In a stationary wave pattern that forms as a result of reflection pf waves from an obstacle the ratio of the amplitude at an antinode and a node is `beta = 1.5.` What percentage of the energy passes across the obstacle ?

Explanation

To solve the problem, we need to determine the percentage of energy that passes across the obstacle when given the ratio of the amplitude at an antinode to that at a node in a stationary wave pattern.

### Step-by-step Solution:

1. **Understanding the Given Ratio**:
The ratio of the amplitude at an antinode (A_max) to that at a node (A_min) is given as β = 1.5. This can be expressed as:
\[ \frac{A_{max}}{A_{min}} = \frac{3}{2} \]

2. **Relating Amplitudes to Energy**:
In wave mechanics, the energy (E) is proportional to the square of the amplitude (A). Therefore, we can write:
\[ E \propto A^2 \]
This means that the energy at the antinode (E_max) and the energy at the node (E_min) can be expressed as:
\[ E_{max} \propto A_{max}^2 \quad \text{and} \quad E_{min} \propto A_{min}^2 \]

3. **Expressing Amplitudes in Terms of Incident and Reflected Waves**:
The amplitude at the antinode is the result of constructive interference of the incident wave and the reflected wave:
\[ A_{max} = A_{incident} + A_{reflected} \]
The amplitude at the node is the result of destructive interference:
\[ A_{min} = A_{incident} - A_{reflected} \]

4. **Setting Up the Equation**:
From the ratio given, we can set up the equation:
\[ \frac{A_{max}}{A_{min}} = \frac{A_{incident} + A_{reflected}}{A_{incident} - A_{reflected}} = \frac{3}{2} \]

5. **Cross Multiplying**:
Cross multiplying gives:
\[ 2(A_{incident} + A_{reflected}) = 3(A_{incident} - A_{reflected}) \]
Expanding this results in:
\[ 2A_{incident} + 2A_{reflected} = 3A_{incident} - 3A_{reflected} \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ 2A_{reflected} + 3A_{reflected} = 3A_{incident} - 2A_{incident} \]
Simplifying leads to:
\[ 5A_{reflected} = A_{incident} \]
Thus:
\[ \frac{A_{reflected}}{A_{incident}} = \frac{1}{5} \]

7. **Calculating the Energy Ratio**:
Since energy is proportional to the square of the amplitude:
\[ \frac{E_{reflected}}{E_{incident}} = \left(\frac{A_{reflected}}{A_{incident}}\right)^2 = \left(\frac{1}{5}\right)^2 = \frac{1}{25} \]

8. **Determining the Percentage of Energy Reflected**:
The percentage of energy reflected is:
\[ \text{Percentage reflected} = \frac{E_{reflected}}{E_{incident}} \times 100 = \frac{1}{25} \times 100 = 4\% \]

9. **Calculating the Percentage of Energy Passed**:
Therefore, the percentage of energy that passes across the obstacle is:
\[ \text{Percentage passed} = 100\% - \text{Percentage reflected} = 100\% - 4\% = 96\% \]

### Final Answer:
The percentage of energy that passes across the obstacle is **96%**.