Question

Two waves are represented by $ {{y}_{1}}=a\sin \left( \omega t+\frac{\pi }{6} \right) $ and $ {{y}_{2}}=a\cos \omega t $ What will be their resultant amplitude?

• A. a
• B. $ \sqrt{2}a $
• C. $ \sqrt{3}a $
• D. 2a

Answer 1

Answer: (C)

Given waves are $ {{y}_{1}}=a\,\sin \left( \omega t+\frac{\pi }{6} \right) $ $ {{y}_{2}}=a\,\cos \,\omega t $ $ =a\,\sin \left( \omega t+\frac{\pi }{2} \right) $ Therefore, phase difference between two waves is $ \text{o }\!\!|\!\!\text{ =}\frac{\pi }{2}-\frac{\pi }{6}=\frac{\pi }{3} $ The resultant amplitude is given by $ \therefore $ $ {{A}^{2}}={{a}^{2}}+{{a}^{2}}+2a.a\,\cos \,\frac{\pi }{3} $ $ ={{a}^{2}}+{{a}^{2}}+2{{a}^{2}}.\frac{1}{2} $ $ =3{{a}^{2}} $ Hence, $ A=\sqrt{3}a $