Chapter 12, Problem 34P

(a)

To determine

To identify: Whether the organ pipe is closed or open.

Answer to Problem 34P

Closed pipe.

Explanation of Solution

Given:

The frequencies = 264 Hz, 440 Hz, and 616 Hz.

Formula used:

Open pipe: $f_1=f_{n+1}-f_n$

Closed pipe: $f_1=\frac{(f_{n+1}-f_n)}{2n}$

Calculation:

Open pipes resonate at frequencies f , 2f , 3f , 4f , 5f , etc.

Closed pipes resonate at frequencies f , 3f , 5f , 7f , etc.

From the given frequencies,

440Hz − 264 Hz = 176 Hz,

616 Hz − 440 Hz = 176 Hz

The given frequencies are not the multiplies of 176. Therefore, it is a closed pipe. And the fundamental frequency f_f is,

  $f_f=\frac{176\text{ Hz}}{2}$

  $f_f=88\text{ Hz}$ .

  $n=\frac{264}{88}=3$

  $n=\frac{440}{88}=5$

The solution has odd integer. Therefore, it is closed pipe.

Conclusion:

The frequencies are in the range of odd integers. Therefore, it is a closed pipe.

(b)

To determine

The fundamental frequency of a pipe.

Answer to Problem 34P

  $f_f=88\text{ Hz}$

Explanation of Solution

Given:

The frequencies = 264 Hz, 440 Hz, and 616 Hz.

Formula used:

Closed pipe: $f_1=\frac{(f_{n+1}-f_n)}{2n}$

Calculation:

Open pipes resonate at frequencies f , 2f , 3f , 4f , 5f , etc.

From the given frequencies,

440Hz − 264 Hz = 176 Hz,

  $n=\frac{440}{176}=2.5$

  $n=\frac{264}{176}=1.5$

616 Hz − 440 Hz = 176 Hz

The given frequencies are not the multiplies of 176. Therefore, it is a closed pipe.

Closed pipes resonate at frequencies f , 3f , 5f , 7f , etc.

And the fundamental frequency f_f is,

  $f_f=\frac{176\text{ Hz}}{2}$

  $f_f=88\text{ Hz}$ .

Conclusion:

The fundamental frequency is 88 Hz, with respect to the concept of open and closed pipe principle.