Two sources of sound are emitting progressive | vedclass

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

Question

(D) The given wave equations are $y_1 = 4 \sin(708 \pi t)$ and $y_2 = 3 \sin(700 \pi t)$.Comparing these with the standard equation $y = A \sin(2 \pi

Vedclass · Accepted Answer

$4, 49:1$

Vedclass · Answer

(D) The given wave equations are $y_1 = 4 \sin(708 \pi t)$ and $y_2 = 3 \sin(700 \pi t)$.Comparing these with the standard equation $y = A \sin(2 \pi f t)$, we get:For $y_1$: $2 \pi f_1 = 708 \pi \implies f_1 = 354 	ext{ Hz}$ and amplitude $A_1 = 4$.For $y_2$: $2 \pi f_2 = 700 \pi \implies f_2 = 350 	ext{ Hz}$ and amplitude $A_2 = 3$.Beat frequency $f_b = |f_1 - f_2| = |354 - 350| = 4 	ext{ beats per second}$.The intensity $I$ is proportional to the square of the amplitude $(I \propto A^2)$.Maximum intensity (waxing) $I_{max} \propto (A_1 + A_2)^2 = (4 + 3)^2 = 7^2 = 49$.Minimum intensity (wanning) $I_{min} \propto (A_1 - A_2)^2 = (4 - 3)^2 = 1^2 = 1$.Therefore, the ratio of intensity between waxing and wanning is $I_{max} : I_{min} = 49:1$.The number of beats per second is $4$ and the intensity ratio is $49:1$.

Two sources of sound are emitting progressive waves $y_1 = 4 \sin 708 \pi t$ and $y_2 = 3 \sin 700 \pi t$. The sources are placed close to each other. The number of beats heard per second and intensity ratio between waxing and wanning are respectively

Similar Questions

$16$ tuning forks are arranged in the order of increasing frequencies. Any two successive forks give $8$ beats per second when sounded together. If the frequency of the last fork is twice the first,then the frequency of the first fork is:

$A$ pair of tuning forks produces $2$ beats in a time interval of $0.4 \ s$. The beat frequency is .... $Hz$.

Two tuning forks of frequencies $320 \,Hz$ and $480 \,Hz$ are sounded together to produce sound waves. The velocity of sound in air is $320 \,ms^{-1}$. The difference between the wavelengths of these waves is nearly: (in $cm$)

$A$ and $B$ are two wires whose fundamental frequencies are $256 \text{ Hz}$ and $382 \text{ Hz}$ respectively. When the third harmonic of $A$ and the second harmonic of $B$ are sounded together, the number of beats heard in two seconds will be:

The frequency of tuning fork $A$ is $256\,Hz$. It produces $4\,beats/sec$ with tuning fork $B$. When wax is applied to tuning fork $B$,$6\,beats/sec$ are heard. By reducing a small amount of wax,$4\,beats/sec$ are heard. The frequency of $B$ is .... $Hz$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module