Question

Equations of motion in the same direction are given by : $y_1-2a\sin (\omega t-kx)$ $y_2-2a\sin (\omega t-kx\theta )$ The amplitude of the medium particle will be :

• A. $2a\cos \theta$
• B. $\sqrt{2a}\cos \theta$
• C. $4a\cos \theta /2$
• D. $\sqrt{2a}\cos \theta /2$

Answer 1

Answer: (C)

The equations of motion are $y_1=2a\sin (\omega t-kx)$ $y_2=2a\sin (\omega t-kx-\theta )$ Now, the equation of resultant wave is given by $y=y_1+y_2$ $=2a\sin (\omega t-kx)+2a\sin (\omega t-kx-\theta )$ $y=2a[2\sin \frac{(\omega t-kx+\omega t-kx-\theta )}{2}$ $\times \cos \frac{\omega t-kx-(\omega t-kx-\theta )}{2}]$ $y=4a\cos \frac{\theta }{2}\sin \left(\omega t-kx-\frac{\theta }{2}\right)$ ...(i) Now, comparing equation (i) with $y=A\sin (\omega t-kx),$ we have Resultant amplitude $A=4a\cos \frac{\theta }{2}$