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\left[ 2\sin \frac{(\omega t-kx+\omega t-kx-\theta )}{2} \right. $ $ \left. \times \cos \frac{\omega t-kx-(\omega t-kx-\theta )}{2} \right] $ $ 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} $