# Homework Help: Oscillating spring problem

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 9, 2006

### mukundpa

Angular velocity of what? What is rotating or circulating here?

That is why it is batter to call this quantity as angular frequency.

V = w sqrt[A^2 - x^2]

3. Jan 9, 2006

### Gamma

Using conservation of energy between the points y=y and y=10 cm,
and again between the points y=y and y=8 cm, should yeild the result. This is a less confusing way as far as I am concerned.

There are 3 types of energy to consider in this problem. Those are Grav. potantial energy, Potantial energy stored in the spring and the KE.