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Two sinusoidal waves with the same amplitude of 4.0mm and the same wavelength traveled together along a string that is stretched along an x axis, the resultant wave due to their interference was recorded on video tape. The curves in Fig. 16-20 represent the resultant wave in two freeze frames, first as the solid curve and then, 1.0 ms later, as the dotted curve. The grid lines along the x axis are 1.0 cm apart, and the string clements oscillated vertically (perpendicular to the x axis) by 6.0 mm as the resultant wave passed through them. That wave moved a distance d =4.20 cm to the right in the 1.0 ms time interval. Write equations for the two interfering waves and for their resultant wave. Figure 16-20 The resultant wave of two sinusoidal string wave traveling along an x axis is shown at two instants, between which the resultant wave travels distance d. |
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Answer» Solution :From Fig. 16-20, we see that the resultant wave is a sinusoidal wave that moves in the positive direction of the x axis. The interfering waves must both move in that direction and we can write them as `y_(1) (x,t) =y_(m) sin (kx-omegat) ` and `y_(2) (x,t) =y_(m) sin (kx-omegat+phi)` where `y_(m)=4.0 mm.` We can now write the resultant wave as `y. (x,t) =[2y_(m)^(.) cos""1/2 phi] sin (kx-omegat+1/2 phi) ` The amplitude `2y_(m) cos 1//2 phi` of the resultant wave is half the total oscillation distance of 6.0 mm, so we have `2y_(m) cos ""1/2 phi=3.0mm` SUBSTITUTING. `y_(m)= 4.0 mm` and solving for a we get `phi=2cos^(-1) (3.0mm)/(2(4.0mm))=136^(@) approx 2.4 rad` Thus, the phase constant in Eq. 16-59 is = 2.4 rad and the phase constant in Eq. 16-60 is 1/2 `phi =1// (2.4+ rad) approx "1.2 rad,"`. To find the angular wave number we use two key ideas. First, the common value of k in Eqs. 16-58 16.59 and 16-60 is related to the common wavelength EQUATION `k= 2pi//lambda`. Second, the wavelength can measured in Fig. 16-20 as the distance (parallel to the axis) between REPETITIONS of the wave shape. Let.s use the solid curve and pick any point at which it crosses the x axis. That curve makes an identical crossing 3.0 cm to the right (or LEFT) from the first point. Thus, `lambda= 3.0 cm,` and `k=(2pi)/(lambda)=(2pi)/(0.030m)=209 m^(-1)` To find the common angular frequency `omega` we use two more key ideas: The common value of `omega` in Eqs. 16-58, 16-59, and 16-60 is related to k and the common wave speed y by equation `v=omega//k`. That wave speed v is the ratio of the distance d traveled by the resultant wave to The time interval At required for that travel. Thus, we have `omega=kv-k d/(triangle t)=(209 m^(-1)) (0.0420 m)/(0.0010 s)=8778 s^(-1) approx 8800 s^(-1)` We can now write Eqs. 16-58 and 16-59 for the interfering waves as `y_(1) (x,t) =(4.0mm) sin (209 x-8807)` and `y_(2) (x,t) =(4.0mm) sin (209x -8800t+2.4 rad)` with x in meters and t in SECONDS. We can also write Eq. 16-60 for the resultant waves as `y. (x,t)=(3.0 mm) sin (209 x-8800 t+1.2 rad)` again with x meters and t in seconds. The phase difference between two waves must be 24 rad although the two wave equations can have any value of initial phase. |
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