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Two sinusoidal waves with the same amplitude and wavelength travel along a string that is stretched along an x axis, the resultant wave due to their superposition was recorded on video tape. The curves in Fig. 16-31 represent the resultant wave in four freeze frames in the sequence of a, b, c, and d, with 1.0 ms elapsing between curves a and d. The grid lines along the x axis are 1.0 cm apart, and the string elements oscillated perpendicular to the r axis by 16.0 mm (between the extreme displacements shown by curves a and d) as the resultant wave passed through them. Write equations for the two superposition waves and for their resultant wave. |
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Answer» Solution :FIGURE 16-31, shows a standing wave and so the two waves must be travelling in opposite directions. The two waves have the forms given by `y_(1)(x,t) =y_(m) sin (kx-omegat)` and `y_(2) (x,t) =y_(m) sin (kx+omega t)` Hence, we can write the RESULTANT wave as `y. (x,t)=(2y_(m), sin kx) cos omega t` We need to find the common values of `y_m, k,` and a in the three equations. Calculations: To find `y_m` we use the key idea that the amplitude of a standing wave is a maximum at an antinode, and there the amplitude is `2y_m.` Here the maximum amplitude of the standing wave is 8.0 mm, so `y_(m)=4.0mm` To find the angular wave number k, we use two more key ideas. First, the common angular wave number k in Eqs. 16-71, 16-72 and 16-73 is related to the common wavelength `lambda` by equation `k= 2pi//lambda.` Second, the nodes in a standing wave arc separated by 0.502. Because the antinodes in Fig. 16-31 are separated by 2.0 cm, we have `lambda=2 xx 2.0 cm =4.0cm` .Then `k=(2pi)/(lambda) =(2pi)/(0.040 m)=157 m^(-1) APPROX 160m^(-1)` To find the common angular frequency `omega` we again use two more key ideas: We can relate the common value of `omega` in Eqs. 16-71, 16-72, and 16-73 to the period T of the standing wave with equation o = 24/7. A string element at an antinode TAKES time T to move through a full oscillation from one extreme and back to the first extreme. According to the given data, the string element at the dot on curve a in Fig. 16-31 takes 10 ms to move to the POSITION of the dot on curve d, which is one half a full oscilations. Thus, T=2.0 ms, and `omega=(2pi)/(T)=(2pi)/(0.0020 s)=3142 s^(-1) approx 3100 s^(-1)` Now we can write Eqs. 16-71 and 16-72 for the interfering waves as `y_(1) (x,t) =(4.0 mm) sin (160x-3100t)` `and y_(2) (x,t)=(4.0mm) sin (160x-3100t)` with x in meters and t in seconds. We can also write the standing wave of Eq. 16-73 as |
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