Consider an asymmetrical nonlinear spring whose force law is given by F=-kx+k2x^
ID: 1916330 • Letter: C
Question
Consider an asymmetrical nonlinear spring whose force law is given by F=-kx+k2x^2+k3x^3. The spring with mass m attached is initially stretched a distance d to the right of the equilibrium position.
a. Describe breifly why this is called an asymmetrical spring.
b. The mass is released from its initial position. Find the speed of the mass when it is at a point 0.25 m to the left of the equilibrium position. Use the following values for the constants,
k=300N/m k2=300 N/m^2 k3 = 200 Nm^3 m = 2.0 kg d = 0.50m
Explanation / Answer
Part a A spring is linear if the force is proportional to the extension or compression – in this case if k2 and k3 were zero If k2 were zero and k3 were positive, then the spring is strong and symmetrical – the force increases at a faster rate than the extension or compression If k2 were zero and k3 negative then the spring is symmetrical and weak, - the force increases at a slower rate than the extension or compression If k3 were zero and k2 were positive, then the spring is asymmetrical – the force increases at a faster rate in extension and a slower rate in compression. Vice versa for k2 negative. Hence it is the fact that there is a square term in the equation that makes the spring asymmetrical. Part b Caution - my integration is rusty, my terminology out of date, and the character set available limited. However I believe the following is correct. F=kx+k2x^2+k3x^3 A=F/m V=A*t Increment in speed dV= dA*dt dV/ dt = dA V=SUM(dA) between limits V=SUM(d(F/m ) between limits V=SUM(d(kx+k2x^2+k3x^3)/m) between limits V=SUM(d (kx/m+k2x^2/m+k3x^3/m)) between limits Integrating V=kx^2/2m+k2x^3/3m+k3x^4/4m between limits Substituting values V= 300*x^2/4+300*x^3/6+200*x^4/8 between the limits .5 to -.25 V= (18.75+6.25+1.5625)-(4.6875-0.78125+.097… V= 22.56 metres per second