Spring problem: compressed and pushed down with a velocity

[ayjakk]

Homework Statement

I'm trying to come up with a problem and solve it. As of right now, I have a vertical spring with an equilibrium length of .38 m hanging. A .61 kg mass is attached to the bottom, and the new equilibrium length is 1.05 m.

If the spring is compressed .1 m upwards then pushed down with a velocity of 4.06 m/s, how would you write position as a function of time, as well as velocity and acceleration?

Homework Equations

Force constant: k = F/x = (mg)/x = (.61 kg * 9.81 m/s^1)/(1.05 m - .38 m) = 8.931 N/M

Period: T = 2pi(m/k)^(1/2)

T = tpi(0.61kg/8.931 N/M)^(1/2) = 1.642 s^-1​

w = 2pif
f = 1/T
x(t) = Acos(wt)

The Attempt at a Solution

I solved for x(t) without any initial velocity. I came up with w = 3.826 s^-1, and an equation of x(t) = -.1cos(3.826t)
This may be a dumb question, but would adding the initial velocity just be x(t) = Acos(wt) + vt, where v = 4.06 m/s? What about units? Would it contribute towards angular velocity? Is the -.1 the correct sign for being pushed upwards?

I feel confident taking derivatives, so velocity and acceleration I have no problem with getting. It's just the position function I'd like to clarify first.

 

[Delta2]

You have made a slight mistake, the general solution for the free harmonic oscillator is x(t)=Acos(wt+\phi). The first derivative will be the velocity v(t). From initial conditions x(0), v(0) you can determine A and \phi. No the initial velocity v(0) doesn't affect w.

 

[Doc Al]

I solved for x(t) without any initial velocity. I came up with w = 3.826 s^-1, and an equation of x(t) = -.1cos(3.826t)

OK, except for that minus sign.

This may be a dumb question, but would adding the initial velocity just be x(t) = Acos(wt) + vt, where v = 4.06 m/s? What about units? Would it contribute towards angular velocity?

The correct approach is outlined by Delta². Start with the basic equation and solve for the new amplitude and phase angle.

Is the -.1 the correct sign for being pushed upwards?

I would choose up as positive, making the initial position +.1. (But the initial velocity is down, and thus negative.)

 

[BvU]

Re relevant equations: that's relevant equations, not to be mixed up with things you bring up in attempt at solution.

Re period: that has the dimension of time, not of time^-1

Re solution without initial velocity: define a coordinate system first. If you let the x-axis point downwards and place the origin at the steady state equilibrium point (0.95 m spring length), then yes, -0.1 is the starting postion.

Living upside down is tiring. I advise strongly to let the position axis point upwards. Your choice.

Re solution without initial velocity: $x(t) = -0.1\; \cos(\omega t)$ is correct. You get the velocity as a function of time by differentiating.

Re solution with initial velocity: you go back to the general solution $x(t) = A\; \cos(\omega t + B)$ and fill in the initial conditions for $x(0)$ and ${dx\over dt}(0)$.

Re

would adding the initial velocity just be x(t) = Acos(wt) + vt

You can not do this: x = vt only applies when $a = {d^2x\over dt^2} = 0$ all the time. That is definitely not the case here !