Question:
The beats are produced by two sound sources of same amplitude and of nearly equal frequencies. The maximum intensity of beats will be __________ when compared to that of one source is
The beats are produced by two sound sources of same amplitude and of nearly equal frequencies. The maximum intensity of beats will be __________ when compared to that of one source is
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A common mistake is to think that if amplitude doubles, intensity also doubles. Remember that intensity is proportional to the square of the amplitude. So, if amplitude becomes \(n\) times, intensity becomes \(n^2\) times. Here, amplitude doubles (from \(A_0\) to \(2A_0\)), so intensity becomes \(2^2 = 4\) times.
Updated On: Jul 1, 2026
- Same
- Double
- Four times
- Eight times
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
We are comparing the maximum intensity during the phenomenon of beats with the intensity of a single sound source. The sources have equal amplitudes.
Step 2: Key Formula or Approach:
The intensity (\(I\)) of a wave is proportional to the square of its amplitude (\(A\)).
\[ I \propto A^2 \]
When two waves interfere, the resultant amplitude depends on the phase difference. For beats, the waves cyclically go in and out of phase.
- At constructive interference (maximum loudness), the amplitudes add up.
- At destructive interference (minimum loudness), the amplitudes subtract.
Step 3: Detailed Explanation:
Let the amplitude of each individual sound source be \(A_0\).
The intensity of a single source, \(I_0\), is proportional to \(A_0^2\).
\[ I_0 = k A_0^2 \] (where k is a proportionality constant)
During the formation of beats, the maximum intensity occurs at points of constructive interference. At these points, the amplitudes of the two waves add.
The maximum resultant amplitude, \(A_{max}\), is:
\[ A_{max} = A_0 + A_0 = 2A_0 \]
The maximum intensity, \(I_{max}\), is proportional to the square of this maximum amplitude:
\[ I_{max} = k (A_{max})^2 = k (2A_0)^2 = k (4A_0^2) \]
Now, let's compare the maximum intensity \(I_{max}\) with the intensity of one source \(I_0\):
\[ I_{max} = 4 (k A_0^2) = 4 I_0 \]
This means the maximum intensity is four times the intensity of a single source.
Step 4: Final Answer:
The maximum intensity of beats will be four times the intensity of one source.
We are comparing the maximum intensity during the phenomenon of beats with the intensity of a single sound source. The sources have equal amplitudes.
Step 2: Key Formula or Approach:
The intensity (\(I\)) of a wave is proportional to the square of its amplitude (\(A\)).
\[ I \propto A^2 \]
When two waves interfere, the resultant amplitude depends on the phase difference. For beats, the waves cyclically go in and out of phase.
- At constructive interference (maximum loudness), the amplitudes add up.
- At destructive interference (minimum loudness), the amplitudes subtract.
Step 3: Detailed Explanation:
Let the amplitude of each individual sound source be \(A_0\).
The intensity of a single source, \(I_0\), is proportional to \(A_0^2\).
\[ I_0 = k A_0^2 \] (where k is a proportionality constant)
During the formation of beats, the maximum intensity occurs at points of constructive interference. At these points, the amplitudes of the two waves add.
The maximum resultant amplitude, \(A_{max}\), is:
\[ A_{max} = A_0 + A_0 = 2A_0 \]
The maximum intensity, \(I_{max}\), is proportional to the square of this maximum amplitude:
\[ I_{max} = k (A_{max})^2 = k (2A_0)^2 = k (4A_0^2) \]
Now, let's compare the maximum intensity \(I_{max}\) with the intensity of one source \(I_0\):
\[ I_{max} = 4 (k A_0^2) = 4 I_0 \]
This means the maximum intensity is four times the intensity of a single source.
Step 4: Final Answer:
The maximum intensity of beats will be four times the intensity of one source.
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