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Springs and Harmonic Motion (Physics)

Date: 5/29/96 at 13:51:41
From: Anonymous
Subject: Physics (springs)

A cart mass 3kg is held between two stretched springs. It is found 
that the cart moves .1m from its equillibrium position when a force of 
60N is applied. The cart is then released and allowed to vibrate.
What is the force constant for the system of springs? Calculate the 
frequency of vibration. What is its velocity as it passes through its 
equillibrium position?

I'm stuck at the beginning. How do I find the force constant; what 
exactly is it?

Date: 5/31/96 at 11:21:7
From: Doctor Anthony
Subject: Re: Physics (springs)

For this question you need to be familiar with simple harmonic motion.  
You will find the definitions and derivation of formulae in any 
appropriate textbook on mechanics.

We use the formula T = kx for the spring system, where k is the 
'stiffness', x is the displacement from equilibrium, and T is the 
force of the system in the direction towards the equilibrium position.

From this formula 60 = k x 0.1, and so k = 600 (= force constant)

The equation of motion when the cart is vibrating about the 
equilibrium position, and is distance x from that position, is

3(d^2(x)/dt^2) = -600x  (Newton's formula:  mass x acceleration = 

   d^2(x)/dt^2 = -200x    (= -w^2.x)

This is the equation of simple harmonic motion.

The period of vibration is given by  2pi/w 
                                = 2pi/sqrt(200)
                                = 2pi/(10sqrt(2))
                                = sqrt(2).pi/10

The speed when x = 0 is found from the expression

  v^2 = w^2(a^2 - x^2)  where a = amplitude = 0.1

 so v^2 = 200(0.01 - 0)

        = 2     and so  v = sqrt(2) m/sec

The answers you require are:  period = sqrt(2).pi/10 secs
                speed through center = sqrt(2) m/sec

-Doctor Anthony,  The Math Forum
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