%% Real Shock to the System: Mtn Biker Problem:

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3 d) Figure 1. Comparison of mountain biker oscillations. e) We see that both Hotdogger and Smallfry undergo underdamped oscillations. Hotdogger s max amplitude of oscillation is larger this is to be expected since Hotdogger has a larger mass. Also, we see the longer period (lower angular) frequency of oscillation inherent with a larger mass (keeping spring and damping the same for both riders). Smallfry s oscillations decay faster. This makes sense each oscillation saps energy from the system; that s what the damper does. Smallfry undergoes more oscillations per second, so the energy is dissipated faster overall. f) The oscillations above are a bit much, but not too dramatic oscillations are basically complete within about 0.5 s. Perhaps the riders could increase their damping factor a bit if they want cushion, but less oscillation. This increases the damping factor (zeta), so they ll get closer to critically damped. Perhaps increase the damping constant such that zeta becomes ~ This is a gut feel recommendation having saddled up quite a bit myself (with ZERO experience hucking off crazy cliffs like that). %% Real Shock to the System: Mtn Biker Problem: % Define physical parameters m_sf = 50; %kg, for Smallfry m_hd = 100; %kg, for Hotdogger m_bike = 10; %kg, mass of bike % total mass for small fry. % could easily update this for hotdogger, as we do below m = m_sf + m_bike;

4 k = ; %N/m c = 1500; %Ns/m wo = sqrt(k/m); %natural freq of oscillation, rad/s zeta = c/(2*m*wo); %damping ratio wd = wo*sqrt(1-zeta^2); %damped frequency of oscillation % more physical parameters for our system g = 9.81; %m/s^2 good ole Earth gravity h = 2; %m, cliff height, big drop off from here % initial conditions zo = m*g/k; % The Laws Smith Special: our spring is uncompressed, % and we are keeping track of oscillations % about static equilibrium position. vo = -sqrt(2*g*h); %The Simon Marland Special: make sure that negative sign is in there % keeping track of z positive upward, but clearly we are moving downward % upon impact D1 = zo; D2 = (vo + zeta*wo*zo)/wd; %% Now let's plot for small fry! figure; % plots z_smallfry vs. time % >> doc fplot to learn more % Basically this plots the desired function vs. time, % Time bounds are [0 1] second % Making lines 2x thick for pretty display fplot(@(t) exp(-zeta*wo*t).*(d1*cos(wd*t) + D2*sin(wd*t)),[0 1],'Linewidth',2); %% Now let's work on Hotdogger. m changes, so we have to recompute w_o, zeta, etc m = m_hd + m_bike; %hotdogger's mass + the bike wo = sqrt(k/m); %natural freq of oscillation, rad/s zeta = c/(2*m*wo); %damping ratio wd = wo*sqrt(1-zeta^2); %damped frequency of oscillation % more physical parameters for our system g = 9.81; %m/s^2 good ole Earth gravity h = 2; %m, cliff height, big drop off from here

5 % initial conditions zo = m*g/k; % The Laws Smith Special: our spring is uncompressed, % and we are keeping track of oscillations % about static equilibrium position. vo = -sqrt(2*g*h); %The Simon Marland Special: make sure that negative sign is in there % keeping track of z positive upward, but clearly we are moving downward % upon impact D1 = zo; D2 = (vo + zeta*wo*zo)/wd; %now add hotdogger's position vs. time to the plot hold on; %add to the previously opened figure %plots z_hotdogger vs. time. Notice all the constants & parameters were %recomputed above. fplot(@(t) exp(-zeta*wo*t).*(d1*cos(wd*t) + D2*sin(wd*t)),[0 1],'Linewidth',2); %make the graph pretty, add labels legend('smallfry', 'Hotdogger') xlabel('time (s)') ylabel('vertical position z(t) (meters) ') title('mountain biker cliff jump') grid on; set(gca, 'fontsize', 14)

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7 Figure 1. Forced solution only. Amplitude is fairly modest at ~ 2 mm. This is favorable for the biker---one does NOT want to feel bouncing just riding on the straight away. Figure 2. Total solution = forced + unforced. We see a huge transient deflection jumping off cliff followed by much smller forced oscillation. %% Pedal Faster; I hear banjos Solution. % This solution sums. the unforced (homogenous) and forced soltuions %to get the total response.

8 %% Define physical parameters k = ; %N/m c = 1500; %Ns/m m_sf = 50; %kg, for Smallfry m_hd = 100; %kg, for Hotdogger m_bike = 10; %kg, mass of bike % more physical parameters for our system g = 9.81; %m/s^2 good ole Earth gravity h = 5; %m, cliff height, big drop off from here %% 1. Compute unforced solution for SmallFry % total mass for small fry. % could easily update this for hotdogger, as we do below m = m_sf + m_bike; wo = sqrt(k/m); %natural freq of oscillation, rad/s zeta = c/(2*m*wo); %damping ratio wd = wo*sqrt(1-zeta^2); %damped frequency of oscillation % initial conditions zo = m*g/k; % The Laws Smith Special: our spring is uncompressed, % and we are keeping track of oscillations % about static equilibrium position. vo = -sqrt(2*g*h); %The Simon Marland Special: make sure that negative sign is in there % keeping track of z positive upward, but clearly we are moving downward % upon impact D1 = zo; D2 = (vo + zeta*wo*zo)/wd; %unforced solution is of form: % exp(-zeta*wo*t).*(d1*cos(wd*t) + D2*sin(wd*t)) %% 2. Compute forced solution for smallfry. This solution comes from Earlier Bat Mobile Problem % on Complex Numbers problem set (Prob Set 5; Number 2) %given constants Fo = 500; %N force w = 10; %rad/s angular frequency r = w/wo; %ratio of forcing frequency to natural frequency xo = (Fo/k)/(sqrt((1- r^2)^2 + (2*zeta*r)^2)); %amplitude of forced oscillations phi_x = -atan(2*zeta*r/(1-r^2));

9 % Total response is: % sn xo*cos(w*t + phi_x); %% 3. Now let's plot for small fry! figure; % plots z_smallfry vs. time % >> doc fplot to learn more % Basically this plots the desired function vs. time, % Time bounds are [0 1] second % Making lines 2x thick for pretty display fplot(@(t) exp(-zeta*wo*t).*(d1*cos(wd*t) + D2*sin(wd*t)) + xo*cos(w*t + phi_x), [0 3],'Linewidth',2); %% 4. Now let's work on Hotdogger. m changes, so we have to recompute w_o, zeta, etc m = m_hd + m_bike; %hotdogger's mass + the bike wo = sqrt(k/m); %natural freq of oscillation, rad/s zeta = c/(2*m*wo); %damping ratio wd = wo*sqrt(1-zeta^2); %damped frequency of oscillation % initial conditions zo = m*g/k; % The Laws Smith Special: our spring is uncompressed, % and we are keeping track of oscillations % about static equilibrium position. vo = -sqrt(2*g*h); %The Simon Marland Special: make sure that negative sign is in there % keeping track of z positive upward, but clearly we are moving downward % upon impact D1 = zo; D2 = (vo + zeta*wo*zo)/wd; %% 5. Compute forced solution for Hotdogger. This solution comes from Earlier Bat Mobile Problem % on Complex Numbers problem set (Prob Set 5; Number 2) %given constants Fo = 500; %N force w = 10; %rad/s angular frequency r = w/wo; %ratio of forcing frequency to natural frequency xo = (Fo/k)/(sqrt((1- r^2)^2 + (2*zeta*r)^2)); %amplitude of forced oscillations phi_x = -atan(2*zeta*r/(1-r^2)); %% 6. now add hotdogger's position vs. time to the plot hold on; %add to the previously opened figure

10 %plots z_hotdogger vs. time. Notice all the constants & parameters were %recomputed above. exp(-zeta*wo*t).*(d1*cos(wd*t) + D2*sin(wd*t)) + xo*cos(w*t + phi_x), [0 3],'Linewidth',2); %% make the graph pretty, add labels legend('smallfry', 'Hotdogger') xlabel('time (s)') ylabel('vertical position x(t) (meters) ') title('forced Solutions') grid on; set(gca, 'fontsize', 14) set(gcf, 'color', 'w')

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14 Figure 1. Transient oscillations. Capacitor discharges; resistor dissipates energy over time, so oscillations die down Figure 2. Unforced solution dominates orignally with 5V charged up capacitor. Long term steady-state oscillations dominate with ~ 2V pk-pk oscillations % RLC oscillations problem R = 50; %Ohm C = 100e-12; % F L = 1e-6; %H wo = sqrt(1/(l*c)); %natural frequency of oscillation zeta = R/2*sqrt(C/L) %damping factor wd = wo*sqrt(1-zeta^2); %damped frequency of oscillation Td = 2*pi/wd; %damped period of oscillations, use this to set time bounds below in fplot() Tau = 1/(zeta*wo); %decay time tau %initial conditions Vo = 5;

15 qo = C*Vo; D1 = qo; D2 = zeta*qo/sqrt(1-zeta^2); fplot(@(t) (1/C)*exp(-zeta*wo*t).*(D1*cos(wd*t) + D2*sin(wd*t)),[0 5*Td],'Linewidth',2); %make the graph pretty, add labels xlabel('time (s)') ylabel(' V_c(t) (Volts) ') title('rlc Circuits Oscillations') grid on; set(gca, 'fontsize', 14) set(gcf, 'color', 'w') %% Now do forced solution Vin = 0.5; w = wo; Q = Vin/sqrt( (1/C - w^2*l)^2 + (w*r)^2 ) ; phi_q = -atan( (w*r)/(1/c - w^2*l)); %% plot total response figure; fplot(@(t) 1/C*(Q*cos(w*t + phi_q) +exp(-zeta*wo*t).*(d1*cos(wd*t) + D2*sin(wd*t))),[0 5*Td],'Linewidth',2); xlabel('time (s)') ylabel(' V_c(t) (Volts) ') title('rlc Circuits Oscillations') grid on; set(gca, 'fontsize', 14) set(gcf, 'color', 'w')

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