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
(b) frequencyat which beats are produced.

Explanation

To find the frequency at which beats are produced when two sound waves are superimposed, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Angular Frequencies**:
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 identify the angular frequencies (\(\omega_1\) and \(\omega_2\)):
\[ \omega_1 = 596\pi \quad \text{and} \quad \omega_2 = 604\pi \]

2. **Convert Angular Frequencies to Frequencies**:
The relationship between angular frequency (\(\omega\)) and frequency (\(f\)) is given by:
\[ \omega = 2\pi f \]
Therefore, we can find the frequencies \(f_1\) and \(f_2\) as follows:
\[ f_1 = \frac{\omega_1}{2\pi} = \frac{596\pi}{2\pi} = 298 \text{ Hz} \]
\[ f_2 = \frac{\omega_2}{2\pi} = \frac{604\pi}{2\pi} = 302 \text{ Hz} \]

3. **Calculate the Beat Frequency**:
The beat frequency (\(f_b\)) is given by the absolute difference between the two frequencies:
\[ f_b = |f_2 - f_1| = |302 - 298| = 4 \text{ Hz} \]

### Final Answer:
The frequency at which beats are produced is \(4 \text{ Hz}\).

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