Question

Two sources S1 and S2, separated by 2.0 m, vibrate according to the equation y1 = 0.03 sin πt and y2 = 0.02 sin πt, where y1, y2 and t are in S.I. units. They send out waves of equal velocity 1.5 m/s. Calculate the amplitude of the resultant motion of the particle co-linear with S1 and S2 and located at a point :
(a) P1 to the right of S2
(b) P2 to the left of S1 and
(c) P3 in the middle of S1 and S2.

Answer 1

Wave source S1:  wave     y1 = 0.03 Sin πt  meters
       Source S2:  wave      y2 = 0.02 Sin πt   m

velocity of the waves = v = 1.5 m/s

Traveling/propagating wave equation :
      y1 = A Sin (ωt - k x) = A Sin ω(t - x/v)

     We deduce that ω = π rad.  so f = 1/2 Hz.  v = 1.5 m/s (given)
 
Case a)  both waves are propagating in +x direction:
     P1 is at x m from S1.  S1 is at x=0 m.

    y1 = 0.03 Sin π{t - x/1.5}  m   = 0.03 Sin[πt - 2πx/3]  m
    y2 = 0.02 Sin [πt - π(x-2)/1.5]  m   = 0.02 Sin[πt - 2πx/3 + 4π/3]  m

=> y1+y2 = 0.01 Sin(πt-2πx/3)+0.02 {sin(πt-2πx/3) + Sin(πt-2πx/3+4π/3)  m
               = 0.01 Sin(πt-2πx/3)+0.02 *2 Sin(πt-2πx/3) Cos(2π/3)  m
               = - 0.01 Sin (πt - 2πx/3)   meters

case (b)  Both waves are traveling in -x direction.
     P1 is at x m from S1.  S1 is at x=0 m.

    y1 = 0.03 Sin π{t + x/1.5}  m   = 0.03 Sin[πt + 2πx/3]  m
    y2 = 0.02 Sin [πt + π(x+2)/1.5]  m   = 0.02 Sin[πt + 2πx/3 + 4π/3]  m

y1+y2 = 0.01 Sin(πt+2πx/3)+0.02 {sin(πt+2πx/3) + Sin(πt+2πx/3+4π/3)  m
          = 0.01 Sin(πt+2πx/3)+0.02 * 2 Sin(πt+2πx/3) Cos(2π/3)  m
          = - 0.01 Sin (πt + 2πx/3)     meters

Case c) y1 is traveling in +x direction & y2 is traveling in -x direction.

     P1 is at x m from S1.  S1 is at x=0 m.

    y1 = 0.03 Sin π{t - x/1.5}  m  = 0.03 Sin[πt - 2πx/3]  m
    y2 = 0.02 Sin [πt - π(2-x)/1.5]  m   = 0.02 Sin[πt + 2πx/3 - 4π/3]  m

y1+y2 = 0.01 Sin(πt-2πx/3)+0.02 {sin(πt-2πx/3) + Sin(πt+2πx/3-4π/3)  m
          = 0.02 Sinπt Cos(2πx/3) - 0.04 Cosπt Sin(2πx/3) 
               - 0.01√3 Cosπt Cos(2πx/3) + 0.01√3 Sin πt Sin(2πx/3)    meters