Two loudspeakers (L_1 and L_2) are placed wit | vedclass

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

Question

(B) Let the distance of point $A$ from the line joining the speakers be $D = 40 \ m$. The speakers $L_1$ and $L_2$ are separated by $10 \ m$, so their

Vedclass · Accepted Answer

$600$

Vedclass · Answer

(B) Let the distance of point $A$ from the line joining the speakers be $D = 40 \ m$. The speakers $L_1$ and $L_2$ are separated by $10 \ m$, so their coordinates relative to the midpoint of the line joining them are $(0, 5)$ and $(0, -5)$. Point $A$ is at $(40, 0)$.When the recorder moves to point $B$ at a distance of $25 \ m$ from $A$, its coordinates are $(40, 25)$.The distances from the speakers to point $B$ are:$L_1B = \sqrt{40^2 + (25-5)^2} = \sqrt{40^2 + 20^2} = \sqrt{1600 + 400} = \sqrt{2000} = 20\sqrt{5} \ m$.Given $\sqrt{5} = 2.23$, $L_1B = 20 	imes 2.23 = 44.6 \ m$.$L_2B = \sqrt{40^2 + (25+5)^2} = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \ m$.The path difference at point $B$ is $\Delta x = L_2B - L_1B = 50 - 44.6 = 5.4 \ m$.Since the recorder undergoes $10$ cycles of minima and maxima, point $B$ corresponds to the $10^{	ext{th}}$ maxima, so $\Delta x = n\lambda$, where $n = 10$.$5.4 = 10 	imes \lambda \implies \lambda = 0.54 \ m$.The frequency $f$ is given by $f = \frac{v}{\lambda} = \frac{324}{0.54} = 600 \ Hz$.

Similar Questions

An observer receives waves directly from a source of sound distant $120\,m$ in a big hall. He also receives waves reflected from the mid-point of a $25\,m$ high ceiling. The wavelength of sound for constructive interference to take place between the two waves must be:

The resultant amplitude of waves ${y_1} = a \sin \left( \omega t + \frac{\pi}{3} \right)$ and ${y_2} = a \sin \omega t$ is:

Two waves of frequency $f$ and amplitude $a$ superimpose with each other. The total intensity is directly proportional to

Three harmonic waves having equal frequency $v$ and same intensity $I_{0}$,have phase angles $0, \frac{\pi}{4}$ and $-\frac{\pi}{4}$ respectively. When they are superimposed,the intensity of the resultant wave is close to

Vedclass Test Series

Exam Paper Generator

Online Exam Module