Chapter 3 Simple Harmonic Motion 3 1 Simple

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Chapter 3 Simple Harmonic Motion

Chapter 3 Simple Harmonic Motion

3. 1 Simple Harmonic Motion Force on the pendulum Equation of motion (Eo. M)

3. 1 Simple Harmonic Motion Force on the pendulum Equation of motion (Eo. M) Analytical solution: constants determined by initial conditions. The period of the SHO is given by

Euler method to solve the Eo. M Run t for a few period (T)

Euler method to solve the Eo. M Run t for a few period (T) to see the oscillatory behavior Set Dt, q 0, w 0, l, g, iterate ti+1=ti + Dt sample code 3. 1. 1

Total energy of SHO You may use the Mathematica command Series[Cos[x], {x, 0, 2}]

Total energy of SHO You may use the Mathematica command Series[Cos[x], {x, 0, 2}] to obtain the Taylor series expansion of cosq at q = 0. This is amount to making the assumption that q 0.

Euler method fails in this case The oscillation amplitude increases with time Euler method

Euler method fails in this case The oscillation amplitude increases with time Euler method is not suitable for oscillatory motion (but may be good for projectile motion which is nonperiodic) The peculiar effect of the numerical solution can also be investigated from the energy of the oscillator when Euler method is applied: E ≈ (1/2)m l 2 [w 2 + (g/l)q 2] (with q 0 ) Descretise the equation by applying Euler method, Ei+1 = Ei + (1/2)mgl 2[wi 2 + (g/l)qi 2](Dt)2 E increases with i this is an undesirable result.

Euler vs. Euler-Cromer method In the Euler method: wi+1 is calculated based on previous

Euler vs. Euler-Cromer method In the Euler method: wi+1 is calculated based on previous values wi and qi So is qi+1 is calculated based on previous values wi and qi In the Euler-Cromer method: wi+1 is calculated based on previous values wi and qi but qi+1 is calculated based on present values wi+1 and previous value qi sample code 3. 2. 2

Why Euler-Cromer works better? Because it conserver energy over each complete period of motion.

Why Euler-Cromer works better? Because it conserver energy over each complete period of motion.

3. 2 Forced SHM Drag force on a moving object, fd = - kv

3. 2 Forced SHM Drag force on a moving object, fd = - kv For a pendulum, instantaneous velocity v = wl = l (dq/dt) Hence, fd = - kl (dq/dt). l The net force on the forced pendulum along the tangential direction - kl (dq/dt). fd r w E. o. M

Analytical solution Underdamped regime (small damping). Still oscillate, but amplitude decay slowly over many

Analytical solution Underdamped regime (small damping). Still oscillate, but amplitude decay slowly over many period before dying totally. Overdamped regime (very large damping), decay slowly over several period before dying totally. q is dominated by exponential term. Critically damped regime, intermediate between under- and overdamping case.

Cromer-Euler descretisation sample code 3. 2. 3

Cromer-Euler descretisation sample code 3. 2. 3

Numerical solution of the damped oscillator for q = 10, 5, 1, l=1. 0

Numerical solution of the damped oscillator for q = 10, 5, 1, l=1. 0 m This is generated with NDSolve sample code 3. 2. 4

Adding driving force - kl (dq/dt) + FD sin(WDt) WD frequency of the applied

Adding driving force - kl (dq/dt) + FD sin(WDt) WD frequency of the applied force

Analytical solution Resonance happens when

Analytical solution Resonance happens when

Euler-Kromer descretisation sample code 3. 2. 5

Euler-Kromer descretisation sample code 3. 2. 5

Second order Runge-Kutta method Consider a generic second order differential equation. It can be

Second order Runge-Kutta method Consider a generic second order differential equation. It can be numerically solved using second order Runge-Kutta method. First, split the second order DE into two first order parts:

Algorithm Set initial conditions: Calculate

Algorithm Set initial conditions: Calculate

RK 2 for SHO

RK 2 for SHO

Use RK 2 to calculate the total energy of SHO We have seen previously

Use RK 2 to calculate the total energy of SHO We have seen previously that energy as calculated using Euler method is unstable. Now try to calculate E again, using the values of wi and qi as calculated using RK 2. Ei+1 You will have to do this in the HA 3.