Consider two waves of equal amplitude A1cm and equal angular

Consider two waves of equal amplitude A=1cm and equal angular frequency 20n radVs traveling in opposite directions along a infinitely tong rope at the speed v=30 m/s Write the equations for the 2 waves and the equation of the resultant wave on the rope: describe in words the resulting wave. Find the location of a point along the rope where the amplitude of the simple harmonic motion is 1.5 cm. Write the equation of the motion for the point located at x = 0.75m

Solution

we can write the two waves in the form

y1 = ASin( t-kx) and

y2= ASin( t-kx)

where A= 0.01 m,    =20   and k = /v=20/30 = 2/3

the resultant wave is givne by

y = y1+y2 = ASin( t-kx) + ASin( t-kx)

                  = 2ACos(kx) Sin( t)

                 putting the values for A, and k

y = 0.02 Cos(2x/3)Sin(20t)

The resultant wave amplitude is 0.02 m = 2 cm, for points at x=0, 3n, the amplitude is maximum , the particle at these points move in +y and –y direction with time when Sin(20t) = +1 ot -1, these points are called anti-nodes.

for vaues of x such that Cos(2x/3) = 0, or x= 3(2n+1)/4 the particle will always be at rest and such points are called nodes

B. the amplitude of the wave is

          0.02 Cos(2x/3) = 0.015

                        Cos(2x/3) = 0.75

                                60x   = 41.41

                                    x =0.69 m = 69 cm

the SHM of these particle is fiven by

Sin( t)

at x = 0.75 Cos(2x/3) = 0

   the equation of motion of the particle is given by

y = 0.02 *0*Sin(20t) =0, the particles are always at rest.