Question:medium

In an organ pipe closed at one end; the sum of the frequencies of first three overtones is $3930 \text{ Hz}$. The frequency of the fundamental mode of organ pipe is

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Closed pipe: $(2n+1)f$ harmonics. 1st overtone = 3rd harmonic ($n=1$).
Updated On: May 14, 2026
  • 256 Hz
  • 262 Hz
  • 320 Hz
  • 384 Hz
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
A pipe closed at one end produces only odd harmonics.
The fundamental mode corresponds to the first harmonic. The overtones refer to the higher resonant frequencies.
Step 2: Key Formula or Approach:
Harmonics for closed pipe: $f_n = n f_0$, where $n = 1, 3, 5, 7, \dots$
$f_0$ is fundamental frequency ($1^{\text{st}}$ harmonic).
First overtone = $3^{\text{rd}}$ harmonic = $3f_0$.
Second overtone = $5^{\text{th}}$ harmonic = $5f_0$.
Third overtone = $7^{\text{th}}$ harmonic = $7f_0$.
Step 3: Detailed Explanation:
The sum of the frequencies of the first three overtones is given as 3930 Hz.
Set up the equation: \[ 3f_0 + 5f_0 + 7f_0 = 3930 \] Add the coefficients: \[ 15f_0 = 3930 \] Solve for $f_0$: \[ f_0 = \frac{3930}{15} \] Performing the division: \[ f_0 = 262 \text{ Hz} \] Step 4: Final Answer:
The fundamental frequency is 262 Hz.
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