On my honor, I have received no help during this test. Your Name ______________________ Your UF ID # (last 4 digits):_________ Midterm Test II, Nov. 3, 2006 PHY3221 Show work on all 4 problems (just the right answer is not enough unless otherwise stated.) Each problem is worth 5 points. The parts to each problem have the points listed. You are allowed two 8.5x11" formula/whatever sheets, plus handouts from class (which may not have additional writing on them.) kHz 1000 Hertz, msec millisecond, kJ kiloJoule 1. A lightly damped oscillator that is driven at 10 kHz (which is near resonance) has Q=1000. a. If the total energy of the system at t=0 is 1 kJ, what is the approximate energy loss after 0.2 msec? (2 points) (3 sig figs for full credit) As shown in your homework, prob. 3-18, Q≈2 x (Total Energy/Energy loss during one period) If the frequency is 10 kHz, then the period is 1/f = 10-4 s, so 0.2 msec is 2 periods, so the energy loss is approximately 2 x 2 x 1000 J/1000 = 4 Joules, or 12. 6 J 1b.) what is the damping parameter β (give units as well) (1 point) (3 sig figs for full credit) Q = R / 2β So 1000 = (≈104 Hz x 2 rad/Hz)/2β so β=10 rad/s = 31.4 rad/s (This is so small that to a very good approximation R = (02 – 2β2)0.5 = 0) 1c.) What is the width of the resonance (distance ∆ between the two frequencies where the amplitude of the oscillations is 1/√2 of the maximum amplitude)? (2 points) Give the units, and the answer to 3 sig figs for full credit. As discussed in class, for a lightly damped oscillator Q = 0/∆, so ∆ = 0/Q ≈ 104 x 2 / 1000 = 20 =62.8 rad/s 2a.) For an under-damped oscillator, write the general equation for the amplitude of motion, x(t). (1 point) x(t)=Ae-βt cos(1t – δ) 2b.) What is 1 equal to in terms of the angular frequency of the oscillator with no damping, 0? (1 point) 12 = 02 – β2 2c.) If the amplitude of an under-damped oscillator decreases by a factor of ½ after 5 oscillations ( 5 periods), find β in terms of 1 to 3 significant figures (2 points). exp(-β*5*(2/1)) = ½ -10β/1 = ln(1/2) = -0.693 β = 0.02206 1 2d.) Approximately what is 1/0, to 5 sig figs (1 point)? Using the answer to 2b., 1/0 = 0.99976 3.) A simple pendulum (mass ‘m’ at the end of a massless rod of length ‘l’) operates in a viscous medium with a retarding force equal to m(gl)1/2(dθ/dt), where g is the acceleration due to gravity and dθ/dt is the first time derivative of the angle theta. a.) (1 point) Write down the equation of motion This is like example 3.3 in the book. I (where the moment of inertia I = ml2) = I d2θ/dt2 = torque = r x F = mglsinθ ≈ mglθ for small angles – without the damping term. Remember that is negative for positive θ, so: ml2 d2θ/dt2 ≈ -mglθ - m(gl)1/2l(dθ/dt) (the last two terms have units of force times a length) b.) What is 0? (1 point) When write down the equation canceling m and dividing thru by l2 , you get d2θ/dt2 + (g/l)θ + (g/l)1/2(dθ/dt) = 0, so 0=(g/l)1/2 c.) What are β and 1 in terms of 0? (2 points) The coefficient of dθ/dt is defined as 2β, so β=(1/2) (g/l)1/2, or ½ 0 12 = 02 – β2, so 1 = (3/4)1/2 0 d.) Discuss the motion of this oscillator (i. e. is it under-, over-, or critically damped?) Discuss why using your answers from above. (1 point) The motion is underdamped, since β2 < 02 4.) Physical System – electrical analog. Consider an LRC (inductor,resistor,capacitance) circuit, when initially the capacitor has charge q0 on it and is isolated from the L and the R by switches as shown. ____________________ | L | / switch 1 | / | | | __|__ | __ __ C | | | | | \ | \ switch 2 R | |_______________/\/\/\/\/\/\___| a.) after both switches are closed simultaneously, write down the 2nd order differential equation for the charge q (1 point.) on the capacitor L dI/dt + q/C +IR = 0 where I is dq/dt, the current L d2q/dt2 + (dq/dt)R +q/C = 0 b.) Using the oscillator analogy, identify β and 02 in terms of R, L, and C (2 points) 02 (the coefficient of q when d2q/dt2 has “one” as a coefficient) = 1/(LC) β = b/2m (for the oscillator) = R/(2L) c.) (2 points) Remember that the proper units for R, L, and C are Ohms (Ω), Henrys, and Farads. If R=100 Ω and L = 1 Henry, what would be the value of C for the charge q stored on the capacitor to decay monotonically (no oscillations) down to 0 within the shortest possible time in this circuit? (give units) β2 = R2/(4L2) If this is equal to 02 = 1/(LC), then we have critical damping, β2 = 02 so R2/(4 L2) = 1/(LC), or C=4L/R2 = 4 Henrys/104 Ω2 = 0.4 milliFarad