2) Assume two speakers that emit sound of wavelength 0.34 m are at positions (-1,0) and (1,0).
a) Use Cartesian distances �(𝑥𝑥 − 𝑥𝑥1)2 + (𝑦𝑦 − 𝑦𝑦1)2. Find the four points along the y=12 m line that are
closest to the x=0 axis that are maxima in intensity. 


We are given two sound sources emitting sound with a wavelength of \( \lambda = 0.34 \, \text{m} \), located at \( (-1, 0) \) and \( (1, 0) \) in a Cartesian coordinate plane. We need to find the points along the line \( y = 12 \, \text{m} \) that result in maxima in sound intensity. 

### Step-by-Step Solution

#### 1. **Distance from each source to a point on the line \( y = 12 \, \text{m} \)**

For a point \( P = (x, 12) \) on the line \( y = 12 \, \text{m} \), the distance from the two sound sources at \( (-1, 0) \) and \( (1, 0) \) are given by:

\[
d_1 = \sqrt{(x + 1)^2 + (12 - 0)^2} = \sqrt{(x + 1)^2 + 144}
\]
\[
d_2 = \sqrt{(x - 1)^2 + (12 - 0)^2} = \sqrt{(x - 1)^2 + 144}
\]

#### 2. **Condition for Maxima in Intensity (Constructive Interference)**

For constructive interference (which gives maxima in sound intensity), the path difference between the two distances \( d_1 \) and \( d_2 \) must be an integer multiple of the wavelength \( \lambda \):

\[
d_1 - d_2 = m \lambda
\]
where \( m \) is an integer (0, ±1, ±2, …) representing different maxima.

#### 3. **Simplifying the Equation**

We can approximate the distances because \( y = 12 \, \text{m} \) is quite large compared to the horizontal distances between the sources. Using a first-order approximation for the difference in distances:

\[
d_1 - d_2 \approx \frac{2x}{12} = \frac{x}{6}
\]

Substitute this into the condition for constructive interference:

\[
\frac{x}{6} = m \lambda
\]

Thus, the values of \( x \) where maxima occur are given by:

\[
x = 6m\lambda = 6m \times 0.34 = 2.04m
\]

#### 4. **Finding the Values of \( x \)**

We are interested in the points along the line \( y = 12 \, \text{m} \) that are closest to the \( x = 0 \) axis. The values of \( x \) closest to 0 occur for small values of \( m \). Let’s calculate \( x \) for several values of \( m \):

- For \( m = 0 \):
  \[
  x = 6(0) \times 0.34 = 0 \, \text{m}
  \]
  
- For \( m = 1 \):
  \[
  x = 6(1) \times 0.34 = 2.04 \, \text{m}
  \]
  
- For \( m = -1 \):
  \[
  x = 6(-1) \times 0.34 = -2.04 \, \text{m}
  \]
  
- For \( m = 2 \):
  \[
  x = 6(2) \times 0.34 = 4.08 \, \text{m}
  \]
  
- For \( m = -2 \):
  \[
  x = 6(-2) \times 0.34 = -4.08 \, \text{m}
  \]

#### 5. **Conclusion**

The four points on the line \( y = 12 \, \text{m} \) that are closest to the \( x = 0 \) axis and correspond to maxima in intensity are:

- \( (0, 12) \)
- \( (2.04, 12) \)
- \( (-2.04, 12) \)
- \( (4.08, 12) \)
- \( (-4.08, 12) \)

Would you like more details or have any further questions?

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Here are 5 related questions to further your understanding:
1. How would the positions of maxima change if the wavelength was reduced to 0.17 m?
2. What happens to the interference pattern if the two speakers are moved further apart?
3. What is the condition for destructive interference in this problem?
4. How do you calculate the exact sound pressure level at each maximum point?
5. How would the interference pattern change if the sources were at different heights?

**Tip:** In interference problems, remember that constructive interference corresponds to path differences being multiples of the wavelength, while destructive interference happens at half-wavelength differences.

Assume two speakers that emit sound of wavelength 0.34 m are at positions (-1,0) and (1,0). a) Use Cartesian distances √((x−x1)^2 + (y−y1)^2). Find the four points along the y=12 m line that are closest to the x=0 axis that are maxima in intensity.

Learn how to find the four maxima points along the y = 12 m line for sound waves emitted by two speakers positioned at (-1,0) and (1,0) with a wavelength of 0.34 m. This problem covers wave interference, constructive interference, and the distance formula. Suitable for students in Grades 10-12.

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