Q: A pipe closed at one end produces a fundamental note of 412 Hz. It is cut into two pieces of equal length. The fundamental frequencies produced in the two pieces are
(a) 206 Hz , 412 Hz
(b) 824 Hz , 1648 Hz
(c) 412 Hz , 824 Hz
(d) 206 Hz , 824 Hz
Ans:(b)
Sol: Here, v/4l = 412
When cut into two equal pieces, frequency of closed pipe of half the length
= v/4(l/2) = 2v/4l
= 2 × 412 = 824 Hz
Frequency of open pipe of half the length
= v/2(l/2)
= 4 × 412 = 1648 Hz