Question

For a certain organ pipe, three successive resonant frequencies are heard as $300\ \text{Hz}$, $420\ \text{Hz}$ and $540\ \text{Hz}$. If the speed of sound in air is $360\ \text{m/s}$, then the pipe is a

• A. open pipe of $1.5\ \text{m}$ length
• B. open pipe of $3\ \text{m}$ length
• C. closed pipe of $3\ \text{m}$ length
• D. closed pipe of $1.5\ \text{m}$ length

Hint:

Equal spacing of successive resonant frequencies indicates a closed pipe when only odd harmonics are present.

Answer 1

The Correct Option is D

Step 1: Find the frequency difference.
\[ 420 - 300 = 120\ \text{Hz}, \quad 540 - 420 = 120\ \text{Hz} \] Hence, successive resonant frequencies differ by $120\ \text{Hz}$.
Step 2: Identify type of pipe.
For a closed organ pipe, successive resonant frequencies differ by: \[ \Delta f = \dfrac{v}{2L} \] Step 3: Substitute given values.
\[ 120 = \dfrac{360}{2L} \] Step 4: Solve for length $L$.
\[ 2L = 3 \Rightarrow L = 1.5\ \text{m} \] Step 5: Final conclusion.
Thus, the pipe is a closed organ pipe of length $1.5\ \text{m}$.