Answer to Question #117770 in Mechanics | Relativity for raymond barakat

Question

The wave functions of two waves traveling in the same direction are given below. The two waves have the same frequency, wavelength, and amplitude, but they differ in their phase constant.

y1 (x,t) = 2 sin⁡(2πx ‒ 20πt), and

y2 (x,t) = 2 sin(2πx ‒ 20πt + φ),

where, y is in centimetres, x is in meters, and t is in seconds.
Which of the following wave functions represents the resultant wave due to the interference between the two waves:
y_result (x,t) = 4 sin⁡(2πx ‒ 20πt + φ)
y_result (x,t) = 2 sin⁡(πx ‒ 10πt + φ/2)
y_result (x,t) = 4 cos⁡(φ/2) cos⁡(πx ‒ 10πt - φ/2)
y_result (x,t) = 4 cos⁡(φ/2) sin⁡(2πx ‒ 20πt + φ/2)
y_result (x,t) = 2 sin⁡(φ/2) sin⁡(2πx ‒ 20πt + φ/2)

Answer 1

Solution: According to the superposition principle $y_{result} (x,t) =y1 (x,t)+y2 (x,t)$ . If we denote $\beta =2πx ‒ 20πt+\frac{\phi}{2}$ then we get

$y_{result} (x,t) =2sin(\beta-\frac{\phi}{2})+2 sin(\beta+\frac{\phi}{2})=\\=2\cdot [sin(\beta)\cdot cos(\frac{\phi}{2})-cos(\beta)\cdot sin(\frac{\phi}{2})]+2\cdot [sin(\beta)\cdot cos(\frac{\phi}{2})+cos(\beta)\cdot sin(\frac{\phi}{2})]=\\=4\cdot sin(\beta)\cdot cos(\frac{\phi}{2})$

Answer: the following wave functions represents the resultant wave due to the interference between the two waves: $y_{result} (x,t) ==4\cdot sin(2πx ‒ 20πt+\frac{\phi}{2})\cdot cos(\frac{\phi}{2})$