A mass of a spring vibrates in simple harmonic motion at a frequency of 8.0 Hz and an amplitude of 3.9 cm. If a timer is started with its displacement is a maximum, what is the speed of the mass when the timer reads 3 seconds

 e = 50.27 give / s

Explanation:

The expression for simple harmonic motion is

    x = A cos (wt + Ф)

in this case they give us the amplitude A = 3.9 cm and frequency f = 8.0 Hz

The angular and linФear variables are related

      e = 2π d

      e = 2π 8

      e = 50.27 give / s

let's look for the constant fi

       so let's find the time to have the maximum displacement

       v = dx / dt

       v = -A w sin (wt +Ф)

for the point of maximum displacement the speed is I think

        0 = - sin (0 + Ф)

therefore fi = 0

Let's put together the equation of motion

          x = 0.039 sin (50.27 t)

          v = 0.039 50.27 sin (50.27 3)

           v = 1.96 50 0.01355

            v = 0.0266 m / s

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 .