If speed of sound in air in 330 m/s then find the number of tones present in an open organ pipe of length 1 m whose frequency if `le1000`.

Options

A. 2
B. 4
C. 8
D. 6

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the Formula for Fundamental Frequency
The fundamental frequency \( f \) of an open organ pipe is given by the formula:
\[ f = \frac{v}{2L} \]
where:
- \( v \) is the speed of sound in air (330 m/s),
- \( L \) is the length of the organ pipe (1 m).

### Step 2: Substitute Values into the Formula
Substituting the given values into the formula:
\[ f = \frac{330 \, \text{m/s}}{2 \times 1 \, \text{m}} = \frac{330}{2} = 165 \, \text{Hz} \]

### Step 3: Determine the Maximum Frequency
We need to find the number of tones whose frequency is less than or equal to 1000 Hz.

### Step 4: Calculate the Number of Harmonics
The harmonics of an open organ pipe are integer multiples of the fundamental frequency. Thus, the frequencies of the harmonics are:
\[ f_n = n \cdot f \]
where \( n \) is the harmonic number (1, 2, 3, ...).

To find the maximum harmonic number \( n \) such that \( f_n \leq 1000 \, \text{Hz} \):
\[ n \cdot 165 \leq 1000 \]
\[ n \leq \frac{1000}{165} \approx 6.06 \]

### Step 5: Determine the Number of Allowed Harmonics
Since \( n \) must be an integer, the maximum integer value for \( n \) is 6. Therefore, the number of tones present in the organ pipe is 6.

### Final Answer
The number of tones present in the open organ pipe of length 1 m, with frequency less than or equal to 1000 Hz, is **6**.

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