Question

When two sound waves having amplitudes 3 and 5 units are superimposed, find the ratio of maximum to minimum intensity of the resultant wave.

Hint:

This is a standard CET problem. Remember the shortcut: Ratio \(= \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2\). For \(3\) and \(5\), the sum is \(8\) and difference is \(2\). \(8/2 = 4\), and \(4^2 = 16\). Result is \(16:1\).

Answer 1

Step 1: Understanding the Question:
Intensity is proportional to the square of amplitude. Resultant maximum amplitude is the sum, and minimum is the difference of individual amplitudes.
Step 2: Key Formula or Approach:
\( \frac{I_{max}}{I_{min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2 \).
Step 3: Detailed Explanation:
Given \( a_1 = 5 \), \( a_2 = 3 \).
\[ \frac{I_{max}}{I_{min}} = \left( \frac{5 + 3}{5 - 3} \right)^2 = \left( \frac{8}{2} \right)^2 = 4^2 = 16 \]
The ratio is \( 16:1 \).
Step 4: Final Answer:
The intensity ratio is \( 16:1 \).