Spring/Mass/Oscillation problem

[loganblacke]

Homework Statement

A mass that can oscillate without friction on a horizontal surface is attached to a horizontal spring that is pulled to the right 12.5cm and is released from rest. The period of oscillation for the mass is 6.60 s. What is the speed of the mass at t = 1.95 s?

Homework Equations

F = kx

x = Acos(2*pi*t/T+\theta₀)

The Attempt at a Solution

\theta₀=cos^-1(X₀/A)

\theta₀=cos^-1(12.5/25)
x = 25*cos(((2\pi(1.95))/6.6)+60) = 12.314352

The correct answer according to webwork is 11.4, but i have no idea where I'm going wrong or if I'm doing anything right for that matter. Please help!

 

[collinsmark]

\theta₀=cos^-1(X₀/A)

\theta₀=cos^-1(12.5/25)

You've plugged in the wrong value for A.

(Hint: The system will oscillate between A and -A. A is the distance from the center of oscillation to its extreme; not the peak-to-peak distance).

x = 25*cos(((2\pi(1.95))/6.6)+60) = 12.314352

Ignoring the mistake mentioned above for the moment, you're supposed to be solving the velocity v. Not the position x. The equation

x = Acos([2π/T]t + θ₀)​

gives you the mass's position, but not its velocity.

If your class is calculus based, you can determine the equation for velocity by noting
v(t) = d/dt{x(t)}.
If you are not in a calculus based class, the velocity formula should have been given as part of the coursework.