The resultant amplitude of the interference o | vedclass

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

Question

(B) The resultant amplitude $A$ of two waves with amplitudes $a_1$ and $a_2$ and phase difference $\phi$ is given by the formula:$A = \sqrt{a_1^2 + a_

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The resultant amplitude $A$ of two waves with amplitudes $a_1$ and $a_2$ and phase difference $\phi$ is given by the formula:$A = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos \phi}$Given $a_1 = 4$,$a_2 = 3$,and $\phi = \frac{\pi}{3}$.Substituting these values into the formula:$A = \sqrt{4^2 + 3^2 + 2(4)(3) \cos(\frac{\pi}{3})}$Since $\cos(\frac{\pi}{3}) = 0.5$:$A = \sqrt{16 + 9 + 24(0.5)}$$A = \sqrt{25 + 12}$$A = \sqrt{37}$$A \approx 6.08$Rounding to the nearest integer,the resultant amplitude is $6$.

The resultant amplitude of the interference of two waves $y_1 = 4 \sin \omega t$ and $y_2 = 3 \sin (\omega t + \frac{\pi}{3})$ is:

Similar Questions

The resultant amplitude in interference with two coherent sources depends upon:

An interference pattern is observed at $P$ due to the superimposition of two rays coming from a source $S$ as shown in the figure. The value of $l$ for which maxima is obtained at $P$ is: ($R$ is a perfectly reflecting surface)

Which of the following is conserved when light waves interfere?

Two light waves of same intensity superpose at point $P$ with a phase difference of $\pi/3$. The resultant intensity at point $P$ will be?

Four light waves are represented by:
$(i)$ $y = a_1 \sin \omega t$
(ii) $y = a_2 \sin (\omega t + \phi)$
(iii) $y = a_1 \sin 2\omega t$
(iv) $y = a_2 \sin 2(\omega t + \phi)$
Interference fringes may be observed due to the superposition of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module