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 = force) 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 Check out our web site! http://mathforum.org/dr.math/
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