Two harmonic waves are described by where A = 3.6 cm, lambda = 2 cm, and v = 0.5

Question

Two harmonic waves are described by where A = 3.6 cm, lambda = 2 cm, and v = 0.5 cm/s. Consider the superposition y(x, t) = y1 + y2 of these two waves. Find the maximum displacement of y(x.t) at x1 = 0.25 cm. Answer in units of cm 005 (part 2 of 4) 10.0 points Determine the maximum displacement of y(x, t) at X2 = 0.5 cm. Answer in units of cm 006 (part 3 of 4) 10.0 points Determine the maximum displacement of y(x, t) at x3 = 1.5 cm. Answer in units of cm 007 (part 4 of 4) 10.0 points Find the smallest positive value of x corresponding to an antinode. Answer in units of cm

Explanation / Answer

sin(A+B) + sin(A-B)=2sin(A)cos(B)

So y=y1+y2 = 2Asin(2pix/lambda)cos(2pivt/lambda)

5.) Max disp at x1=0.25 cm is

y=2*3.6*sin(2*pi*0.25/2)= 5.09 cm

6.)

Max disp at x1=0.5 cm is

y=2*3.6*sin(2*pi*0.5/2)= 7.2 cm

7.)

Max disp at x1=1.5 cm is

y=2*3.6*sin(2*pi*1.5/2)= -7.2 cm

8.)

Antinodes occur at x=n*lambda/4

Smallest means n=1

So x=lambda/4 = 2/4 = 0.5 cm

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