# Oscillation of object on spring - Find speed

1. Jan 8, 2006

### apchemstudent

A massless spring hangs from the ceiling with a small object attached to its lower end. the object is initially held at rest in a position y such that the spring is at its rest length(not stretched). The object is then released from y and oscillates up and down, with its lowest position being 10 cm below y. What is the speed of the object when it is 8cm below the initial position?

The book says 56 cm/s, but I don't agree with it. First of all, the amplitude will be 5 cm, which, we can use to figure out the angular velocity.

K(.05) = Fg = mg

K = 196m

angular velocity = sqrt(K/m)

So when the object reaches the point, 8cm from y, it would've displaced only 3 cm from the "equilibrium point".

so V = angular velocity * displacement
= 14 * 3 = 42 cm/s

Am I correct here? Or is there something I'm missing? Thanks.

2. Jan 8, 2006

### Tide

Yes, you are missing something. The object is not undergoing simple circular motion.

Even though you can solve for the correct motion I suggest using energy conservation instead.