Two sources of sound are emitting progressive | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given equations are $y_1 = 4 \sin(710 \pi t)$ and $y_2 = 3 \sin(702 \pi t)$.Comparing these with $y = A \sin(2 \pi f t)$,we get:$2 \pi f_1 = 7

Vedclass · Accepted Answer

$4, 49:1$

Vedclass · Answer

(C) The given equations are $y_1 = 4 \sin(710 \pi t)$ and $y_2 = 3 \sin(702 \pi t)$.Comparing these with $y = A \sin(2 \pi f t)$,we get:$2 \pi f_1 = 710 \pi \implies f_1 = 355 	ext{ Hz}$$2 \pi f_2 = 702 \pi \implies f_2 = 351 	ext{ Hz}$The number of beats per second is $n = |f_1 - f_2| = |355 - 351| = 4 	ext{ beats/s}$.The amplitudes are $A_1 = 4$ and $A_2 = 3$.The maximum intensity (waxing) is proportional to $(A_1 + A_2)^2 = (4 + 3)^2 = 7^2 = 49$.The minimum intensity (waning) is proportional to $(A_1 - A_2)^2 = (4 - 3)^2 = 1^2 = 1$.Therefore,the intensity ratio is $49:1$.The correct answer is $4, 49:1$.

Two sources of sound are emitting progressive waves $y_1 = 4 \sin(710 \pi t)$ and $y_2 = 3 \sin(702 \pi t)$. The sources are placed close to each other. The number of beats heard per second and the intensity ratio between waxing and waning are respectively:

Similar Questions

$56$ tuning forks are arranged such that each fork produces $4$ beats per second with its previous one. If the frequency of the last fork is twice that of the first, the frequency of the $19^{\text{th}}$ fork is . . . . . . (in $\text{Hz}$)

$A$ wire vibrates at a fundamental frequency of $500 \,Hz$. $A$ second identical wire produces $5$ beats per second with it when the tension in the first wire is slightly decreased. The ratio of the tension in the second wire to the tension in the first wire is approximately equal to

The frequency of a tuning fork is $256 \, Hz$. The velocity of sound in air is $344 \, ms^{-1}$. The distance travelled (in $meters$) by the sound during the time in which the tuning fork completes $32$ vibrations is

Beats are produced by waves $y_1 = a \sin(2000 \pi t)$ and $y_2 = a \sin(2008 \pi t)$. The number of beats heard per second is

Two sound waves of slightly different frequencies propagating in the same direction produce beats due to

Vedclass Test Series

Exam Paper Generator

Online Exam Module