A gas spring is a spring mechanism that, rather than using a physical coil to ge

Question

A gas spring is a spring mechanism that, rather than using a physical coil to generate the spring behaviour, uses the properties of a gas to achieve this behaviour. Here is a simplified representation of a gas spring: The force exerted by the gas spring is dictated by the pressure of the gas within it. As the temperature of the gas remains constant, and no gas escapes, the pressure is dictated by Boyle's Law, p_1V_1 = p_2 V_2 = k where p is pressure and V is volume. The force due to the gas is given by F = pA where A is the area of the surface the pressure, and thus the force, applies to (in our case, this area is the cross-sectional area). Note that the air outside the gas spring will also exert a force, but that the pressure for the air will remain constant. As such, the total pressure producing the force is the pressure inside the gas spring minus the air pressure. By considering the forces involved - the force due to pressure in the gas spring and the force due to gravity, as well as air resistance (linear) - show that the motion of mass m is described by md^2x/st^2 + c dx/dt - k/x + k/x_0 = mg Determine the resting length of the spring, x_infinity, when the mass is attached. That is, the length for which all forces balance. Using MATLAB's ode45 function, solve the ODE if m = 0.13, c = 0.27, k = 0.35, 9 = 9.81, and x_0 = 0.1, with x(0) = 0.03 and x'(0) = 0, for 0

Explanation / Answer

a) From FBD
mg + F*x - cv = ma
mg + F - cx' = mx"
now, F = pA
At some displacement x
p*x*A = k ( constant)
so, mg + pA - cx' = mx"
mg + k/x - cx' = mx"
using initial conditions
mx" + cx' - k(1/x - 1/xo) = mg ( where x0 is the initial position of the mass)

b) for all forces balanced, x" = 0
cx' - k(1/x - 1/xo) = mg
cx'= mg - k/xo + k/x
integrating
cx = mgt - kt/xo + kln(x) [ from x = xo to x = xinf]
here xinf is the length required

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