A `3 m` long organ pipe open at both ends is driven to third harmonic standing wave. If the ampulitude of pressure oscillations is `1` per cent of mean atmospheric pressure `(p_(o) = 10^(5) Nm^(2))`. Find the ampulited of particle displacement and density oscillations. Speed of sound `upsilon = 332 m//s` and density of air `rho = 1.03 kg//m^(3)`.

Explanation

To solve the problem step by step, we will follow the outlined procedure to find the amplitude of particle displacement and density oscillations in a 3 m long organ pipe open at both ends, driven to the third harmonic standing wave.

### Step 1: Determine the Wavelength
The length of the organ pipe \( L \) is given as \( 3 \, \text{m} \). For an organ pipe open at both ends, the relationship between the length of the pipe and the wavelength \( \lambda \) for the \( n \)-th harmonic is given by:

\[ L = \frac{n \lambda}{2} \]

For the third harmonic (\( n = 3 \)):

\[ 3 = \frac{3 \lambda}{2} \]

Solving for \( \lambda \):

\[ \lambda = \frac{2 \cdot 3}{3} = 2 \, \text{m} \]

### Step 2: Calculate the Pressure Amplitude
The amplitude of pressure oscillations \( \Delta P \) is given as \( 1\% \) of the mean atmospheric pressure \( p_0 = 10^5 \, \text{N/m}^2 \):

\[ \Delta P = 0.01 \times p_0 = 0.01 \times 10^5 = 10^3 \, \text{N/m}^2 \]

### Step 3: Calculate the Wave Number \( k \)
The wave number \( k \) is given by:

\[ k = \frac{2\pi}{\lambda} \]

Substituting \( \lambda = 2 \, \text{m} \):

\[ k = \frac{2\pi}{2} = \pi \, \text{rad/m} \]

### Step 4: Calculate the Amplitude of Particle Displacement
Using the relationship between pressure amplitude \( \Delta P \), bulk modulus \( B \), displacement amplitude \( A \), and wave number \( k \):

\[ \Delta P = B \cdot A \cdot k \]

The bulk modulus \( B \) can be expressed as:

\[ B = \rho v^2 \]

where \( \rho = 1.03 \, \text{kg/m}^3 \) (density of air) and \( v = 332 \, \text{m/s} \) (speed of sound).

Calculating \( B \):

\[ B = 1.03 \cdot (332)^2 = 1.03 \cdot 110224 = 113,000 \, \text{N/m}^2 \]

Now substituting into the pressure amplitude equation:

\[ 10^3 = 113000 \cdot A \cdot \pi \]

Solving for \( A \):

\[ A = \frac{10^3}{113000 \cdot \pi} \approx \frac{1000}{355,000} \approx 0.0028 \, \text{m} \text{ or } 2.8 \, \text{cm} \]

### Step 5: Calculate the Density Oscillation Amplitude
The amplitude of density oscillations \( \Delta \rho \) can be calculated using the formula:

\[ \Delta \rho = \frac{\Delta P}{v^2} \]

Substituting the values:

\[ \Delta \rho = \frac{10^3}{(332)^2} = \frac{1000}{110224} \approx 0.0091 \, \text{kg/m}^3 \]

### Final Answers
- Amplitude of particle displacement \( A \approx 0.0028 \, \text{m} \) or \( 2.8 \, \text{cm} \)
- Amplitude of density oscillations \( \Delta \rho \approx 0.0091 \, \text{kg/m}^3 \)