Question

The vertical displacement of a transverse wave on a rope as a function of horizontal displacement and time is given by:

D(x,t)= (0.12m)sin[(3.0m^-1)x-(15.0s^-1)t]

a) calculate the wavelength and frequency

b) what is the vertical displacement and velocity of the point on the rope at t=0.20s and x = 0.6m.

c) the original rope is connected to a thinner rope with smaller mass per unit length. Describe what happens when the transverse wave is incident upon the interface between the two ropes.

Answer 1

(a) As the displacement is given by:

As waveequation has the form:

hence, by comparision,

m

is the required wavelength.

and Frequency,

(b) At t = 0.20 s and x = 0.6m

is the required displacement.

as

hence, for x=0.6m and t = 0.20 s

is the required velocity.

(c) As the wave goes to a thinner rope, its frequency of vibration changes and the transmitted wave has different wave equation. As the new wave has lower mass per unit lenght, its frequency increases, as frequency is: