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?
Show Hint
The number of beats per second is simply the absolute difference between the frequencies of the two interfering waves.
Step 1: Understanding the Question:
The number of times a listener hears maximum intensity in one second is equal to the beat frequency, which is the absolute difference between the frequencies of the two sound waves. Step 2: Key Formula or Approach:
A standard wave equation is $y = A \sin(kx - \omega t)$, where $\omega = 2\pi f$.
Beat frequency $b = |f_1 - f_2|$. Step 3: Detailed Explanation:
From $y_1 = a \sin(0.2\pi x - 50\pi t)$:
$\omega_1 = 50\pi \implies 2\pi f_1 = 50\pi \implies f_1 = 25$ Hz.
From $y_2 = a \sin(0.15\pi x - 46\pi t)$:
$\omega_2 = 46\pi \implies 2\pi f_2 = 46\pi \implies f_2 = 23$ Hz.
Beat frequency:
\[ b = |25 - 23| = 2 \text{ beats/second} \]
This means the maximum intensity is heard 2 times in one second. Step 4: Final Answer:
The listener hears maximum intensity 2 times in one second.