How Do Oscillations in Spring-Mass Systems and Rotating Rods Work?

[bodensee9]

Hello:

I am wondering if someone can help with the following?

1. 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(init) such that the spring is at its rest length. The eobject is released from y(init) and oscillates up and down, with its lowest position being .01m below y(init). Find (1) frequency of the oscillation and (2) the speed of the hobject when it's .08m below y(init).

Wouldn't the maximum displacemenet be .01m. So wouldn't I have m*g = m*w^2*x, where x is the maximum displacement, w is the angular frequency, and x is the maximum displacement. Then I can solve for w, and use 2*pi*f = w, where f is the frequency? But then I seem to have not taken into account the force from the spring?

2. A long uniform rod of mass 0.6kg is free to rotate in a horizontal plane about a vertical axis through its center. A spring with force constatnt k = 1850 N/m is connected horizontallly between 1 end of the rod and a fixed wall. When the rod is in equilibrium, it is parallel to the wall. What's the period of the small oscillations that result when the rod is rotated slightly and released?

So it's like

------------------------------------wall
|
| (spring)
|
------------rod in equilibrium position (rod moves up and down)

But how does the spring come into it? There seems to be no compression of the spring by the direction of the rod's movements?

Many Thanks.

 

[tiny-tim]

1. … So wouldn't I have m*g = m*w^2*x, where x is the maximum displacement, w is the angular frequency, and x is the maximum displacement. Then I can solve for w, and use 2*pi*f = w, where f is the frequency? But then I seem to have not taken into account the force from the spring?

2. … when the rod is rotated slightly and released
… But how does the spring come into it? There seems to be no compression of the spring by the direction of the rod's movements?

Hi bodensee9!

1. It's probably easier if you use the spring constant k, and Newton's second law with energy = kx²/2.

2. There's no compression in equilibrium. But the question specifically says that the rod is rotated slightly away from equilibrium.