Question

The difference between the fundamental frequencies of an open pipe and a closed pipe of the same length is 100 Hz. The difference between the frequencies of the second harmonic of the open pipe and the third harmonic of the closed pipe is:

• A. \( 100 \, \text{Hz} \)
• B. \( 150 \, \text{Hz} \)
• C. \( 200 \, \text{Hz} \)
• D. \( 250 \, \text{Hz} \) \bigskip

Hint:

The fundamental frequency of an open pipe is given by: \[ f_o = \frac{v}{2L} \] The fundamental frequency of a closed pipe is: \[ f_c = \frac{v}{4L} \] Since the difference between these frequencies is given as 100 Hz, the relationship can be used to find higher harmonics. The second harmonic of the open pipe is \( 2 f_o \), and the third harmonic of the closed pipe is \( 3 f_c \). By substituting the given difference, the final frequency difference comes out to 100 Hz.

Answer 1

The Correct Option is A

Let the length of the pipe be \( L \), and the speed of sound in air be \( v \). For the open pipe, the fundamental frequency \( f_1 \) is given by: \[ f_1 = \frac{v}{2L} \] For the closed pipe, the fundamental frequency \( f_1' \) is given by: \[ f_1' = \frac{v}{4L} \] The difference between the fundamental frequencies is given as 100 Hz: \[ f_1 - f_1' = 100 \] \[ \frac{v}{2L} - \frac{v}{4L} = 100 \] \[ \frac{v}{4L} = 100 \] \[ v = 400L \] Now, the second harmonic frequency of the open pipe is: \[ f_2 = \frac{v}{L} \] And the third harmonic frequency of the closed pipe is: \[ f_3' = \frac{3v}{4L} \] The difference between \( f_2 \) and \( f_3' \) is: \[ f_2 - f_3' = \frac{v}{L} - \frac{3v}{4L} = \frac{v}{4L} = 100 \, \text{Hz} \] Thus, the difference is \( 100 \, \text{Hz} \).