The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is $20\,cm$, the length of the open organ pipe is

Question

A steel wire with mass per unit length $ 7.0 \times 10^{-3} \, \text{kg/m} $ is under a tension of $ 70 \, \text{N} $. The speed of transverse waves in the wire will be:

Answer 1

To solve this problem, we need to understand the relationship between the harmonics of open and closed organ pipes.

For a closed organ pipe, the harmonics are odd multiples of the fundamental frequency. The fundamental frequency is the first harmonic, the third harmonic is three times the fundamental frequency, and so on.

For an open organ pipe, the harmonics are all integer multiples of the fundamental frequency.

Given:

The length of the closed organ pipe, $ L_c = 20 \,\text{cm} $.

The fundamental frequency of the open organ pipe is equal to the third harmonic of the closed organ pipe.

Calculation:

Fundamental frequency of open pipe: $ f_{\text{open}} = \dfrac{v}{2L_o} $
Third harmonic of closed pipe: $ f_{\text{closed}} = \dfrac{3v}{4L_c} $

According to the question:

$ \dfrac{v}{2L_o} = \dfrac{3v}{4L_c} $

Cancel $ v $:

$ \dfrac{1}{2L_o} = \dfrac{3}{4L_c} $

Cross-multiplying:

$ 4L_c = 6L_o $

Simplifying:

$ L_o = \dfrac{2}{3}L_c $

Substitute $ L_c = 20 \,\text{cm} $:

$ L_o = \dfrac{2}{3} \times 20 = 13.33 \,\text{cm} $

Thus, the length of the open organ pipe is approximately 13.3 cm.

Answer 2