Date: _15 April (1) The only reference material you may use is one 8½x11 crib sheet and a calculator.

Transcription

1 PH1140: Oscillations and Waves Name: SOLUTIONS Conference: Date: _15 April 2005 EXAM #2: D2005 INSTRUCTIONS: (1) The only reference material you may use is one 8½x11 crib sheet and a calculator. (2) Show your work clearly and give numerical results to at least three significant figures. (3) Attach the proper units to all quantities in your answers. (4) Use the back of the preceding page for scratch paper. PROBLEMS: 1. (40) 2. (40) 3. (20) Total: (100)

2 1. A 4 Kg block rests on top of a spring, which rests on a solid platform. The system has a natural frequency of 1.5 Hz. a) Assuming no damping, determine the spring constant of the spring. b) Now, damping is introduced with b = 0.2 Kg / s. Find the γ, Q and new frequency of this system. c) For the same configuration as part (b), find the number of periods it takes for the amplitude to be reduced by a factor of 2. d) The platform is now driven up and down with amplitude of 2 mm. What is the driving frequency of oscillations if the phase lag is π/4?

3 2. A series LRC-circuit has an undamped frequency of 100 KHz. The capacitor has a capacitance of 2.5 µf. a) Determine a value for the self-inductance L. b) The resistance of the circuit is R = 3.5 Ω. Find γ and Q. c) The driving voltage has a peak value of 0.1 V. Determine the peak voltage across the capacitor for (i) a frequency 100 times lower than the natural frequency and for (ii) a frequency equal to the natural frequency. d) Calculate the peak charge on the capacitor for the second situation (ii) in part c).

4 3. The transverse displacement of a traveling wave along a long line of coupled oscillators is given by (in SI units): u(x, t) = 0.5 cos( (150π) t + (π / 5) x + 0.1π ) a) Find the frequency f, wavelength λ, and the wave velocity v of this wave. Which way, relative to x, is the wave propagating? b) Find two consecutive (neighboring) points along this line of coupled oscillators that have u(x, t) = 0 at t = 0.1 sec.

5 PH 1140: EXAM #2 Solutions 1) Given: mass = m = 4 Kg undamped frequency = f o = 1.5 Hz ideal angular frequency = ω o = 2πf o = 3π rad / s = rad / s a) The spring constant and mass are related to the ideal frequency by ω o 2 = k / m rewrite as k = m ω o 2 So, k = 36π 2 N / m = Kg / s 2 b) Now, the dampening coefficient is b = 0.2 Kg / s dampening frequency is γ = b / m = 0.05 rad / s Quality Factor is Q = ω o / γ = 60π = new frequency is f = f o ( 1 1 / (4Q 2 ) ) ½ = Hz = 1.5 Hz c) With dampening and initial amplitude of A o, amplitude of the oscillations vary with time as A(t) = A o exp( γ t / 2 ) solving for t gives the general relation t = -2 ln(a / A o ) / γ For A / A o = ½ the elapsed time is t = sec This time is equivalent to t = nt, where n = number of periods. n = t / T = t f = periods

6 d) The amplitude of the platform oscillation is unimportant here, only that the platform is now oscillating and so driving the mass-spring system. The tangent of the phase lag is +π/4 and the relationship is given by tan( φ ) = γω / (ω ο 2 ω 2 ) = tan( π/4 ) = 1 Rearranging gives γω = ω ο 2 ω 2 or ω 2 + γω ω ο 2 = 0 This is a quadratic expression like ax 2 + bx + c = 0 which has two solutions x = [ b ± (b 2 4ac) ½ ] / 2a Here, x = ω and the coefficients are a = 1, b = γ, and c = ω ο 2. The two frequencies are ω = rad /s or ω = rad /s Since ω must be positive, the answer is f = ω / 2π = Hz This checks out since φ = +π/4 < π/2, the driving frequency f must be less than the ideal (or natural) frequency, f o.

7 2) Given f o = 100 KHz = Hz ω o = 2πf o = 2π 10 5 rad / s C = 2.5 µf = F a) For an ideal LC oscillator, the inductance is given by ω o 2 = 1 / (LC) rewriting gives L = 1 / (C ω 2 o ) = H µh b) Damping is turned on with R = 3.5 Ω γ = R / L = rad / s Q = ω o / γ = ω o L / R = ( over-damped ) f = f o ( 1 1 / (4Q 2 ) ) ½ = 0 ( does not oscillate when free! ) c) The driving voltage has a peak value (amplitude) of V o = 0.1 V. Although the system will not oscillate freely, it is forced to do so now by the driver. The amplitude of the voltage across the capacitor is given by V(ω) = V o ( (1 (ω/ω o ) 2 ) 2 + (ω/ω o ) 2 /Q 2 ) ½ i. ω = ω o / 100 gives V = V ii. ω = ω o gives V = Q V o = V d) For part c.ii), the peak charge is given by the definition of capacitance ( C = q / V ), which gives q(ω ο ) = C V(ω ο ) = Coulombs = 45.5 nc

8 3) The traveling wave is given by u(x, t) = 0.5 cos( (150π) t + (π / 5) x + 0.1π ) a) Comparing to the general form u(x, t) = A cos( ω t k x + φ ) identifies ω = 150π rad / s or f = ω / 2π = 75 Hz k = 2π / λ or λ = 2π / k = 10 m and the wave speed v = λ f = 750 m / s Since the + sign appears in front of the k x term in θ(x,t), the wave moves in the negative x-direction. b) u(x, t) = 0 ( at t = 0.1 sec ) when the argument of the cosine function is an odd multiple of π/2 So, let (150π) (0.1) + ( π / 5) x + 0.1π = π/2 Solving for x gives the first zero point x 1 = 5 [ 1/ ] = 73 m Since all subsequent zeros occur twice every 2π phase angle rotation, they are all a distance λ / 2 = 5 m apart. The next 3 zeros are 68 m, 63 m, 58 m, and so on