In a closed organ pipe the frequency of fundamental note is 50 Hz . The note of which of the following frequencies will not be emitted by it

Options

A. `50 Hz`
B. `100 Hz`
C. `150 Hz`
D. None of the above

Answer

To solve the problem, we need to determine the frequencies that can be emitted by a closed organ pipe and identify which frequency will not be emitted.

### Step-by-Step Solution:

1. **Understanding the Closed Organ Pipe:**
A closed organ pipe has one end closed (node) and the other end open (antinode). The fundamental frequency (first harmonic) corresponds to the situation where there is one quarter of a wavelength (λ/4) in the length of the pipe (L).

2. **Given Information:**
The fundamental frequency (f1) is given as 50 Hz.

3. **Finding Wavelength for Fundamental Frequency:**
The relationship between frequency (f), speed of sound (v), and wavelength (λ) is given by:
\[ f = \frac{v}{\lambda} \]
For the fundamental frequency in a closed pipe:
\[ \lambda = 4L \]
Therefore, we can express the fundamental frequency as:
\[ f_1 = \frac{v}{\lambda} = \frac{v}{4L} \]
Given that \( f_1 = 50 \, \text{Hz} \), we can rearrange this to find the speed of sound:
\[ v = 50 \times 4L = 200L \]

4. **Finding Overtone Frequencies:**
The frequencies of the harmonics in a closed organ pipe can be expressed as:
\[ f_n = \frac{(2n - 1)v}{4L} \]
where \( n \) is the harmonic number (1 for fundamental, 2 for first overtone, etc.).

5. **Calculating Frequencies:**
- For \( n = 1 \) (fundamental frequency):
\[ f_1 = 50 \, \text{Hz} \]
- For \( n = 2 \) (first overtone):
\[ f_2 = \frac{(2 \times 2 - 1) \cdot v}{4L} = \frac{3v}{4L} = 3 \times 50 \, \text{Hz} = 150 \, \text{Hz} \]
- For \( n = 3 \) (second overtone):
\[ f_3 = \frac{(2 \times 3 - 1) \cdot v}{4L} = \frac{5v}{4L} = 5 \times 50 \, \text{Hz} = 250 \, \text{Hz} \]
- For \( n = 4 \) (third overtone):
\[ f_4 = \frac{(2 \times 4 - 1) \cdot v}{4L} = \frac{7v}{4L} = 7 \times 50 \, \text{Hz} = 350 \, \text{Hz} \]

6. **Identifying Frequencies That Will Not Be Emitted:**
The frequencies emitted by a closed organ pipe are given by the formula \( f_n = (2n - 1) \cdot f_1 \). Therefore, the emitted frequencies are:
- 50 Hz (fundamental)
- 150 Hz (first overtone)
- 250 Hz (second overtone)
- 350 Hz (third overtone)

Frequencies that will not be emitted are those that do not fit the form \( (2n - 1) \cdot 50 \). For example, 100 Hz or 200 Hz will not be emitted as they do not correspond to any \( n \) value.

### Conclusion:
The note of 100 Hz will not be emitted by the closed organ pipe.