Question #117770
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)
1
Expert's answer
2020-05-25T10:47:58-0400

Solution: According to the superposition principle yresult(x,t)=y1(x,t)+y2(x,t)y_{result} (x,t) =y1 (x,t)+y2 (x,t) . If we denote β=2πx20πt+ϕ2\beta =2πx ‒ 20πt+\frac{\phi}{2} then we get

yresult(x,t)=2sin(βϕ2)+2sin(β+ϕ2)==2[sin(β)cos(ϕ2)cos(β)sin(ϕ2)]+2[sin(β)cos(ϕ2)+cos(β)sin(ϕ2)]==4sin(β)cos(ϕ2)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: yresult(x,t)==4sin(2πx20πt+ϕ2)cos(ϕ2)y_{result} (x,t) ==4\cdot sin(2πx ‒ 20πt+\frac{\phi}{2})\cdot cos(\frac{\phi}{2})


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