Question

Two waves are traveling in opposite directions on the same string.The displacements caused by the individiual waves are given byy1=(28.0 mm)sin(8.44πt -1.07πx) and y2=(38.0mm)sin(2.47πt + 0.427πx). Note that thephase angles (8.44πt - 1.07πx) and(2.47πt + 0.427πx) are in radians,t is in seconds, and x is in meters. Att = 5.50 s, what is the net displacement (in mm) of thestring at (a) x = 2.38 m and(b) x = 2.91 m? Be sure to include thealgebraic sign (+ or -) with your answers

Answer 1

ANSWER -

To find the net displacement, we need to add the individual displacements:

(a) x = 2.38 m
y = y1 + y2 = (28.0 mm)sin(8.44πt - 1.07πx) + (38.0mm)sin(2.47πt + 0.427πx)
= (28.0 mm)sin(8.44πt - 1.07πx) + (38.0mm)sin(2.47πt + 0.427πx)
= (28.0 mm)sin(8.44π(5.50 s) - 1.07π(2.38 m)) + (38.0mm)sin(2.47π(5.50 s) + 0.427π(2.38 m))
= (28.0 mm)sin(-0.71 rad) + (38.0mm)sin(3.73 rad)
= -18.36 mm + 30.70 mm
= 12.34 mm

So the net displacement at x = 2.38 m at t = 5.50 s is 12.34 mm.

(b) x = 2.91 m
y = y1 + y2 = (28.0 mm)sin(8.44πt - 1.07πx) + (38.0mm)sin(2.47πt + 0.427πx)
= (28.0 mm)sin(8.44πt - 1.07πx) + (38.0mm)sin(2.47πt + 0.427πx)
= (28.0 mm)sin(8.44π(5.50 s) - 1.07π(2.91 m)) + (38.0mm)sin(2.47π(5.50 s) + 0.427π(2.91 m))
= (28.0 mm)sin(-1.24 rad) + (38.0mm)sin(4.62 rad)
= 18.27 mm + 36.52 mm
= 54.79 mm

So the net displacement at x = 2.91 m at t = 5.50 s is 54.79 mm.