In case of closed organ pipe, which harmonin the `p^(th)` overtone will be

Options

A. 2p+1
B. 2p-1
C. p+1
D. p-1

Answer

To determine which harmonic corresponds to the \( p^{th} \) overtone in a closed organ pipe, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Closed Organ Pipe**: A closed organ pipe has one end closed and the other end open. The closed end is a displacement node (where the air does not move), and the open end is a displacement antinode (where the air moves freely).

2. **Identify the Overtone**: The \( p^{th} \) overtone refers to the \( p^{th} \) harmonic that is produced by the pipe. In a closed organ pipe, the harmonics are odd multiples of the fundamental frequency.

3. **Number of Loops**: For the \( p^{th} \) overtone, there are \( p \) loops. Each overtone adds one additional loop to the standing wave pattern.

4. **Length of the Pipe**: The length of the closed organ pipe can be expressed in terms of the wavelength \( \lambda \). For the \( p^{th} \) overtone, the relationship can be given as:
\[ L = \frac{p\lambda}{2} + \frac{\lambda}{4} \]
Here, \( \frac{\lambda}{4} \) accounts for the closed end of the pipe.

5. **Simplifying the Length Equation**: Rearranging the equation gives:
\[ L = \frac{p\lambda + \lambda/2}{2} = \frac{(2p + 1)\lambda}{4} \]
Therefore, we can express \( \lambda \) as:
\[ \lambda = \frac{4L}{2p + 1} \]

6. **Finding the Frequency**: The frequency \( f \) of the wave can be calculated using the wave speed \( V \) and wavelength \( \lambda \):
\[ f = \frac{V}{\lambda} = \frac{V}{\frac{4L}{2p + 1}} = \frac{V(2p + 1)}{4L} \]

7. **Identifying the Harmonic**: The harmonic corresponding to the \( p^{th} \) overtone is given by the formula:
\[ \text{Harmonic} = 2p + 1 \]

### Final Answer:
Thus, the harmonic corresponding to the \( p^{th} \) overtone in a closed organ pipe is \( 2p + 1 \).

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