21 1 Model The principle of superposition comes into play whenever the waves overlap Visualize The graph at t 1 0 s differs from the graph at t 0 0 s in that the left wave has moved to the right by 1 0 m and the right wave has moved to the left by 1 0 m This is because the distance covered by the wave pulse in 1 0 s is 1 0 m The snapshot graphs at t 2 0 s 3 0 s and 4 0 s are a superposition of the left and the right moving waves The overlapping parts of the two waves are shown by the dotted lines 21 2 Model The principle of superposition comes into play whenever the waves overlap Visualize The snapshot graph at t 1 0 s differs from the graph t 0 0 s in that the left wave has moved to the right by 1 0 m and the right wave has moved to the left by 1 0 m This is because the distance covered by each wave in 1 0 s is 1 0 m The snapshot graphs at t 2 0 s 3 0 s and 4 0 s are a superposition of the left and the right moving waves The overlapping parts of the two waves are shown by the dotted lines 21 3 Model The principle of superposition comes into play whenever the waves overlap Visualize At t 4 0 s the shorter pulses overlap and cancel At t 6 0 s the longer pulses overlap and cancel 21 4 Model The principle of superposition comes into play whenever the waves overlap Solve 4 s a As graphically illustrated in the figure below the snapshot graph of Figure EX21 5b was taken at t b 21 5 Model A wave pulse reflected from the string wall boundary is inverted and its amplitude is unchanged Visualize The graph at t 2 s differs from the graph at t 0 s in that both waves have moved to the right by 2 m This is because the distance covered by the wave pulse in 2 s is 2 m The shorter pulse wave encounters the boundary wall at 2 0 s and is inverted on reflection This reflected pulse wave overlaps with the broader pulse wave as shown in the snapshot graph at t 4 s At t 6 s only half of the broad pulse is reflected and hence inverted the shorter pulse wave continues to move to the left with a speed of 1 m s Finally at t 8 s both the reflected pulse waves are inverted and they are both moving to the left 21 6 Model Reflections at both ends of the string cause the formation of a standing wave Solve Figure EX21 6 indicates 5 2 wavelengths on the 2 0 m long string Thus the wavelength of the standing wave is The frequency of the standing wave is 5 2 0 m 0 80 m 2 f v 40 m s 0 80 m 50 Hz 21 7 Model Reflections at the string boundaries cause a standing wave on the string Solve Figure EX21 7 indicates two full wavelengths on the string Hence Thus 1 2 60 cm 30 cm 0 30 m v f 0 30 m 100 Hz 30 m s 21 8 Model Reflections at the string boundaries cause a standing wave on the string 2 Solve f the wavelength is halved 02 f 1 0 a When the frequency is doubled wavelength will increase the number of antinodes to six b Increasing the tension by a factor of 4 means This halving of the v v T T 4 T 2 v For the string to continue to oscillate as a standing wave with three antinodes means 0 Hence v 2 v f 2 f 0 0 f 0 2 f 0 0 f 2 f 0 That is the new frequency is twice the original frequency a We have fa 24 Hz mf1 where f1 is the fundamental frequency that corresponds to m 1 The next 21 9 Model A string fixed at both ends supports standing waves Solve successive frequency is fb 36 Hz m 1 f1 Thus 1 f 1 mf 1 36 Hz 24 Hz m 1 1 5 m 1 5 m m f f 1 b a The wave speed is b The frequency of the third harmonic is 36 Hz For m 3 the wavelength is v f 1 1 f 1 2 0 m 12 Hz 24 m s L 2 1 m 2 L m 2 1 m 3 m 2 3 m m 2 1 f 12 Hz 24 Hz 2 21 10 Model A string fixed at both ends supports standing waves Solve a A standing wave can exist on the string only if its wavelength is m 2 L m m 1 2 3 The three longest wavelengths for standing waves will therefore correspond to m 1 2 and 3 Thus 2 2 40 m 2 2 2 40 m 3 2 2 40 m 1 2 40 m 4 80 m 1 60 m 2 3 1 b Because the wave speed on the string is unchanged from one m value to the other f 2 2 f 3 3 f 3 f 2 2 3 50 Hz 2 40 m 1 60 m 75 Hz 21 11 Model A string fixed at both ends forms standing waves Solve a The wavelength of the third harmonic is calculated as follows m 3 2 L m L 2 3 2 42 m 3 0 807 m 0 81 m b The speed of waves on the string is v 0 807 m 180 Hz 145 3 m s The speed is also given by f 3 3 v T S so the tension is T S 2 v 2 v m L 0 004 kg 1 21 m 145 3 m s 2 69 7 N 70 m 21 12 Model For the stretched wire vibrating at its fundamental frequency the wavelength of the standing wave is 1 Visualize 2 L Solve The wave speed on the steel wire is f f v wire 2 L 80 Hz 2 0 90 m 144 m s and is also equal to T where S mass length 3 5 0 10 kg 0 90 m 5 555 10 kg m 3 The tension TS in the wire equals the weight of the sculpture or Mg Thus v wire M Mg 2 v wire g 5 555 10 kg m 144 m s 3 2 12 kg 9 8 m s 2 21 13 Model The laser light forms a standing wave inside the cavity Solve The wavelength of the laser beam is m 2 L m 100 000 2 0 5300 m 100 000 10 60 m f 100 000 c 100 000 8 3 000 10 m s 10 60 10 m 6 2 830 10 Hz 13 The frequency is 21 14 Solve possible wavelengths for this case are a For …
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