Question

You are calculating the number of beats produced per second by two sound waves with different wavelengths. What's given in the problem The wavelength of the first sound wave is $\lambda _{1}=49\text{\ cm}$. The wavelength of the second sound wave is $\lambda _{2}=50\text{\ cm}$. The velocity of sound in air at $30^{\circ }\text{C}$ is $v=332\text{\ m/s}$.

Answer 1

The number of beats produced per second is approximately $13.55 \text{ Hz}$.

Answer 2

Explanation of the solution:

1. Convert Wavelengths to Meters: The given wavelengths are in centimeters, and the velocity is in meters per second. To ensure unit consistency, convert the wavelengths to meters: $\lambda_1 = 49 \text{ cm} = 0.49 \text{ m}$ $\lambda_2 = 50 \text{ cm} = 0.50 \text{ m}$

2. Calculate Frequencies: The relationship between velocity ($v$), frequency ($f$), and wavelength ($\lambda$) is $v = f\lambda$, which implies $f = v/\lambda$. Frequency of the first wave: $f_1 = \frac{v}{\lambda_1}$ Frequency of the second wave: $f_2 = \frac{v}{\lambda_2}$

3. Calculate Beat Frequency: The number of beats produced per second (beat frequency) is the absolute difference between the frequencies of the two waves: $f_{\text{beat}} = |f_1 - f_2|$ Substitute the expressions for $f_1$ and $f_2$: $f_{\text{beat}} = \left|\frac{v}{\lambda_1} - \frac{v}{\lambda_2}\right| = v \left|\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right|$

4. Substitute Values and Calculate: Given $v = 332 \text{ m/s}$, $\lambda_1 = 0.49 \text{ m}$, $\lambda_2 = 0.50 \text{ m}$. $f_{\text{beat}} = 332 \left|\frac{1}{0.49} - \frac{1}{0.50}\right|$ $f_{\text{beat}} = 332 \left|\frac{0.50 - 0.49}{0.49 \times 0.50}\right|$ $f_{\text{beat}} = 332 \left|\frac{0.01}{0.245}\right|$ $f_{\text{beat}} = \frac{3.32}{0.245}$ $f_{\text{beat}} \approx 13.551 \text{ Hz}$