For two sound waves `lamda_(1) = 100 cm, lamda_(2) = 110 cm` and velocity of sound is 330 m/s. When `lamda_(1) and lamda_(2)` super impose, the number of beats produced per second is

Options

A. 10
B. 20
C. 30
D. 45

Explanation

To find the number of beats produced per second when two sound waves with wavelengths \( \lambda_1 = 100 \, \text{cm} \) and \( \lambda_2 = 110 \, \text{cm} \) superimpose, we can follow these steps:

### Step 1: Convert Wavelengths to Meters
The given wavelengths are in centimeters, so we need to convert them to meters.

\[ \lambda_1 = 100 \, \text{cm} = 100 \times 10^{-2} \, \text{m} = 1.00 \, \text{m} \]

\[ \lambda_2 = 110 \, \text{cm} = 110 \times 10^{-2} \, \text{m} = 1.10 \, \text{m} \]

### Step 2: Calculate Frequencies
The frequency \( f \) of a wave can be calculated using the formula:

\[ f = \frac{v}{\lambda} \]

where \( v \) is the velocity of sound. Given \( v = 330 \, \text{m/s} \):

For \( \lambda_1 \):

\[ f_1 = \frac{330 \, \text{m/s}}{1.00 \, \text{m}} = 330 \, \text{Hz} \]

For \( \lambda_2 \):

\[ f_2 = \frac{330 \, \text{m/s}}{1.10 \, \text{m}} = 300 \, \text{Hz} \]

### Step 3: Calculate the Number of Beats
The number of beats produced per second is given by the difference in frequencies:

\[ \text{Number of beats} = |f_1 - f_2| \]

Substituting the values we found:

\[ \text{Number of beats} = |330 \, \text{Hz} - 300 \, \text{Hz}| = 30 \, \text{beats per second} \]

### Final Answer
The number of beats produced per second is \( 30 \, \text{beats per second} \).

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