Answer to Question #162252 in Optics for Onel

Question

Two traveling sinusoidal waves are described by the wave function y1 = 5.00 sin ([pie(4.00x = 1200t)] , y2 = 5.00 sin [pie(4.00x - 1200t - 0.250)] where x, y1, and y2 are in meters and t is in seconds.

(a) what is the amplitude of the resultant wave function y1 + y2?

(b) what is the frequency of the resultant wave function?

Answer 1

(a) Let $y_1=Asin(kx-\omega t)$ and $y_2=Asin(kx-\omega t-\phi)$.

Then, the resultant wave function can be written as follows:

$$y=Asin(kx-\omega t)+Asin(kx-\omega t-\phi)=2Acos(\dfrac{\phi}{2})sin(kx-\omega t+\dfrac{\phi}{2}).$$

Finally, the resultant amplitude can be calculated as follows:

$$A_{res}=2Acos(\dfrac{\phi}{2})=2\cdot5.0\ m\cdot cos(\dfrac{0.25\pi}{2})=9.24\ m.$$

b) We can find the frequency of the resultant wave function as follows:

$$\omega=2\pi f,$$ $$f=\dfrac{\omega}{2\pi}=\dfrac{1200\pi\ \dfrac{rad}{s}}{2\pi}=600\ Hz.$$