An organ pipe of `(3.9 pi) m`long, open at both ends is driver to third harmonic standing wave. If the amplitude of pressure oscillation is `1%` of mean atmospheric pressure `[p_(o) = 10^(5) N//m^(2)]`. The maximum displacement of particle from mean position will be [Given, velocity of sound `= 200 m//s` and density of air `= 1.3 kg//m^(3)]`

Options

A. `2.5 cm`
B. `5 cm`
C. `1 cm`
D. `2 cm`

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant physics concepts.

### Step 1: Understand the given data
- Length of the organ pipe, \( L = 3.9 \pi \, \text{m} \)
- The organ pipe is open at both ends and is in the third harmonic.
- Amplitude of pressure oscillation is \( 1\% \) of mean atmospheric pressure, \( p_0 = 10^5 \, \text{N/m}^2 \).
- Velocity of sound, \( v = 200 \, \text{m/s} \).
- Density of air, \( \rho = 1.3 \, \text{kg/m}^3 \).

### Step 2: Calculate the pressure amplitude
The pressure amplitude \( \Delta P_{\text{max}} \) can be calculated as:
\[ \Delta P_{\text{max}} = p_0 \times \frac{1}{100} = 10^5 \times 0.01 = 10^3 \, \text{N/m}^2 \]

### Step 3: Relate pressure amplitude to displacement amplitude
The relationship between pressure amplitude and displacement amplitude is given by:
\[ \Delta P_{\text{max}} = \rho v \omega A \]
where:
- \( \omega = 2 \pi f \) (angular frequency),
- \( A \) is the displacement amplitude.

### Step 4: Determine the wavelength for the third harmonic
For an organ pipe open at both ends, the length of the pipe is related to the wavelength \( \lambda \) by:
\[ L = \frac{3}{2} \lambda \quad \Rightarrow \quad \lambda = \frac{2L}{3} \]
Substituting \( L = 3.9 \pi \):
\[ \lambda = \frac{2 \times (3.9 \pi)}{3} = 2.6 \pi \, \text{m} \]

### Step 5: Calculate the frequency
The speed of sound is related to frequency and wavelength by:
\[ v = f \lambda \quad \Rightarrow \quad f = \frac{v}{\lambda} \]
Substituting the values:
\[ f = \frac{200}{2.6 \pi} \approx \frac{200}{8.168} \approx 24.5 \, \text{Hz} \]

### Step 6: Calculate angular frequency
The angular frequency \( \omega \) is given by:
\[ \omega = 2 \pi f = 2 \pi \times 24.5 \approx 154.3 \, \text{rad/s} \]

### Step 7: Substitute values to find displacement amplitude
Now we can rearrange the equation for pressure amplitude to find the displacement amplitude \( A \):
\[ A = \frac{\Delta P_{\text{max}}}{\rho v \omega} \]
Substituting the known values:
\[ A = \frac{10^3}{1.3 \times 200 \times 154.3} \]
Calculating the denominator:
\[ 1.3 \times 200 \times 154.3 \approx 40000 \]
Thus,
\[ A = \frac{1000}{40000} = 0.025 \, \text{m} \]

### Step 8: Convert to centimeters
To convert meters to centimeters:
\[ A = 0.025 \, \text{m} = 2.5 \, \text{cm} \]

### Final Answer
The maximum displacement of the particle from the mean position is \( 2.5 \, \text{cm} \).

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