Two waves Y_1 = A_1 sin(omega t - beta_1) and | vedclass

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

Question

(A) The two waves are given by $Y_1 = A_1 \sin(\omega t - \beta_1)$ and $Y_2 = A_2 \sin(\omega t - \beta_2)$.The phase difference between the two wave

Vedclass · Accepted Answer

$\sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\beta_1 - \beta_2)}$

Vedclass · Answer

(A) The two waves are given by $Y_1 = A_1 \sin(\omega t - \beta_1)$ and $Y_2 = A_2 \sin(\omega t - \beta_2)$.The phase difference between the two waves is $\phi = (\omega t - \beta_1) - (\omega t - \beta_2) = \beta_2 - \beta_1$.The resultant amplitude $A$ of two superimposing waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by the formula:$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}$.Substituting $\phi = \beta_2 - \beta_1$ into the formula,and noting that $\cos(\beta_2 - \beta_1) = \cos(\beta_1 - \beta_2)$:$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\beta_1 - \beta_2)}$.

Two waves $Y_1 = A_1 \sin(\omega t - \beta_1)$ and $Y_2 = A_2 \sin(\omega t - \beta_2)$ superimpose to form a resultant wave whose amplitude is

Similar Questions

Two waves each of intensity $I$ interfere with a phase difference of $120^o$. The resultant intensity of the waves will be:

If two waves represented by $y_1 = 4 \sin \omega t$ and $y_2 = 3 \sin (\omega t + \pi / 3)$ interfere at a point,find the amplitude of the resultant wave.

Two waves $Y_1 = a \sin \omega t$ and $Y_2 = a \sin (\omega t + \delta)$ produce interference. The resultant intensity is ......

Two waves represented by $y_1 = 10 \sin(200\pi t)$ and $y_2 = 20 \sin(200\pi t + \pi/2)$ are superimposed at any point at a particular instant. The amplitude of the resultant wave is:

If the phase difference between two waves is $2\pi$ during superposition,then the resultant amplitude is

Vedclass Test Series

Exam Paper Generator

Online Exam Module