Chapter 21

Chapter 21 - 21.1. Model: The principle of superposition...

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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.
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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.
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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.
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21.4. Model: The principle of superposition comes into play whenever the waves overlap. Solve: (a) As graphically illustrated in the figure below, the snapshot graph of Figure EX21.5b was taken at t = 4 s. (b)
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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.
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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 () 2 5 2.0 m 0.80 m λ == . The frequency of the standing wave is 40 m/s 50 Hz 0.80 m v f =
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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 1 2 (60 cm) 30 cm 0.30 m. λ == = Thus ( )( ) 0.30 m 100 Hz 30 m/s vf =
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21.8. Model: Reflections at the string boundaries cause a standing wave on the string. Solve: (a) When the frequency is doubled ( ) 0 2 f f ′ = , the wavelength is halved ( ) 1 0 2 λ ′ = . This halving of the wavelength will increase the number of antinodes to six.
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This note was uploaded on 11/26/2010 for the course MTH 232 taught by Professor Smith during the Spring '10 term at UConn.

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Chapter 21 - 21.1. Model: The principle of superposition...

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