Two waves represented by y_1 = 10 sin(2000 pi | vedclass

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

Question

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

Vedclass · Accepted Answer

$14.1$

Vedclass · Answer

(C) The resultant amplitude $A$ of two waves with amplitudes $a_1$ and $a_2$ and a phase difference $\phi$ is given by the formula:$A = \sqrt{a_1^2 + a_2^2 + 2 a_1 a_2 \cos \phi}$Given the equations of the waves:$y_1 = 10 \sin(2000 \pi t + 2x)$$y_2 = 10 \sin(2000 \pi t + 2x + \frac{\pi}{2})$Here,$a_1 = 10$,$a_2 = 10$,and the phase difference $\phi = \frac{\pi}{2} = 90^\circ$.Substituting these values into the formula:$A = \sqrt{10^2 + 10^2 + 2(10)(10) \cos(90^\circ)}$Since $\cos(90^\circ) = 0$:$A = \sqrt{100 + 100 + 0} = \sqrt{200} = 10\sqrt{2} \approx 14.14 \ unit$.Thus,the resultant amplitude is approximately $14.1 \ unit$.

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 ..... $unit$.

Similar Questions

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

When two progressive waves $y_1=4 \sin (2 x-6 t)$ and $y_2=3 \sin \left(2 x-6 t-\frac{\pi}{2}\right)$ are superimposed,the amplitude of the resultant wave is

Equations of motion for two waves traveling in the same direction are given by $y_1 = 2a \sin(\omega t - kx)$ and $y_2 = 2a \sin(\omega t - kx - \theta)$. The resultant amplitude of the medium particle will be:

Two waves of same frequency and intensity superimpose with each other in opposite phases,then after superposition the

Write the equation of displacement of the resultant wave for two superposed waves with an initial phase difference.

Vedclass Test Series

Exam Paper Generator

Online Exam Module