Two waves are propagating to point P along a | vedclass

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

Question

(D) Given: Amplitude of each wave = $a$. Wavelength $\lambda = 1 \ m$. Path difference $\Delta x = AP - BP = 50 \ cm = 0.5 \ m$.The phase difference d

Vedclass · Accepted Answer

$a$

Vedclass · Answer

(D) Given: Amplitude of each wave = $a$. Wavelength $\lambda = 1 \ m$. Path difference $\Delta x = AP - BP = 50 \ cm = 0.5 \ m$.The phase difference due to path difference is $\phi_{path} = \frac{2\pi}{\lambda} 	imes \Delta x = \frac{2\pi}{1} 	imes 0.5 = \pi$.Since source $A$ is ahead of $B$ by $\frac{\pi}{3}$,the total phase difference $\Delta \phi$ between the two waves at point $P$ is $\Delta \phi = \phi_{path} - \frac{\pi}{3} = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.The resultant amplitude $R$ is given by $R = \sqrt{a^2 + a^2 + 2a^2 \cos(\Delta \phi)}$.Substituting $\Delta \phi = \frac{2\pi}{3}$:$R = \sqrt{2a^2 + 2a^2 \cos(120^\circ)} = \sqrt{2a^2 + 2a^2(-0.5)} = \sqrt{2a^2 - a^2} = \sqrt{a^2} = a$.

Similar Questions

$A$ sound wave of wavelength $32 \ cm$ enters the tube at $S$ as shown in the figure. Then the smallest radius $r$ so that a minimum of sound is heard at detector $D$ is ... $cm$.

Two waves have equations $x_1 = a \sin(\omega t + \phi_1)$ and $x_2 = a \sin(\omega t + \phi_2)$. If the frequency and amplitude of the resultant wave remain equal to those of the superimposing waves,then the phase difference between them is:

The similarity between sound waves and light waves is:

Four harmonic waves of equal frequencies and equal intensities $I_0$ have phase angles $0, \pi / 3, 2 \pi / 3$ and $\pi$. When they are superposed,the intensity of the resulting wave is $nI_0$. The value of $n$ is

Two waves are represented by $y_1 = a \sin(\omega t + \frac{\pi}{6})$ and $y_2 = a \cos(\omega t)$. What will be their resultant amplitude?

Vedclass Test Series

Exam Paper Generator

Online Exam Module