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424: Oscillations & Waves 1 PH424: Homework 30%; Laboratory reports 35%; Final 35%. Mon Tue 13 14 Simple Harmonic Motion - Free motion of an - 4 representations oscillator -Free damped oscillations Feb & March 2017 ~ Wed 15 -PS1a due Fri 16 17 Upload data Lab & Discussion: -PS1b due the LCR circuit Forced motion of a damped oscillator LCR circuit resonance 23 24 -PS2b due (Math Methods) (Math Methods) 20 Forced motion of a damped oscillator -data workshop o phase shifts o resonance 27 (Math Methods) Formal LCR Lab Report Due (in class) 21 6 -Multiple Driving Frequencies 7 8 9 Fourier coefficients & Wave Mechanics transform -PS4a due Fast Fourier Methods Fast Fourier Transform & Impulse Demo Lab 10 14 17 13 22 -PS2a due Thu Forced motion & resonances (Math Methods) 28 (Math Methods) 1 -PS3a due -The Fourier Series & Transform 2 Lab & Discussion: Coax Cable Lab Intro to Wave Reflection Workshop & Transmission Pre-Lab 15 16 3 -PS3b due Lab Data Workshop Upload data -PS4b due 2 Introducction to Formal Technical Writing • Two “formal” lab reports (35%) are required. Good technical writing is very similar to writing an essay with sub-headings. We want to hear a convincing scientific story, not a shopping list of everything you did. • Check out course web-site • We will have data & write-up workshops 3 Are oscillations ubiquitous or are they merely a paradigm? Superposition of 4 brain neuron activity for our purposes …. Oscillations modulations in time Waves modulations in space Superposition of 5 brain neuron activity REPRESENTING SIMPLE HARMONIC MOTION simple not simple 6 http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/shm.gif Simple Harmonic Motion y(t) Watch as time evolves 7 x(t) = A cos (w 0t + f ) phase angle amplitude position 1 T= = w0 f angular freq (cyclic) freq x0 -0.25 -A 2p 1 A f t=w0 period 0.25 0.75 1.25 1.75 -1 time determined by initial conditions determined by physical system 8 1 0.5 Position (cm) x(t) = A cos (w 0t + f ) 0 -0.5 x -1 0 t 1 2 0 t 1 2 0 t 1 2 6 4 2 Velocity (cm/s) 0 x(t) = Aw 0 cos (w 0t + f + p2 )v -2 -4 -6 40 20 Acceleration (cm/s2) 0 x(t) = Aw cos (w 0t + f + p )a 2 0 -20 -40 time (s) 9 These representations of the position of a simple harmonic oscillator as a function of time are all equivalent - there are 2 arbitrary constants in each. Note that A, f, Bp and Bq are REAL; C and D are COMPLEX. x(t) is real-valued variable in all cases. A: x(t) = Acos(w0t + f ) B: x(t) = B p cos w 0 t + Bq sin w 0 t C: x(t) = C exp(iw 0 t ) + C * exp(-iw 0 t ) D: x(t) = Re[ D exp(iw 0 t )] Engrave these on your soul - and know how to derive the relationships among A & f; Bp & Bq; C; and D . 10 Example: initial conditions k m x(t) = Acos(w0t + f ) k x(t) = B p cos w 0 t + Bq sin w 0 t m x m = 0.01 kg; k = 36 Nm-1. At t = 0, m is displaced 50mm to the right and is moving to the right at 1.7 ms-1. Express the motion in form A form B 11 x(t) = Acos(w0t + f ) ( ) x(t) = 57.5 cos éë 60s ùû t - 0.516 mm -1 x(t) = B p cos w 0 t + Bq sin w 0 t ( ) ( x(t) = 50mm cos éë 60s -1 ùû t + 28.3mmsin éë 60s -1 ùû t Bp = Acos f Bq = -Asin f A = Bp + Bq 2 tan f = - ) 2 Bq Bp 12 Using complex numbers: initial conditions. Same example as before, but now use the "C" and "D" forms k k m x(t) = C exp(iw 0 t ) + C * exp(-iw 0 t ) x(t) = Re[ D exp(iw 0 t )] m x m = 0.01 kg; k = 36 Nm-1. At t = 0, m is displaced 50mm to the right and is moving to the right at 1.7 ms-1. Express the motion in form C form D 13 x(t) = C exp(iw 0 t ) + C * exp(-iw 0 t ) x(t ) (25 14.16i)e i 60s 1t (25 14.16i)e i 60s 1t mm x(t) = Re[ D exp(iw 0 t )] x(t ) Re 50 28.3i e i 60s 1t mm Acos f = Bp = 2Re[C ] = Re[ D] Asin f = -Bq = 2Im[C] = Im[ D] D = 2C = A Im[ D] Im[C ] tan f = = Re[ D] Re[C14] Differential Equation Representation our equation is a second-order ODE our basis can be {cos(wt), sin(wt)} But all linear combinations are solutions, i.e., Acos(wt) + Bsin(wt) Consider all the equivalent solutions what about Asin(wt+f) ? Asin(wt+f) =A{sin(wt) cos(f) + cos(wt) sin(f)} =Acos(f) • sin(wt) + Asin(f)•cos(wt) =A' sin(wt) + A' tan(f)•cos(wt) = A' sin(wt) + B' cos(wt) so Asin(wt+f) gives all linear combinations of basis solutions … spans the space Clicker Questions 17 A particle executes simple harmonic motion. When the velocity of the particle is a maximum which one of the following gives the correct values of potential energy and acceleration of the particle. (a)potential energy is maximum and acceleration is maximum. (b)potential energy is maximum and acceleration is zero. (c)potential energy is minimum and acceleration is maximum. (d)potential energy is minimum and acceleration is zero. 18 A particle executes simple harmonic motion. When the velocity of the particle is a maximum which one of the following gives the correct values of potential energy and acceleration of the particle. (a)potential energy is maximum and acceleration is maximum. (b)potential energy is maximum and acceleration is zero. (c)potential energy is minimum and acceleration is maximum. (d)potential energy is minimum and acceleration is zero. Answer (d). When velocity is maximum displacement is zero so potential energy and acceleration are both zero. 19 A mass vibrates on the end of the spring. The mass is replaced with another mass and the frequency of oscillation doubles. The mass was changed by a factor of (a) 1/4 (b) ½ (c) 2 (d) 4 20 A mass vibrates on the end of the spring. The mass is replaced with another mass and the frequency of oscillation doubles. The mass was changed by a factor of (a) 1/4 (b) 1/2 (c) 2 (d) 4 Answer (a). Since the frequency has increased the mass must have decreased. Frequency is inversely proportional to the square root of mass, so to double frequency the mass must change by a factor of 1/4. 21 A mass vibrates on the end of the spring. The mass is replaced with another mass and the frequency of oscillation doubles. The maximum acceleration of the mass: (a) (b) (c) (d) remains the same. is halved. is doubled. is quadrupled. 22 A mass vibrates on the end of the spring. The mass is replaced with another mass and the frequency of oscillation doubles. The maximum acceleration of the mass: (a) (b) (c) (d) remains the same. is halved. is doubled. is quadrupled. Answer (d). Acceleration is proportional to frequency squared. If frequency is doubled than acceleration is quadrupled. 23 A particle oscillates on the end of a spring and its position as a function of time is shown below. At the moment when the mass is at the point P it has (a) positive velocity and positive acceleration (b) positive velocity and negative acceleration (c) negative velocity and negative acceleration (d) negative velocity and positive acceleration 24 A particle oscillates on the end of a spring and its position as a function of time is shown below. At the moment when the mass is at the point P it has (a) positive velocity and positive acceleration (b) positive velocity and negative acceleration (c) negative velocity and negative acceleration (d) negative velocity and positive acceleration Answer (b). The slope is positive so velocity is positive. Since the slope is getting smaller with time the acceleration is negative. 25 Optional Review of Complex Numbers 26 i = -1 Complex numbers z = a + ib Re ( z ) = a üï ý real numbers Im ( z ) = b ïþ z = ze if Imag |z| b f a z = a +b 2 2 Real Argand diagram b tan f = a 27 Euler’s relation exp(if) = cos f + isin f exp(iw 0 t ) = cos(w 0 t ) + i sin(w 0 t ) 28 Consistency argument z = z e if z = a + ib If these represent the same thing, then the assumed Euler relationship says: a + ib = z cos f + i z sin f Equate real parts: Equate imaginary parts: z = a +b 2 2 a = z cos f b = z sin f b tan f = a 29 x(t) = Re[ Ae e if iw 0 t ] = Re[ Ae i (w 0 t+f ) ] Imag t = T0/4 p +f 2 PHASOR t = 0, T0, 2T0 f Real w0t + f t=t 30 Adding complex numbers is easy in rectangular form z = a + ib w = c + id Imag z + w = [ a + c] + i [b + d ] b a c Real d 31 Multiplication and division of complex numbers is easy in polar form z = ze if iq w= we Imag i [f + q ] zw = z w e |z| qf q |w| f Real 32 Another important idea is the COMPLEX CONJUGATE of a complex number. To form the c.c., change i -> -i z = a + ib z* = a - ib Imag z + z* = [ a + a ] + i [ b - b ] = 2a b z = ze if z* = z e if zz* = z e z e -if -if = z |z| a f Real 2 The product of a complex number and its complex conjugate is REAL. We say “zz* equals mod z squared” 33 And finally, rationalizing complex numbers, or: what to do when there's an i in the denominator? a + ib z= c + id a + ib c - id z= ´ c + id c - id z= ac + bd + i ( bc - ad ) c +d ac + bd ( bc - ad ) = 2 +i 2 2 c +d c + d2 2 Re ( z ) 2 Im ( z ) 34