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17 Pages

### Chapter 15 SHM (Part 2)

Course: PCS 213, Winter 2011
School: Ryerson
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Word Count: 720

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= 121 K mv = m 2 A2 sin 2 ( t + ) 2 2 1212 2 U = kx = kA cos ( t + ) 2 2 12 E = K + U = kA sin 2 ( t + ) + cos2 ( t + ) 2 12 E = kA 2 [ ] The total mechanical energy of a s.h.o. is a constant of the motion and is proportional to the square of the amplitude. At x = A , E = max U stored in the spring. At x = 0, E = max K Energy is continuously being transformed betwn U stored in the spring and K of the...

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= 121 K mv = m 2 A2 sin 2 ( t + ) 2 2 1212 2 U = kx = kA cos ( t + ) 2 2 12 E = K + U = kA sin 2 ( t + ) + cos2 ( t + ) 2 12 E = kA 2 [ ] The total mechanical energy of a s.h.o. is a constant of the motion and is proportional to the square of the amplitude. At x = A , E = max U stored in the spring. At x = 0, E = max K Energy is continuously being transformed betwn U stored in the spring and K of the block. 1 21212 E = K + U = mv + kx = kA 2 2 2 k2 2 2 2 (A x ) = (A x ) v= m v is max at x = 0 and 0 at x = A A projection of the circular motion of a rotating ball matches the SHM of an object on a spring Uniform circular motion projected onto one dimension is simple harmonic motion. A particle in uniform circular motion with radius A and angular velocity . When the particle is at angle , x = Acos If the particle starts from 0 = 0 at t = 0, its angle at a later time t is = t , where is the particles angular velocity. The particles x-component can be expressed as x(t) = A cos(t + ) The x - component of a particle in uniform circular motion is SHM Fig. 15-14, p. 430 A particle rotates counterclockwise in a circle of radius 3.00m with a constant angular speed of 8.00 rad/s. At t = 0, the particle has an x coordinate of 2.00 m and is moving to the right. (a) Determine the x-coordinate of the particle as a function of time. (b) Find the x component of the particles velocity and acceleration at any time t. x = A cos( t + ) = (3.00m) cos(8.00t + ) 2.00m = (3.00m ) cos(0 + ) = cos1 2.00m 0 = 48.2 = 0.841 rad 3.00m x = (3.00m) cos(8.00t 0.841) dx = ( 3.00m)(8.00rad / s) sin(8.00t 0.841) dt = ( 24.0m / s ) sin(8.00t 0.841) dv a x = x = ( 24.0m / s )(8.00rad / s ) cos(8.00t 0.841) dt = (192m / s ) cos(8.00t 0.841) vx = Pendulum d 2s Ft = mg sin = m 2 dt d 2 g s = L = sin 2 dt L g small = L = g L 2 L T= = 2 g The period and frequency depend only on the length of string and acceleration due to gravity. Fig. 15-16, p. 432 Damped Oscillations Nonconservative force retards the motion. Mechanical energy of the system in time and motion is damped. r r R = bv Retarding force b: damping coefficient F x r r R bv = = kx bv x = ma x d 2x dx kx b = m 2 dt dt x = Ae ( b 2 m ) t cos( t + ) 2 = 0 = k b b = 0 m 2m 2m k m : natural 2 frequency Fig. 15-20, p. 436 Damped oscillator When the retarding force is small, the oscillatory character of the motion is preserved but the amplitude decreases in time, with the result that the motion ultimately ceases. Amplitude decays exponentially with time. Oscillations dampen more rapidly for larger retarding force. overdamped critically damped underdamped Magnitude of the retarding force small, underdamped b/2m < 0, As b, amplitude of the oscillations more rapidly. When b bc such that bc/2m = 0, system does not oscillate, critically damped If b/2m > 0, system is overdamped What is the period of the motion of a damped oscillator given that m = 250g, k = 85N/m, and b = 70g/s. How long does it take for the amplitude of the damped oscillations to drop to half its initial value? m 0.25kg T = 2 = 2 = 0.34 s k 85 N / m 1 1 bt / 2 m xm e = xm ln = ln(e bt / 2 m ) = bt / 2m 2 2 2m ln 1 2 ( 2)(0.25kg )(ln 1 / 2) t= = = 5.0 s b 0.070kg / s Forced Oscillations energy added periodically to maintain oscillatory behaviour F ( t ) = F0 sin t : angular frequency of driving force dx d 2x F = ma F0 sin t b dt kx = m dt 2 x = A cos( t + ) F0 / m A= steady-state 2 b 2 22 ( 0 ) + m 0 = k m natural frequency of undamped oscillator For small damping, A is large when 0 Resonance 0 resonance frequency r r F is in phase with v rr Power delivered to oscillator, F v max when in phase A with damping (b 0), and resonance curve broadens as b . b = 0, steady state amplitude infinity as 0 A block weighing 40.0 N is suspended from a spring that has a force constant of 200 N/m. The system is undamped and is subjected to a harmonic driving force of frequency 10.0 Hz, resulting in a force-motion amplitude of 2.00 cm. Determine the maximum value of the driving force. A= F0 m ( 2 ) 22 0 ( = 2 f = 20.0 s 1 ( 2 F0 = mA 2 0 ( ) 2 0 = ) ) 200 k = 40.0 = 49.0 s 2 m 9.80 () 40.0 2 F0 = 2.00 10 ( 3 950 49.0) = 318 N 9.80
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