two wave on a string are given by the followingfunctions: Y1(x,t) = 4cos(20t - 3

Question

two wave on a string are given by the followingfunctions:     Y1(x,t) = 4cos(20t - 30x)     Y2(x,t) =-4cos(20t - 30x) x is in centimeters , when the two waves interfere , |Ys| =| Y1 + Y2 | a) At t = (pi/50) s , at what location do the two wavesinterfere constructively , and what is the corresponding value ofYs. b) At t = (pi/50) s , at what location do the two wavesinterfere destructively , and what is the corresponding value ofYs. two wave on a string are given by the followingfunctions:     Y1(x,t) = 4cos(20t - 30x)     Y2(x,t) =-4cos(20t - 30x) x is in centimeters , when the two waves interfere , |Ys| =| Y1 + Y2 | a) At t = (pi/50) s , at what location do the two wavesinterfere constructively , and what is the corresponding value ofYs. b) At t = (pi/50) s , at what location do the two wavesinterfere destructively , and what is the corresponding value ofYs.

Explanation / Answer

a)when the two waves interfere constructively we get Ys = Y1 + Y2 where Y1 = 4cos(20t - 30x) and Y2 =-4cos(20t + 30x) or Ys = 4cos(20t - 30x) - 4cos(20t + 30x) = 4* (cos(20t - 30x) - cos(20t + 30x)) = 4 * 2 * sin(20t - 30x + 20t + 30x/2) * sin(20t - 30x -20t - 30x/2) = 8 * sin(20t) * sin(-30x) = -8 * sin(20t) *sin(15x) when t = (/50) s we have Ys = -8 * sin(20 * (/50)) * sin(15x) = -8 *sin(72o) * sin(15x) = -7.6 * sin(15x) b)when the two waves interfere destructively the displacementof the wave is Ys = Y1 - (-Y2) =Y1 + Y2 or Ys = 4cos(20t - 30x) + (4cos(20t + 30x)) =[4cos(20t - 30x)  + 4cos(20t + 30x)] = 4 * [cos(20t - 30x)  + cos(20t + 30x)] = 4 *2 * cos((20t - 30x) + (20t + 30x)/2) * cos((20t - 30x) -(20t + 30x)/2) = 8 * cos(20t) * cos(-30x) = 8 * cos(20t) * cos(30x) when t = (/50) s we have Ys = 8 * cos(20 * (/50)) * cos(30x) = 7.99 *cos(30x) where Y1 = 4cos(20t - 30x) and Y2 =-4cos(20t + 30x) or Ys = 4cos(20t - 30x) - 4cos(20t + 30x) = 4* (cos(20t - 30x) - cos(20t + 30x)) = 4 * 2 * sin(20t - 30x + 20t + 30x/2) * sin(20t - 30x -20t - 30x/2) = 8 * sin(20t) * sin(-30x) = -8 * sin(20t) *sin(15x) when t = (/50) s we have Ys = -8 * sin(20 * (/50)) * sin(15x) = -8 *sin(72o) * sin(15x) = -7.6 * sin(15x) b)when the two waves interfere destructively the displacementof the wave is Ys = Y1 - (-Y2) =Y1 + Y2 or Ys = 4cos(20t - 30x) + (4cos(20t + 30x)) =[4cos(20t - 30x)  + 4cos(20t + 30x)] = 4 * [cos(20t - 30x)  + cos(20t + 30x)] = 4 *2 * cos((20t - 30x) + (20t + 30x)/2) * cos((20t - 30x) -(20t + 30x)/2) = 8 * cos(20t) * cos(-30x) = 8 * cos(20t) * cos(30x) when t = (/50) s we have Ys = 8 * cos(20 * (/50)) * cos(30x) = 7.99 *cos(30x)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.