Question

Two waves are traveling in opposite directions on the same string. The displacements caused by the individiual waves are given by y1=(25.0 mm)sin(8.50πt - 1.24πx) and y2=(38.0 mm)sin(3.43πt + 0.267πx). Note that the phase angles (8.50πt - 1.24πx) and (3.43πt + 0.267πx) are in radians, t is in seconds, and x is in meters. At t = 5.80 s, what is the net displacement (in mm) of the string at (a) x = 2.02 m and (b) x = 2.85 m? Be sure to include the algebraic sign (+ or -) with your answers.

Answer 1

In this question we are given two waves y2 38 sin(3.43Tt0.267Tx) here the maximum amplitudes of the waves are in mm we will convert them into meters 31 = 0.025 sin(8.50Trtー1.24TI) y2 0.038 sin(3.43rt 0.267Ta) now, if there was only wave y1, then its displacements at time, t 5.80s and x 2.02 meters would be 31-0.025 sin(8.50π × 5.80 _ 1.2472.02) 0.03 similarly if there was only wave y2, then its displacements at time, t 5.80s andx 2.02 meters would be Y2-0.038 sin(3.43π × 5.80 + 0.26Tm × 2.02) 0.0371 now we can find the total displacement of the superimposed wave by adding both the waves as D-0.0371 +0.030.0671 Similarly as in the above part we can find the net displacement at x 2.85 The single amplitudes of both the waves would be y1 = 0.025 sin(8.50π × 5.80-1.24π × 2.85) =-0.336 y2 0.038 sin(3.437 x 5.800.267T x 2.85) 0.336 so the net displacement will be y1 y20.336 - 0.3360