A pipe closed at one end produces a fundamental note of frequency 412 Hz. If it is cut into two pieces of equal lengths , then the fundamental frequencies produced by the two pieces would be

Options

A. 206Hz, 412Hz
B. 824Hz, 1684Hz
C. 412Hz, 824Hz
D. 206Hz, 824Hz

Answer

B

Explanation

As the pipe is cut into two equal lengths . The frequency of a closed pipe of half the length is
`=(V)/(4(l//2))=(2V)/(4l) = 2xx412=824Hz`
`therefore` frequency of open pipe of half the length is
`=(V)/(2(l//2)) = 4xx412 = 1648Hz`