Question

Two sound waves travelling in the same direction have displacement $y_1 = a \sin(0.2\pi x - 50\pi t)$ and $y_2 = a \sin(0.15\pi x - 46\pi t)$. How many times, a listener can hear sound of maximum intensity in one second?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

The number of beats per second is simply the absolute difference between the frequencies of the two interfering waves.

Answer 1

The Correct Option is B

Step 1: Identify Frequencies
Standard wave equation: $y = a \sin(kx - \omega t)$.
For $y_1$: $\omega_1 = 50\pi \Rightarrow 2\pi n_1 = 50\pi \Rightarrow n_1 = 25 \text{ Hz}$.
For $y_2$: $\omega_2 = 46\pi \Rightarrow 2\pi n_2 = 46\pi \Rightarrow n_2 = 23 \text{ Hz}$.
Step 2: Beat Frequency
The number of times maximum intensity (waxing) is heard per second is the beat frequency: $n_b = |n_1 - n_2|$.
Step 3: Calculation
$n_b = |25 - 23| = 2 \text{ beats per second}$.
Step 4: Conclusion
The listener hears maximum intensity 2 times in one second.
Final Answer:(B)