Question 419: Waves

Question

Two waves $$\mathrm{Y}_1=0.25 \sin 316 \mathrm{t}$$ and $$\mathrm{Y}_2=0.25 \sin 310 \mathrm{t}$$ are propagation same direction. The number of beats produced per second are

Accepted Answer

To determine the number of beats produced per second when two waves propagate in the same direction, we must first understand the concept of beat frequency. The beat frequency is given by the absolute difference between the frequencies of the two waves.
Given the two waves:
$$ \mathrm{Y}_1 = 0.25 \sin(316 \mathrm{t}) $$
$$ \mathrm{Y}_2 = 0.25 \sin(310 \mathrm{t}) $$
The general form of a sine wave is:
$$ \mathrm{Y} = A \sin(2 \pi f t) $$
Here, $$\mathrm{Y}_1$$ and $$\mathrm{Y}_2$$ follow the form $$\mathrm{Y} = A \sin(\omega t)$$, where $$\omega$$ is the angular frequency.
For wave $$\mathrm{Y}_1$$, the angular frequency $$\omega_1$$ is 316, thus the frequency $$f_1$$ is given by:
$$ \omega_1 = 2 \pi f_1 \implies f_1 = \frac{316}{2 \pi} $$
For wave $$\mathrm{Y}_2$$, the angular frequency $$\omega_2$$ is 310, thus the frequency $$f_2$$ is given by:
$$ \omega_2 = 2 \pi f_2 \implies f_2 = \frac{310}{2 \pi} $$
To find the beat frequency $$f_{	ext{beat}}$$, we take the absolute difference between the frequencies $$f_1$$ and $$f_2$$:
$$ f_{	ext{beat}} = |f_1 - f_2| $$
Substituting the respective frequencies:
$$ f_{	ext{beat}} = \left| \frac{316}{2 \pi} - \frac{310}{2 \pi} ight| $$
Simplifying this:
$$ f_{	ext{beat}} = \left| \frac{316 - 310}{2 \pi} ight| = \frac{|6|}{2 \pi} = \frac{6}{2 \pi} = \frac{3}{\pi} $$
Therefore, the number of beats produced per second is:
$$ \boxed{\frac{3}{\pi}} $$
The correct answer is Option A: $$\frac{3}{\pi}$$.

Answer

$$\frac{3}{\pi}$$

Answer

$$\frac{\pi}{3}$$

Answer

$$\frac{\pi}{2}$$

Answer

$$\frac{2}{\pi}$$

Waves

Two waves $\mathrm{Y}_1=0.25 \sin 316 \mathrm{t} and \mathrm{Y}_2=0.25 \sin 310 \mathrm{t}$ are propagation same direction. The number of beats produced per second are

Choose the correct answer: