The fourth overtone of a closed pipes is 900Hz, its fundamental frequency is

1. Analyze the question

We are given the frequency of the fourth overtone of a closed pipe (900 Hz) and asked to find the fundamental frequency ($f_0$).

2. Explain the concept

In a closed pipe (a pipe closed at one end and open at the other), the air column vibrates in odd harmonics only. The relationship for the frequencies of a closed pipe is given by:

Where:

• $f_n$ is the frequency of the $n^{th}$ overtone.
• $n$ is the overtone number ($n = 0, 1, 2, 3, \dots$).
• $f_0$ is the fundamental frequency.

For a closed pipe:

• Fundamental frequency ($f_0$) = $1f_0$
• First overtone = $3f_0$
• Second overtone = $5f_0$
• Third overtone = $7f_0$
• Fourth overtone = $9f_0$

3. Show the solution

• Step 1: Identify the formula for the fourth overtone. Based on the sequence above, the $n^{th}$ overtone of a closed pipe is $(2n + 1)f_0$. For the 4th overtone ($n=4$): $$f_4 = (2(4) + 1)f_0 = 9f_0$$

• Step 2: Substitute the given value into the equation. We are told the fourth overtone is 900 Hz: $$900 = 9f_0$$

• Step 3: Solve for $f_0$. $$f_0 = \frac{900}{9}$$ $$f_0 = 100 \text{ Hz}$$

4. Identify the answer

The fundamental frequency is 100 Hz. This corresponds to option B.

5. Tip

Memory Trick: For closed pipes, remember the sequence of multipliers is always odd numbers: 1, 3, 5, 7, 9...

• Fundamental = 1st harmonic (1)
• 1st overtone = 3rd harmonic (3)
• 2nd overtone = 5th harmonic (5)
• ...and so on. Just add 1 to the overtone number, multiply by 2, and subtract 1 to get the harmonic number! Or simply remember: Overtone $n$ = Harmonic $(2n+1)$.