Vibrations and Waves MP205, Assignment 4 Solutions

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1 Vibrations and Waves MP205, Assignment Solutions 1. Verify that x = Ae αt cos ωt is a possible solution of the equation and find α and ω in terms of γ and ω 0. [20] dt 2 + γ dx dt + ω2 0x = 0, Given x = Ae αt cos ωt we differentiate it using the product rule to find its first and second derivatives [20] dx dt = ωae αt sin(ωt) αae αt cos(ωt) dt 2 = ω2 Ae αt cos(ωt) + αωae αt sin(ωt) + αωae αt sin(ωt) + α 2 Ae αt cos(ωt) = (α 2 ω 2 )Ae αt cos(ωt) + 2αωAe αt sin(ωt) To show its a solution we sub these derivatives back into the original equation to obtain (α 2 ω 2 )Ae αt cos(ωt) + 2αωAe αt sin(ωt) γωae αt sin(ωt) γαae αt cos(ωt) + ω 2 0Ae αt cos(ωt) = 0 (α 2 ω 2 γα + ω 2 0)Ae αt cos(ωt) + (2αω γω)ae αt sin(ωt) = 0 For this to be true for all values of t the coefficients of the cosine and sine functions must be 0. This means we must have that: Looking first at (2): And using this in (1): α 2 ω 2 + ω 2 0 γα = 0 (1) 2αω γω = 0 (2) 2αω γω = 0 α = γ 2 α 2 ω 2 + ω 2 0 γα = 0 γ 2 ω2 + ω 2 0 γ γ 2 = 0 ω 2 + ω 2 0 γ2 = 0 ω 2 = ω0 2 γ2 ω = ω0 2 γ2 So x = Ae αt cos ωt is a solution when α = γ 2 and ω = ω0 2 γ2. * 2. An object of mass 0.2 kg is hung from a spring whose spring constant is 80 N/m. The object is subject to a resistive force given by bv, where v is its velocity in meters per second. (a) Set up the differential equation of motion for free oscillations of the system.

2 (b) If the damped frequency is 3/2 of the undamped frequency, what is the value of the constant b? (c) What is the Q of the system? (d) After 2 seconds, what does the average energy decay to? average energy) (in terms of the initial (a) We have a damped oscillator where the damping term is bv. So the sum of the forces is F = kx bv. By Newton s second law the sum of the forces must equal ma so we have ma = kx bv. ma + bv + kx = 0 m d2 x dt 2 + bdx dt + kx = 0 dt 2 + b dx m dt + k m x = 0 dt 2 + γ dx dt + ω2 0x = 0 where γ = b/m and ω 0 = k/m. Using our values for m and k from the question we can write: γ = b m = b 0.2 k 80 ω 0 = m = 0.2 = 00 = 20 (b) ω 0 is the undamped angular frequency and ω is the damped angular frequency. We have the relation ω = 3/2ω 0. From the previous question we know that ω 2 = ω 2 0 γ2 /. Therefore 3 ω2 0 = ω 2 0 γ 2 / (c) Q is given by the ratio of the constants ω 0 /γ. ω 0 = γ = b m b = ω 0 m = 20(0.2) = (d) Q = ω 0 γ = = 1 Ē(t) = γt γ = ω 0 = 20 Ē(2) = (20)(2) = Ē 0 3. A pendulum on which an object of mass 15g swings with natural angular frequency ω 0 = 20rads 1. The object is subject to a resistive force given by 0.5v, where v is its velocity in meters per second.

3 (a) Set up, with explanations, the differential equation of motion for free oscillations of the system. [10] (b) Find the damped frequency.[10] (c) What is the Q of the system? [5] (a) The force on the mass is given by: F = bv kx where bv is the resistive force, and kx is the force due to SHM. Using F = ma we can rewrite this as: We know ω 0 = k m : ma + 0.8v + kx = 0 m d2 x dt dx dt dx dt + dt + kx = 0 k x = 0 dt dx dt + ω2 0x = 0 dt dx dt + 00x = 0 (b) ω = ω0 2 γ2 = = = 175 = (c) Q = ω 0 γ = = *. The motion of a linear oscillator may be represented by means of a graph in which x is shown on the x-axis and dx/dt on the y-axis. (ie a point on the graph will be of the form (x, dx /dt)) The history of the oscillator is then a curve. (a) Show that for an undamped oscillator this curve is an ellipse. (b) Show (at least qualitatively) that if a damping term is introduced one gets a curve spiraling into the origin. (a) For an undamped oscillator we have x = A cos(ωt + α). x = A cos(ωt + α) dx = ωa sin(ωt + α) dt

4 We want to show that x = A cos(ωt+α) and y = dx dt = ωa sin(ωt+α) satisfy an ellipse equation. Ellipse equation is given by: ( x 2 ( a) + y ) 2 b = 1 In our case x = A cos(ωt + α) and y = dx dt = ωa sin(ωt + α) Using: We can write: as required. sin 2 (x) + cos 2 (x) = 1 x 2 y2 A 2 + ω 2 A 2 = 1 ( x ) 2 ( y ) 2 + = 1 A ωa the curve is an ellipse x 2 = A 2 cos 2 (ωt + α) x 2 A 2 = cos2 (ωt + α) y 2 = ω 2 A 2 sin 2 (ωt + α) y 2 ω 2 A 2 = sin2 (ωt + α) (b) If we think of our ellipse in (a) in terms of a pendulum: If we release the pendulum from the point 1, it s velocity is 0 and it has maximum (positive) displacement When the pendulum goes through 2 it has maximum (negative, as it is going in the negative direction) velocity and its displacement is 0 When the pendulum reaches 3 it has maximum (negative) displacement, and 0 velocity Finally, when the pendulum returns through it will have maximum (positive) velocity, and again 0 displacement We can see this corresponds to the points on the ellipse to the right

5 If we release the pendulum from the point 1, it s velocity is 0 and it has maximum (positive) displacement, as before. When the pendulum goes through 2 it has its maximum (negative, as it is going in the negative direction) velocity for that swing (although this is less than the maximum velocity before due to damping) and its displacement is 0 When the pendulum reaches 3 it has its maximum (negative) displacement for this swing, but due to damping this displacement is not as great as the original maximum displacement, and 0 velocity When the pendulum returns through it will have its maximum (positive) velocity for this swing, which is less than the velocity when it went through it on 2, and again 0 displacement Continuing in this vein we can see the graph that corresponds to this is a curve spiralling into the origin. 5. Many oscillatory systems, although the loss or dissipation mechanism is not analogous to viscous damping, show an exponential decrease in their stored average energy with time Ē = γt. A Q for such oscillators may be defined using the definition Q = ω 0 γ, where ω 0 is the natural angular frequency. (a) When the note middle C on the piano is struck, its energy of oscillation decreases to one half its initial value in about 1 sec. The frequency of middle C is 256 Hz. What is the Q of the system? [10] (b) If the note an octave higher (512 Hz) takes about the same time for its energy to decay, what is its Q? [10] (c) A free, damped harmonic oscillator, consisting of a mass m = 0.1kg moving in a viscous liquid of damped coefficient b (F viscous = bv), and attached to a spring of spring constant k = 0.9Nm 1, is observed as it performs oscillatory motion. Its average energy decays to 1 e of its initial value in sec. What is the Q of the oscillator? What is the value of b? [10] (a) We know that Ē = γt. Using this: Ē = γt

6 after 1 second; t = 1 we have: Ē 0 γ(1) = = e γ ( ) 1 ln = γ 2 ln ( 2 1) = γ ln (2) = γ γ = ln(2) = 0.69 To find Q we also need to find ω 0 ω 0 = 2π T = 2πf = 2π(256) = 512π Now we can use Q = ω 0/γ Q = ω 0 γ = 512π 0.69 = (b) Our change in energy is the same so we still have γ = 0.69 ω 0 = 2πf = 2π(512) = 2π(256)(2) = 2π(2f) = 2(2πf) where f is the frequency from part (a), this gives us: ω 0 = 2ω 0 Using this we can find our Q for this note: Q = ω 0 γ = 2 ω0 γ (c) To find b, we know that Ē = γt. Using this: = 2Q = 2( ) = after seconds; t = we have: Using γ = b /m: To find Q: Ē = γt Ē 0 γ() = e Ē 0 e 1 γ() = 1 = γ γ = 1 1 = b m b = m = 0.1 = Q = ω 0 k γ = m γ = = 9 = 3() = 12 1/

Oscillations. Simple Harmonic Motion of a Mass on a Spring The equation of motion for a mass m is attached to a spring of constant k is

Oscillations. Simple Harmonic Motion of a Mass on a Spring The equation of motion for a mass m is attached to a spring of constant k is Dr. Alain Brizard College Physics I (PY 10) Oscillations Textbook Reference: Chapter 14 sections 1-8. Simple Harmonic Motion of a Mass on a Spring The equation of motion for a mass m is attached to a spring