How Long Until Mass Reaches 0.05m in Oscillatory Motion?

[scan]

ok here's the problem:

When a mass of 0.2kg is suspended from a spring, it stretches 0.04m. The mass is pulled down an additional distance 0.1m from its equilibrium position and released.

Question:

how long after being released is the posiiton of the mass equal to 0.05m

 

[Tide]

Use the first statement to find the spring constant. Divide it by the mass and find its square root. That gives you the angular frequency and you should be able to take it from there.

 

[scan]

Use the first statement to find the spring constant. Divide it by the mass and find its square root. That give you the angular frequency and you should be able to take it from there.

i've found all of those. there are several parts to this question. I've done them up to this question which I'm stuck on.

the constant k=49
period T= .4
frequency f= 2.5

 

[HallsofIvy]

If you have those then you should know that, taking x to be the distance from the equilibrium position, positive above, negative below, x= -0.1 cos(2pi t/.4). To answer the question "how long after being released is the posiiton of the mass equal to 0.05m", set x= 0.05 and solve for t.

By the way, do you understand that saying "period T= .4 frequency f= 2.5" doesn't mean anything unless you give the units?