Question

An apparatus similar to the one used in lab uses an oscillating motor at one end to vibrate a long rope with frequency f = 40 Hz and amplitude A = 0.25 m. The rope is held at constant tension by hanging a mass on the other end. The rope has mass denstiy μ = 0.02 kg/m, and tension T = 20.48 N. Assume that at t = 0 the end of the rope at x = 0 has zero y-displacement and is moving downward. What is the y-displacement of the piece of rope at x1 = 0.5 m when t = 0? y1 =

Answer 1

Answer:

The displacement in t = 0,  

y (0) = - 0.18 m

Explanation:

Given f = 40 Hz , A = 0.25m , μ = 0.02 kg / m, T = 20.48 N

v = √ T / μ

v = √20.48 N / 0.02 kg /m = 32 m/s

λ = v / f

λ = 32 m/s / 40 Hz = 0.8

K = 2 π / λ

K = 2π / 0.8 = 7.854

φ = X * 360 / λ

φ = 0.5 * 360 / 0.8 = 225 °

Using the model of y' displacement

y (t) = A* sin ( w * t - φ )

When t = 0

y (0) = 0.25 m *sin ( w*(0) - 225 )

y (0) = 0.25 * -0.707

y (0) = - 0.18 m