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 =

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

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we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .