If sound travels in air with the speed of 340 m/s, then number of tones present in an open organ pipe of length 2 m and its maximum frequency 1200 Hz, are

Options

A. 17
B. 11
C. 9
D. 14

Explanation

To solve the problem step by step, we need to find the number of tones present in an open organ pipe of length 2 m, given that the speed of sound in air is 340 m/s and the maximum frequency is 1200 Hz.

### Step 1: Calculate the Fundamental Frequency
The fundamental frequency (\(F_0\)) of an open organ pipe can be calculated using the formula:

\[ F_0 = \frac{v}{2L} \]

where:
- \(v\) is the speed of sound (340 m/s),
- \(L\) is the length of the pipe (2 m).

Substituting the values:

\[ F_0 = \frac{340 \, \text{m/s}}{2 \times 2 \, \text{m}} = \frac{340}{4} = 85 \, \text{Hz} \]

### Step 2: Calculate the Number of Tones
The number of tones (or harmonics) present in the organ pipe can be calculated using the formula:

\[ \text{Number of Tones} = \frac{F}{F_0} \]

where:
- \(F\) is the maximum frequency (1200 Hz),
- \(F_0\) is the fundamental frequency (85 Hz).

Substituting the values:

\[ \text{Number of Tones} = \frac{1200 \, \text{Hz}}{85 \, \text{Hz}} \approx 14.12 \]

Since the number of tones must be a whole number, we round it off to 14.

### Final Answer
Thus, the number of tones present in the open organ pipe is **14**.

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