Two sources of sound placed close to each oth | vedclass

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

Question

(D) Given equations are $y_1 = 4 \sin(600\pi t)$ and $y_2 = 5 \sin(608\pi t)$.Comparing with $y = A \sin(2\pi 
u t)$,we get:For the first source: $A_

Vedclass · Accepted Answer

$4$ beats per second with intensity ratio $81 : 1$ between waxing and waning.

Vedclass · Answer

(D) Given equations are $y_1 = 4 \sin(600\pi t)$ and $y_2 = 5 \sin(608\pi t)$.Comparing with $y = A \sin(2\pi 
u t)$,we get:For the first source: $A_1 = 4$ and $2\pi 
u_1 = 600\pi \implies 
u_1 = 300 	ext{ Hz}$.For the second source: $A_2 = 5$ and $2\pi 
u_2 = 608\pi \implies 
u_2 = 304 	ext{ Hz}$.The number of beats heard per second is the difference in frequencies: $	ext{Beat frequency} = 
u_2 - 
u_1 = 304 - 300 = 4 	ext{ beats/sec}$.The ratio of maximum intensity to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2}$.Substituting the values: $\frac{I_{\max}}{I_{\min}} = \frac{(4 + 5)^2}{(4 - 5)^2} = \frac{9^2}{(-1)^2} = \frac{81}{1}$.Thus,the observer hears $4$ beats per second with an intensity ratio of $81 : 1$.

Two sources of sound placed close to each other are emitting progressive waves given by $y_1 = 4 \sin(600\pi t)$ and $y_2 = 5 \sin(608\pi t)$. An observer located near these two sources of sound will hear:

Similar Questions

Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce $4 \text{ beats/sec}$. The highest frequency is twice the lowest. The possible highest and lowest frequencies are:

For the formation of beats,two sound notes must have:

When a tuning fork produces sound waves in air,which one of the following remains the same in the material of the tuning fork as well as in the air?

Two identical piano wires,kept under the same tension $T$,have a fundamental frequency of $600\, Hz$. The fractional increase in the tension of one of the wires which will lead to the occurrence of $6\, beats/s$ when both the wires oscillate together would be:

The wavelengths of two sound notes in air are $\frac{40}{195} \,m$ and $\frac{40}{193} \,m$. Each note produces $9$ beats per second separately with a third note of fixed frequency. The velocity of sound in air in $m/s$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module