The resultant amplitude due to the superposit | vedclass

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

Question

(A) Given waves are $y_1 = 5 \sin (\omega t - kx)$ and $y_2 = -5 \cos (\omega t - kx - 150^{\circ})$.We know that $-\cos(	heta) = \sin(	heta - 90^{\

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Given waves are $y_1 = 5 \sin (\omega t - kx)$ and $y_2 = -5 \cos (\omega t - kx - 150^{\circ})$.We know that $-\cos(	heta) = \sin(	heta - 90^{\circ})$.So,$y_2 = 5 \sin (\omega t - kx - 150^{\circ} - 90^{\circ}) = 5 \sin (\omega t - kx - 240^{\circ})$.Alternatively,$y_2 = 5 \sin (\omega t - kx - 240^{\circ} + 360^{\circ}) = 5 \sin (\omega t - kx + 120^{\circ})$.The resultant amplitude $A$ of two waves $A_1 \sin(\phi_1)$ and $A_2 \sin(\phi_2)$ is given by $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\Delta\phi)}$.Here $A_1 = 5$,$A_2 = 5$,and the phase difference $\Delta\phi = 120^{\circ} - 0^{\circ} = 120^{\circ}$.$A = \sqrt{5^2 + 5^2 + 2(5)(5) \cos(120^{\circ})}$.Since $\cos(120^{\circ}) = -0.5$,we have $A = \sqrt{25 + 25 + 50(-0.5)} = \sqrt{50 - 25} = \sqrt{25} = 5$.

The resultant amplitude due to the superposition of two waves $y_1 = 5 \sin (\omega t - kx)$ and $y_2 = -5 \cos (\omega t - kx - 150^{\circ})$ is:

Similar Questions

When two waves of the same amplitude $a$ and the same frequency $f$ superimpose,the total intensity is proportional to which of the following?

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

Two waves of the same frequency and same amplitude $a$ are superimposed. If the resultant amplitude is also $a$,what is the phase difference between the two waves?

Two waves represented by $y_1 = 10 \sin(2000 \pi t + 2x)$ and $y_2 = 10 \sin(2000 \pi t + 2x + \frac{\pi}{2})$ are superposed at any point at a particular instant. The resultant amplitude is ..... $unit$.

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.

Vedclass Test Series

Exam Paper Generator

Online Exam Module