Question

Two waves are described by $y_1=0.30\sin \left[\pi \right(5x-200\left)t\right]$ and $y_2=0.30\sin \left[\pi \right(5x-200t)+\pi /3]$ where $y_1-y_2$ and x are in meters and t is in seconds. When these two waves are combined, a travelling wave is produced. What are the (a) amplitude, (b) wave speed, and (c) wave length of that travelling wave?

• A. $0.62m,40m/s,0.40m$
• B. $0.52m,80m/s,0.60m$
• C. $0.52m,40m/s,0.40m$
• D. $0.92m,80m/s,0.40m$

Answer 1

The correct option is C $0.52m,40m/s,0.40m$

$y_1=0.3\left[\pi \right(5x-200t\left)\right]$ and $y_2=0.3\sin \left[\pi \right(5x-200t)+\pi /3]$

$\sin \alpha +\sin (\alpha +\beta )=2\sin (\alpha +\beta /2)\cos \frac{\beta }{2}$

Hence $y=y_1+y_2=0.6\sin \left[\pi \right(5x-200t)+\pi /6]\cos \frac{\pi }{6}$

$⟹y=0.3\sqrt{3}\sin \left[\pi \right(5x-200t)+\pi /6]$

Amplitude $A=0.3\sqrt{3}=0.52m$

$v=\frac{\omega }{k}=\frac{200\pi }{5\pi }=40m/s$

$\lambda =\frac{2\pi }{k}=\frac{2\pi }{5\pi }=0.4m$

Answer-(C)