Two waves passing through a region are represented by y=(1.0cm) sin [(3.14 cm^(-1))x - (157s^(-1) x - (157s^(-1))t] and y = (1.5 cm) sin [(1.57 cm^(-1))x- (314 s^(-1))t]. Find the displacement of the particle at x = 4.5 cm at time t = 5.0 ms.
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Answer:According to the principle of superpositon each wave produces its distrubance independent of the other and the resultant distrubance is eqal to the vector sum of the indicidual disturbances, The displacemts of the particle at x = 4.5 cm at time t= 5.0 ms due to the two waves are. y1=(1.0cm)sin[(3.14cm-1)(4.5cm)
-(157s ^(-1) (5.0xx10^(-9)s)]=(1.0cm)sin[4.5 pi - (pi)/(4)]=(1.0 cm) sin [4pi + pi//4] = (1.0cm)/(sqrt2)andy_2 = (1.5 cm) sin [1.57cm^(-1)) (4.5 cm)- (314s^(-1)) (5.0xx10^(-3)s)]=(1.5 cm) sin[ 2.25 pi - (pi)/(2)]=(1.5 cm) sin[2pi - pi/4]= - (1.5cm) sin(pi)/(4) = -(1.5cm)/(sqrt2)The≠tdisplacmentisy =y_1 +y_2 = (-0.5cm)/(sqrt2) = -0.35cm.`
Explanation:
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