pipes and v the speed of sound. The fundamental frequencies are f_o = v 2L for the open pipe and f_c = v 4L for the closed pipe. The initial beat frequency is the absolute difference: f_ beat1 = |f_o - f_c| = | v 2L - v 4L | = v 4L . Given f_ beat1 = 2 Hz, this yields v 4L = 2 , so v L = 8 Hz. After modification, the open pipe length becomes L 2 and the closed pipe length 2L . The new frequencies are f_o' = v 2(L/2) = v L and f_c' = v 4(2L) = v 8L . Substituting v L = 8 gives f_o