Question:
A closed organ pipe has fundamental frequency \( 100 \text{ Hz} \). Length of pipe is:
A closed organ pipe has fundamental frequency \( 100 \text{ Hz} \). Length of pipe is:
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To avoid mixing up pipes, remember that a closed organ pipe acts like a quarter-wavelength resonator (\( L = v / 4f \)) because it alternates boundary types (one open, one closed). An open organ pipe has identical boundaries at both ends, making it a half-wavelength resonator (\( L = v / 2f \))!
Updated On: May 20, 2026
- \( \frac{v}{200} \)
- \( \frac{v}{400} \)
- \( \frac{v}{100} \)
- \( \frac{v}{50} \)
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The Correct Option is B
Solution and Explanation
Concept:
A closed organ pipe is cylindrical in shape and closed at one end while remaining open at the other. When air inside is set into vibration, standing waves are formed under specific boundary constraints:
• At the closed end, the air molecules are fixed, creating a displacement node (zero amplitude).
• At the open end, air molecules can vibrate freely with maximum freedom, creating a displacement antinode (maximum amplitude). For the simplest mode of vibration (known as the fundamental mode or first harmonic), the standing wave pattern consists of exactly one node at the closed base and one antinode at the open top. The distance between an adjacent node and antinode is equal to one-quarter of the wavelength (\( \lambda / 4 \)). Therefore, the relationship between the physical length \( L \) of the pipe and the wavelength \( \lambda \) is: \[ L = \frac{\lambda}{4} \implies \lambda = 4L \] Using the wave relationship \( v = f \cdot \lambda \) (where \( v \) is the speed of sound and \( f \) is the frequency): \[ f = \frac{v}{\lambda} = \frac{v}{4L} \]
Step 1: Setting up the fundamental frequency expression using the given numerical value.
We are given that the fundamental frequency of the closed organ pipe is: \[ f = 100 \text{ Hz} \] Substituting this value into the structural formula: \[ 100 = \frac{v}{4L} \quad \cdots (1) \]
Step 2: Rearranging equation (1) to solve for the length \( L \) of the pipe.
Cross-multiply to isolate \( L \): \[ 100 \cdot 4L = v \] \[ 400L = v \quad \Rightarrow \quad L = \frac{v}{400} \] This precisely matches option (B).
• At the closed end, the air molecules are fixed, creating a displacement node (zero amplitude).
• At the open end, air molecules can vibrate freely with maximum freedom, creating a displacement antinode (maximum amplitude). For the simplest mode of vibration (known as the fundamental mode or first harmonic), the standing wave pattern consists of exactly one node at the closed base and one antinode at the open top. The distance between an adjacent node and antinode is equal to one-quarter of the wavelength (\( \lambda / 4 \)). Therefore, the relationship between the physical length \( L \) of the pipe and the wavelength \( \lambda \) is: \[ L = \frac{\lambda}{4} \implies \lambda = 4L \] Using the wave relationship \( v = f \cdot \lambda \) (where \( v \) is the speed of sound and \( f \) is the frequency): \[ f = \frac{v}{\lambda} = \frac{v}{4L} \]
Step 1: Setting up the fundamental frequency expression using the given numerical value.
We are given that the fundamental frequency of the closed organ pipe is: \[ f = 100 \text{ Hz} \] Substituting this value into the structural formula: \[ 100 = \frac{v}{4L} \quad \cdots (1) \]
Step 2: Rearranging equation (1) to solve for the length \( L \) of the pipe.
Cross-multiply to isolate \( L \): \[ 100 \cdot 4L = v \] \[ 400L = v \quad \Rightarrow \quad L = \frac{v}{400} \] This precisely matches option (B).
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