Explanation Given data: The two waves are given as, Formula used: The mathematical expression for a wave is written as, y = A cos [ 2 π f ( t − x λ f ) ] Here, f is the frequency of the wave, λ is the wavelength of the wave, A is the amplitude of the wave, t is time interval to complete one vibration, y is displacement of wave and x is the position of vibrational disturbance. Explanation: For first wave: From the graph for first wave, observe that the frequency of the first graph is 200 Hz and amplitude is 0.002 N/m 2 . Consider the wave length is λ . Recall the expression for a wave. y 1 = A cos [ 2 π f ( t − x λ f ) ] Substitute 0.002 N/m 2 for A and 200 Hz for f y 1 = ( 0.002 N/m 2 ) cos [ 2 π ( 200 Hz ) ( t − x λ ( 200 Hz ) ) ] …… (1) From the graph for first wave, observe that that the wave disturbance is zero at x = 0 . Substitute 0 for x in equation (1). y 1 = ( 0.002 N/m ) cos [ 2 π ( 200 Hz ) t ] By setting the cosine argument at π 2 for x = 0 , find t . [ 2 π ( 200 Hz ) t ] = π 2 t = 1 800 s Thus, the first wave can be represented as, y 1 = ( 0.002 N/m 2 ) cos [ 2 π ( 200 Hz ) ( ( 1 800 s ) − x λ ( 200 Hz ) ) ] …… (2) For second wave: From the graph for second wave, observe that frequency of second wave is 600 Hz, amplitude is 0.001 N/m 2 and the wavelength is λ / 3 . Recall the expression for a wave. y 2 = A cos [ 2 π f ( t − x λ f ) ] Substitute 0.001 N/m 2 for A and 600 Hz for f . y 2 = ( 0.001 N/m 2 ) cos [ 2 π ( 600 Hz ) ( t − 3 x λ ( 600 Hz ) ) ] …… (3) From the graph of the second wave, observe that the wave disturbance is zero at x = 0

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ISBN: 9780134862897

Author: ETKINA

Publisher: PEARSON

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Chapter 11, Problem 31P

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The resultant wave at the instant the two waves, shown below, are travelling through the same medium.

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Chapter 11, Problem 30P

Chapter 11, Problem 32P

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The resultant wave at the instant the two waves, shown below, are travelling through the same medium.

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The resultant wave at the instant the two waves, shown below, are travelling through the same medium.

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