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 Concept:
The intensity of a wave is proportional to the square of its amplitude (\(I \propto a^2\)). When two waves interfere, the resultant amplitude varies between \((a_1 + a_2)\) and \(|a_1 - a_2|\).

Step 2: Key Formula or Approach:
\[ \frac{I_{max}}{I_{min}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} \]
Step 3: Detailed Explanation:
Given amplitudes: \(a_1 = 3\) and \(a_2 = 5\).
1. Maximum Amplitude (\(A_{max}\)) = \(a_1 + a_2 = 3 + 5 = 8\).
2. Minimum Amplitude (\(A_{min}\)) = \(|a_1 - a_2| = |3 - 5| = 2\).
3. Maximum Intensity (\(I_{max}\)) \(\propto (A_{max})^2 = 8^2 = 64\).
4. Minimum Intensity (\(I_{min}\)) \(\propto (A_{min})^2 = 2^2 = 4\).
Ratio:
\[ \frac{I_{max}}{I_{min}} = \frac{64}{4} = \frac{16}{1} \]
Step 4: Final Answer:
The ratio of maximum to minimum intensity is \(16 : 1\).