The frequency of oscillation of a mass attached to a spring is 3.0 Hz. At time t = 0, the mass has an initial displacement of 0.20 m and an initial velocity of 4.0 m/s. What is the position of the mass as a function of time?
we are given the amplitude, I found the angular frequency this way:
$$f = 3$$ $$T = \frac{1}{3}$$ $$\omega = \frac{2\pi}{1/3} = 6 \pi$$ the formula for the moment is this $$x = 0.20 \cos(6\pi t)$$ I am missing the phase shift and also if a I take the derivate of x at time t is not 4 how can complete this eq?
update the answer was using the formula that express x in terms of his initial vel and position
$$0.20 cos(6\pi t )+ \frac{4}{6\pi}sin(6\pi t)$$
If you're just looking for an answer, then an answer is $$x(t)=\sqrt{x_0^2 + \frac{v_0^2}{\omega^2}}\cos\left(\omega t-\arctan\left(\frac{v_0}{\omega x_0}\right)\right)$$ where $x(0)=x_0$ and $\dot{x}(0)=v_0$ and the frequency $\omega=\sqrt{k/m}$ is the natural frequency of the mass-spring system. Let me know if you want me to show how to derive the equation.