e) Consider two cylindrical pipes of equal length. One of these acts as a closed organ pipe and the other as open organ pipe. The frequency of the third harmonic in the closed pipe is 200 Hz higher than the first harmonic of the open pipe. Calculate the fundamental frequency of the closed pipe.

Introduction

This problem involves harmonics in open and closed cylindrical organ pipes of equal length. We are given a relation between harmonics and are required to find the fundamental frequency of the closed pipe.

Understanding the Harmonics

In an open organ pipe, all harmonics are present: f₁, 2f₁, 3f₁, …
In a closed organ pipe, only odd harmonics are present: f₁, 3f₁, 5f₁, …

Let L be the length of both pipes. Let v be the speed of sound in air (≈ 343 m/s at room temperature).

Step 1: Represent Harmonics

Let f_oc be the fundamental frequency of the closed pipe.
The third harmonic of the closed pipe = 3 × f_oc
Let f_oo be the fundamental frequency of the open pipe = v / (2L)
The first harmonic of the open pipe = f_oo

According to the question:

3 × f_oc = f_oo + 200 …(1)

Step 2: Express f_oo and f_oc in Terms of L

f_oc = v / (4L) (since closed pipe fundamental)

f_oo = v / (2L) (open pipe fundamental)

Step 3: Plug Into Equation (1)

3 × (v / 4L) = (v / 2L) + 200

(3v / 4L) − (v / 2L) = 200

Common denominator = 4L:

((3v − 2v) / 4L) = 200 → (v / 4L) = 200

So:

f_oc = v / 4L = 200 Hz

Final Answer

Fundamental frequency of the closed pipe = 200 Hz

Conclusion

Using the harmonic properties of open and closed organ pipes and the given difference in frequencies, we were able to calculate that the fundamental frequency of the closed pipe is 200 Hz.