A closed pipe of length 22 cm, when excited by a 1875 Hz source forms standing waves. The number of pressure nodes formed in the pipe are [velocity of sound in air = `330 ms^(-1)`]

Options

A. 1
B. 2
C. 4
D. 3

Answer

To solve the problem, we need to find the number of pressure nodes formed in a closed pipe of length 22 cm when excited by a 1875 Hz source.

### Step-by-Step Solution:

1. **Understand the Closed Pipe Configuration**:
- A closed pipe has one end closed and the other end open. In this configuration, the closed end is a displacement node (pressure antinode), and the open end is a displacement antinode (pressure node).

2. **Determine the Length of the Pipe**:
- Given the length of the closed pipe, \( L = 22 \, \text{cm} = 0.22 \, \text{m} \).

3. **Find the Speed of Sound**:
- The speed of sound in air is given as \( v = 330 \, \text{m/s} \).

4. **Use the Frequency Formula for Closed Pipes**:
- The frequency of the standing wave in a closed pipe can be expressed as:
\[ f = \frac{(2n + 1)v}{4L} \]
- Here, \( n \) is the mode number (0, 1, 2, ...).

5. **Substitute the Given Frequency and Solve for \( n \)**:
- We know \( f = 1875 \, \text{Hz} \). Plugging in the values:
\[ 1875 = \frac{(2n + 1) \cdot 330}{4 \cdot 0.22} \]
- Rearranging gives:
\[ 1875 \cdot 4 \cdot 0.22 = (2n + 1) \cdot 330 \]
- Calculating the left side:
\[ 1875 \cdot 0.88 = 1650 \]
- Thus:
\[ 1650 = (2n + 1) \cdot 330 \]
- Dividing both sides by 330:
\[ 5 = 2n + 1 \]
- Solving for \( n \):
\[ 2n = 5 - 1 = 4 \implies n = 2 \]

6. **Calculate the Number of Pressure Nodes**:
- The number of pressure nodes in a closed pipe is given by \( n + 1 \):
\[ \text{Number of pressure nodes} = n + 1 = 2 + 1 = 3 \]

### Final Answer:
The number of pressure nodes formed in the pipe is **3**.