This Assignment has to be done with MATLAB
MATH 2500 Group Project: Forced Vibrations Due: Thursday November 12, 2020 1. Consider our model for the displacement of a mass m, attached to a spring with stiffness k, acted on by a driving force f(t) = F0 cos(ωt), my′′ + dy′ + ky = F0 cos(ωt). (1) If we assume that there is no damping in the system (i.e d = 0) then we reduce the above equation to simply y′′ + ω20y = F0 m cos(ωt) (2) where ω0 = √ k m . 1a. Show that this equation has a general solution of the form y(t) = c1 cos(ω0t) + c2 sin(ω0t) + F0 m(ω20 − ω2) cos(ωt). (3) The solution is the sum of two periodic motions with different frequencies ω 6= ω0. We call the first two terms the transient solution yT = c1 cos(ω0t) + c2 sin(ω0t), which depends on the initial conditions of the system. In the case where damping occurs, the transient solution always satisfies the following limit lim t→∞ yT (t)→ 0. Thus the motion of the mass governed by yT eventually will die out. When there is no damping, then lim t→∞ yT 6= 0 and the motion governed by yT is purely oscillatory and always remains present as time evolves. The last term we call the steady state solution yss(t) = F0 m(ω20 − ω2) cos(ωt) which depends solely on the forcing function f(t). This solution satisfies the limit lim t→∞ y(t) → yss(t) with the assumption that d > 0. Eventually the mo- tion of the mass should reach its steady state. 2. Assume the mass in equation (2) is initially at rest and has no initial displace- ment. That is, y(0) = y′(0) = 0. 2a. Show that the solution in (3) becomes y(t) = F0 m(ω20 − ω2) ( cos(ωt)− cos(ω0t) ) . (4) 2b. Next use the trigonometric identity 2 sin θ sinϕ = cos(θ−ϕ)− cos(θ+ϕ) to convert (4) to y(t) = ( 2F0 m(ω20 − ω2) sin (ω0 − ω)t 2 ) sin (ω0 + ω)t 2 . (5) In Equation (5), if we let the circular frequency of the driving force be ap- proximately equal to the natural circular frequency of the mass-spring system, that is ω ≈ ω0, we encounter a phenomenon called beats. The factor sin ( (ω0 + ω)t 2 ) oscillates much faster than sin ( (ω0 − ω)t 2 ) , so Equation (5) describes a rapidly oscillating function with circular frequency (ω0 + ω) 2 oscillating inside a slower sinusoidal function. The faster oscillating function has Sinusoidal Amplitude = ( 2F0 m(ω20 − ω2) ) sin ( (ω0 − ω)t 2 ) (6) and period 4π ω0 + ω . When beats occur, we say that the faster oscillating wave is amplitude modu- lated by the slower varying wave. The slower wave has amplitude 2F0 m|ω20 − ω2| 2 and period 4π |ω0 − ω| . 2c. Create a script file in MATLAB that will generate a single graph containing the curve described in Equation (5) and also the curves y(t) = ± 2F0 m(ω20 − ω2) sin (ω0 − ω)t 2 using the values ω = 1, ω0 = 1.1, F0 = 0.5 and m = 0.3. Be sure to include one full period of the slower varying wave and also label your axes and title the graph. Your result should resemble the image in Figure 1. Interpret the plot you produced. Figure 1: Plot of beats. 3. When the external circular frequency ω is tuned to exactly match the natural circular frequency ω0 a phenomenon called pure resonance occurs. 3a Show that the differential equation y′′ + ω20y = F0 m cos(ω0t) (7) has general solution y(t) = c1 cos(ω0t) + c2 sin(ω0t) + F0 2mω0 t sin(ω0t). (8) 3 Because there is no damping present in the system it is assumed that the solution in (8) is unbounded as t→∞ and its amplitude grows linearly. The steady-state portion of the solution continually oscillates between the lines y(t) = ± F0 2mω0 t (9) with circular frequency ω0 and period 2π w0 . 3b. Create a graph in MATLAB that contains the steady state solution yss(t) = F0 2mω0 t sin(ω0t) and the lines described in (9) using the same values as before: ω0 = 1.1, m = 0.3 and F0 = 0.5. Make sure your graph contains four full periods of yss and is completely labeled. If done correctly, you should anticipate a result like the one in Figure 2. Interpret the plot you produced. Figure 2: Steady state solution. 4. The notion of pure resonance is easy to understand both mathematically and physically but is impossible to achieve in the physical world. We all know that the value d > 0 in every real mass-spring system. It can be shown that the 4 solution to the differential equation described in (1) is y(t) = yT (t) + F0√ m2(ω20 − ω2)2 + d2ω2 cos(ωt− δ) (10) where the transient solution yT (t) will depend on the roots of the character- istic equation mr2+dr+k = 0 and δ = arctan ( c2 c1 ) is called the phase angle. The steady state portion of this solution has the same circular frequency as the forcing function f(t) = F0 cos(ωt) but is out of phase by an angle of δ radians. Its amplitude is changed by the coefficient M(ω) = 1√ m2(ω20 − ω2)2 + d2ω2 (11) which we call the amplification factor. The graph of M(ω) vs. ω is called the frequency response curve and represents the gain in amplitude of the steady state response as a function of the frequency of the forcing term. 4a. Create a graph of M(ω) in MATLAB for the damping values d = 1 4 , 1 2 , 1, 3 2 , and 2. Use the values m = k = ω0 = 1. You should fit your window to [0, 2] × [0, 4] to accurately capture what is happening. See Figure 3 to get an idea of what the frequency response curves should look like. Practical resonance occurs when the external circular frequency ω has been tuned to maximize M(ω), i.e to produce the largest possible steady state am- plitude. 4b. Differentiate the amplification factor with respect to frequency and set this equal to zero, M ′(ω) = 0, to show that the frequency at which this maximum occurs is: ωd = √ k m − d 2 2m2 (12) so long as k m − d 2 2m2 > 0. If this condition is not met, then there are no values of ω ∈ (0,∞) that will generate a maximum for M(ω), meaning that we cannot generate resonance- like features in the damped mass-spring system. 5 4c. Using this information, on the same graph as above, plot the points (wd,M(ωd)) which are defined for ωd ∈ R. These points represent the maximum gain for each damping factor d. Notice as d → 0, ωd → ω0 and we end up back in the pure resonance case. Interpret the plot you produced. Figure 3: Response frequency. 5. There are several applications of resonance and beats in real life. 5a. Do some research and find an application of resonance and/or beats. Describe a physical phenomenon in which resonance and/or beats are observed. Use as many terms as you can from the list to explain the phenomenon mathematically. • Transient solution • Steady state solution • Beats • Pure resonance • Phase angle • Practical resonance 6