Two waves are given by y_1 = a sin(omega t - | vedclass

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

Question

(D) The first wave is given by $y_1 = a \sin(\omega t - kx)$.The second wave is given by $y_2 = a \cos(\omega t - kx)$.Using the trigonometric identit

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) The first wave is given by $y_1 = a \sin(\omega t - kx)$.The second wave is given by $y_2 = a \cos(\omega t - kx)$.Using the trigonometric identity $\cos(	heta) = \sin(	heta + \frac{\pi}{2})$,we can rewrite the second wave as:$y_2 = a \sin(\omega t - kx + \frac{\pi}{2})$.Comparing the phase of the two waves,the phase of $y_1$ is $(\omega t - kx)$ and the phase of $y_2$ is $(\omega t - kx + \frac{\pi}{2})$.Therefore,the phase difference $\Delta \phi$ is $(\omega t - kx + \frac{\pi}{2}) - (\omega t - kx) = \frac{\pi}{2}$.

Two waves are given by $y_1 = a \sin(\omega t - kx)$ and $y_2 = a \cos(\omega t - kx)$. The phase difference between the two waves is

Similar Questions

The frequency of the sinusoidal wave $y = 0.40\cos(2000t + 0.80x)$ is

Wave equations of two particles are given by $y_1 = a \sin(\omega t - kx)$ and $y_2 = a \sin(kx + \omega t)$. Then:

$A$ transverse wave travelling along a stretched string has a speed of $30 \ m/s$ and a frequency of $250 \ Hz$. The phase difference between two points on the string $10 \ cm$ apart at the same instant is

The equation of a sound wave is $y = 0.0015 \sin (62.4x + 316t)$. The wavelength of this wave is ..... $unit$.

The phase difference between two points separated by $0.8 \ m$ in a wave of frequency $120 \ Hz$ is $\frac{\pi}{2}$. The velocity of the wave is ..... $m/s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module