Question

If two progressive sound waves represented by \( y_1 = 3 \sin 250 \pi t \) and \( y_2 = 2 \sin 260 \pi t \) (where displacement is in metre and time is in second) superimpose, then the time interval between two successive maximum intensities is

• A. \( 0.1 s} \)
• B. \( 0.4 s} \)
• C. \( 0.5 s} \)
• D. \( 0.2 s} \)

Hint:

When two waves superimpose to produce beats, the time interval between two successive maximum intensities (or minimum intensities) is equal to the beat period. The beat period is the reciprocal of the beat frequency, which is the absolute difference between the individual frequencies of the waves (\( f_b = |f_1 - f_2| \)). Ensure that frequencies are in Hertz (Hz) before calculating the beat frequency.

Answer 1

The Correct Option is D

Step 1: Identify the angular frequencies of the two waves.
The equations of the two progressive sound waves are given as:
\( y_1 = 3 \sin 250 \pi t \)
\( y_2 = 2 \sin 260 \pi t \)
Comparing these with the general form of a progressive wave \( y = A \sin(\omega t) \), we can identify the angular frequencies:
For \( y_1 \): \( \omega_1 = 250 \pi \, \text{rad/s} \)
For \( y_2 \): \( \omega_2 = 260 \pi \, \text{rad/s} \)
Step 2: Calculate the frequencies of the two waves.
The angular frequency \( \omega \) is related to the frequency \( f \) by \( \omega = 2\pi f \).
For \( y_1 \): \( f_1 = \frac{\omega_1}{2\pi} = \frac{250 \pi}{2\pi} = 125 \, \text{Hz} \)
For \( y_2 \): \( f_2 = \frac{\omega_2}{2\pi} = \frac{260 \pi}{2\pi} = 130 \, \text{Hz} \)
Step 3: Determine the beat frequency.
When two sound waves of slightly different frequencies superimpose, they produce beats. The beat frequency (\( f_b \)) is the difference between the frequencies of the two waves.
\( f_b = |f_1 - f_2| \)
\( f_b = |125 \, \text{Hz} - 130 \, \text{Hz}| = |-5 \, \text{Hz}| = 5 \, \text{Hz} \)
Step 4: Calculate the time interval between two successive maximum intensities (beat period).
The time interval between two successive maximum intensities (or minimum intensities) is the beat period (\( T_b \)). The beat period is the reciprocal of the beat frequency.
\( T_b = \frac{1}{f_b} \)
\( T_b = \frac{1}{5 \, \text{Hz}} = 0.2 \, \text{s} \)