1. Second Order Homogeneous Linear Differential Equations With Constant Coefficients Part 2 Characteristic Equations with Purely Imaginary Roots Simple Harmonic Motion

2. In this part we consider those 2nd order, homogeneous, linear differential equataions with constant coefficients whose characteristic equations have purely imaginary roots.

3. Consider a mass suspended at the end of a spring. The mass stretches the spring, hanging in equilibrium, with the weight force of the object balanced by the restoring force provided by the stretched spring. If the extension is not too great, it is found that the restoring force is proportional to the extension. Unstretched length l X Positive direction downwards

4. Consider a mass suspended at the end of a spring. The mass stretches the spring, hanging in equilibrium, with the weight force of the object balanced by the restoring force provided by the stretched spring. If the extension is not too great, it is found that the restoring force is proportional to the extension. Unstretched length Extension due to mass l a

5. Consider a mass suspended at the end of a spring. The mass stretches the spring, hanging in equilibrium, with the weight force of the object balanced by the restoring force provided by the stretched spring. If the extension is not too great, it is found that the restoring force is proportional to the extension. Unstretched length Extension due to mass l a Restoring Force -ka Weight Force mg

6. Consider a mass suspended at the end of a spring. The mass stretches the spring, hanging in equilibrium, with the weight force of the object balanced by the restoring force provided by the stretched spring. If the extension is not too great, it is found that the restoring force is proportional to the extension. If the system is in equilibrium Unstretched length Extension due to mass l a Restoring Force -ka Weight Force mg

7. The mass is then pulled down a further distance x and released Equilibrium stretched length Extension l+a x Restoring Force -kx The effect of the weight stretching the spring the initial amount, a, is to cancel out the weight force so the extra extension causes an unbalanced restoring force of -kx to act on the mass which thus experiences an acceleration

8. Technical result can be written in the form kt sin)sin(,cos)cos( 

9. Technical result can be written in the form This means that, if we have purely imaginary conjugate root to a characteristic equation the general solution can be written in trigonometric instead of exponential form

10. Example 1. Write down and solve characteristic equation 2. Write down the general solution (in Acos + B sin form) 3. Find the constants There are 3 steps: Find the solution of

11. What do initial conditions mean if the equation describes the oscillating weight on gthe end of the string? Mass is initially pulled down 1m below the equilibrium point No initial velocity

12. 1. Write down and solve characteristic equation 2. Write down the general solution in trig form 3. Find the constants Hence the solution is

13. Example 1. Write down and solve characteristic equation 2. Write down the general solution (in Acos + B sin form) 3. Find the constants There are 3 steps: Find the solution of

14. What do initial conditions mean if the equation describes the oscillating weight on the end of the string? Mass is initially at the equilibrium point Initial velocity of 2ms -1 upwards

15. 1. Write down and solve characteristic equation 2. Write down the general solution in trig form 3. Find the constants Hence the solution is

16. Second Order Homogeneous Linear Differential Equations With Constant Coefficients General Notes

17. This year you are expected to be able to solve these equations in the following two cases: (a) 2 real distinct roots and (b) purely imaginary roots The methods of solution only differ in the form of the general solutions (a) x = C 1 e kt + C 2 e -kt (b) x = Acoskt + Bsinkt Summary