If two sound waves, `y_(1) = 0.3 sin 596 pi[t - x//300]` and` y_(2) = 0.5 sin 604 pi [t - x//330]` are superimposed, what will be the
(c) Tlie ratio ofmaximum and minimum intensities of beats

Explanation

To find the ratio of maximum and minimum intensities of beats produced by the superposition of two sound waves, we can follow these steps:

### Step 1: Identify the amplitudes of the waves
The given sound waves are:
- \( y_1 = 0.3 \sin(596 \pi [t - \frac{x}{300}]) \)
- \( y_2 = 0.5 \sin(604 \pi [t - \frac{x}{330}]) \)

From these equations, we can extract the amplitudes:
- \( A_1 = 0.3 \)
- \( A_2 = 0.5 \)

### Step 2: Calculate the maximum amplitude
The maximum amplitude \( A_{max} \) when the waves constructively interfere (i.e., when the phase difference \( \phi = 0 \)) is given by:
\[ A_{max} = A_1 + A_2 = 0.3 + 0.5 = 0.8 \]

### Step 3: Calculate the minimum amplitude
The minimum amplitude \( A_{min} \) when the waves destructively interfere (i.e., when the phase difference \( \phi = \pi \)) is given by:
\[ A_{min} = |A_1 - A_2| = |0.3 - 0.5| = 0.2 \]

### Step 4: Relate intensity to amplitude
The intensity \( I \) of a wave is proportional to the square of its amplitude:
\[ I \propto A^2 \]
Thus, we can express the maximum and minimum intensities as:
\[ I_{max} \propto (A_{max})^2 = (0.8)^2 = 0.64 \]
\[ I_{min} \propto (A_{min})^2 = (0.2)^2 = 0.04 \]

### Step 5: Calculate the ratio of maximum to minimum intensity
Now, we can find the ratio of maximum intensity to minimum intensity:
\[ \frac{I_{max}}{I_{min}} = \frac{0.64}{0.04} = 16 \]

### Conclusion
The ratio of maximum and minimum intensities of the beats is:
\[ \frac{I_{max}}{I_{min}} = 16 \]

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