Question 272: Waves

Question

Two sources of sound are emitting progressive waves $\mathrm{y}_1=4 \sin 710 \pi \mathrm{t}$ and $\mathrm{y}_2=3 \sin 702 \pi \mathrm{t}$. The sources are placed close to each other. The number of beats heard per second and intensity ratio between waxing and waning are respectively

Accepted Answer

Given:
$\displaystyle y_1 = 4\sin(710\pi t)$
$\displaystyle y_2 = 3\sin(702\pi t)$
Let's find:
Number of beats per second (beat frequency)
Ratio of intensities during waxing and waning
1. Number of beats per second
The general form of the wave equation is $y = A\sin(2\pi f t)$.
Compare with given:
For $y_1$: $2\pi f_1 = 710\pi \implies f_1 = 355\ 	ext{Hz}$
For $y_2$: $2\pi f_2 = 702\pi \implies f_2 = 351\ 	ext{Hz}$
Number of beats per second:
The beat frequency is the absolute difference of the two frequencies:
$ 	ext{Beat frequency} = |f_1 - f_2| = |355 - 351| = 4\ 	ext{Hz} $
2. Intensity ratio between waxing and waning
For two waves:
$A_1 = 4$, $A_2 = 3$
Waxing
(maximum intensity): When waves are in phase, the resultant amplitude $= A_1 + A_2 = 4 + 3 = 7$.
Waning
(minimum intensity): When waves are out of phase, the resultant amplitude $= |A_1 - A_2| = |4 - 3| = 1$.
Intensity is proportional to the square of amplitude:
$ I \propto A^2 $
So,
$ \frac{I_{	ext{waxing}}}{I_{	ext{waning}}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = \frac{(7)^2}{(1)^2} = \frac{49}{1} $
Final answer
Number of beats per second: $4$
Intensity ratio during waxing and waning: $49:1$
Correct option:
Option C
$4,\ 49:1$

Answer

$4,16: 9$

Answer

$8,16: 9$

Answer

$4,49: 1$

Answer

$8,49: 1$

Waves

Two sources of sound are emitting progressive waves \mathrm{y}_1=4 \sin 710 \pi \mathrm{t} and \mathrm{y}_2=3 \sin 702 \pi \mathrm{t}. The sources are placed close to each other. The number of beats heard per second and intensity ratio between waxing and waning are respectively

Choose the correct answer: