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Simple Harmonic Motion(SHM)

Simple Harmonic Motion(SHM). Vibration (oscillation) Equilibrium position – position of the natural length of a spring Amplitude – maximum displacement. Period and Frequency. Period (T) – Time for one complete cycle (back to starting point) Frequency (Hz) – Cycles per second

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Simple Harmonic Motion(SHM)

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  1. Simple Harmonic Motion(SHM) • Vibration (oscillation) • Equilibrium position – position of the natural length of a spring • Amplitude – maximum displacement

  2. Period and Frequency • Period (T) – Time for one complete cycle (back to starting point) • Frequency (Hz) – Cycles per second • Angular Velocity/Frequency (rad/s), w – FIND THIS FIRST f = 1 T = 1 w = 2pf T f

  3. Period and Frequency A radio station has a frequency of 103.1 M Hz. What is the period of the wave? 103.1 M Hz 1X106 Hz = 1.031 X 108 Hz 1M Hz T = 1/f = 1/(1.031 X 108 Hz) = 9.700 X 10-9 s

  4. Cosines and Sines • Imagine placing a pen on a vibrating mass • Draws a cosine wave

  5. Starting at the Amplitude • Used if spring is pulled out (or compressed) to full position x(t) = A cos2pt = A cos2pft = A coswt T v(t) = -vmaxsinwt vmax = wA A = Amplitude t = time T = period f = frequency

  6. An air-track glider is attached to a spring. It is pulled 20.0 cm to the right and makes 15 oscillations in 10.0s • Calculate the period • Calculate the angular velocity (w) • Calculate the maximum speed • Calculate the speed and position at t = 0.800s

  7. A mass starts at x=A. Using only variables, calculate at what time as a fraction of T that the object passes through ½ A. x(t) = A cos2pt T

  8. Velocity is the derivative of position x = A coswt v = -vosinwt a = -aocoswt Acceleration is the derivative of velocity

  9. A loudspeaker vibrates at 262 Hz (middle C). The amplitude of the cone of the speaker is 1.5 X 10-4 m. • Write the equation to describe the position of the cone over time. (x = (1.5 X 10-4 m) cos(1650 rad/s)t) • Calculate the position at t = 1.00 ms. (-1.2 X 10-5 m) • Calculate the maximum velocity and acceleration (0.25 m/s, 410 m/s2 )

  10. F = ma kx = ma a = kx/m But we don’t know k or m a = k x Solve for k/m m T = 2pm k T2 = (2p)2m k k = (2p)2= (2p)2f2 m T2

  11. a = k x m ao = (2pf)2x = (2pf)2A ao = [(2p)(262 Hz)]2(1.5 X 10-4 m) = 410 m/s2

  12. Find the amplitude, frequency and period of motion for an object vibrating at the end of a spring that follows the equation: x = (0.25 m)cos p t 8.0 (0.25 m, 1/16 Hz, 16 s)

  13. Find the position of the object after 2.0 seconds. x = (0.25 m)cos p t 8.0 x = (0.25 m)cos p 4.0 x = 0.18 m

  14. The Phase Constant • Do not always start at the Amplitude • Can start your observations at any time • o is the starting angle using the circle model x(t) = A cos(wt + o) v(t) = -vmaxsin(wt + o) vmax = wA w = 2pf = 2p/T

  15. An object on a spring oscillates with a period of 0.800 s and an amplitude of 10.0 cm. At t=0, it is 5.0 cm to the left of equilibrium and moving to the left. • Calculate the phase constant (in radians) from the initial conditions. (2/3prad) • Calculate the position at t = 2.0 s (5.0 cm) • Calculate the velocity at t = 2.0 s (68 cm/s) • What direction is the object moving at 2 s? (right)

  16. A mass of 4.00 kg is attached to a horizontal spring with k = 100 N/m. It is displaced 10.0 cm from equilibrium and released. • Calculate the period. (1.25 s) • Calculate the angular velocity w. (5.00 rad/s) • Calculate the phase angle (in radians) from the initial conditions. (p/2rad) • Calculate the maximum velocity (0.500 m/s) • Calculate the velocity when x = 5.0 cm (0.433 m/s)

  17. The initial position and velocity of a block moving in SHM with period T=0.25 s are x(0) = 5.0 cm and v(0) = 218 cm/s. • Calculate the angular velocity w. (25.1 rad/s) • Calculate the amplitude (10.0 cm) • Calculate the phase constant (in radians) from the initial conditions. (p/6rad)

  18. Forces on a Spring • Extreme Position (Amplitude) • Force at maximum • Velocity = 0 • Equilibrium position • Force = 0 • Velocity at maximum

  19. The Equation of Motion F = ma F = -kx ma = -kx a = -kx also a = dv = d2x m dt dt2 d2x = -kx dt2 m

  20. d2x + kx= 0 Equation of Motion dt2 m x(t) = A cos(wt + j) dx/dt = -wA sin(wt + j) d2x/dt2 = -w2A cos(wt + j) -w2A cos(wt + j) + k A cos(wt + j) = 0 m

  21. Energy and Springs • KE = ½ mv2 • PE = ½ kx2 • Maximum PE = ½ kA2 Law of conservation of Energy ½ kA2 = ½ mv2+ ½ kx2 Also w = (k/m)½

  22. All PE All KE All PE Some KE and Some PE

  23. A 0.50 kg mass is connected to a light spring with a spring constant of 20 N/m. • Calculate the total energy if the amplitude is 3.0 cm. (9 X 10-3 J) • Calculate the maximum speed of the mass (0.19 m/s) • Calculate the potential energy and kinetic energy at x = 2.0 cm (U = 4 X 10-3 J, K = 5 X 10-3 J) • At what position is the speed 0.10 m/s? (+ 2.6 cm)

  24. A spring stretches 0.150 m when a 0.300 kg mass is suspended from it. • Find the spring constant. (19.6 N/m) • The spring is now stretched an additional 0.100 m and allowed to oscillate (diagram c). What is the maximum velocity? (0.808 m/s) • Calculate the velocity at x = 0.0500 m (0.700 m/s) • What is the maximum acceleration? (6.53 m/s2)

  25. A 500 g block is pulled 20 cm on a spring and released. It has a period of 0.800 s . At what positions is the block’s speed 1.0 m/s? (Hint: use w = \/k/m )

  26. Trigonometry and SHM • Ball rotates on a table • Looks like a spring from the side • One rev(diameter) = 2pA T = 2pm w = \/k/m k f = 1 T

  27. Period depends only on mass and spring constant • Amplitude does not affect period vmax = 2pAf or vmax = 2pA T w = \/k/m

  28. What is the period and frequency of a 1400 kg car whose shocks have a k of 6.5 X 104 N/m after it hits a bump? w = k = 6.81 rad/s m w = 2pf ANS: 1.09 Hz

  29. An insect (m=0.30 g) is caught in a spiderweb that vibrates at 15 Hz. • What is the spring constant of the web? (2.7 N/m) • What would be the frequency for a lighter insect, 0.10 g? Would it be higher or lower? (26 Hz)

  30. At t = 0, a 5000 g block is moving to the right and is at 15 cm. Its maximum displacement is 25 cm at 0.30 s. • Calculate the phase constant • Calculate the angular velocity • Calculate the time and velocity when the mass is at x = 20 cm • Sketch a graph of the motion, including the phase constant and period

  31. Vertical Motion of a Spring • Gravity is ALWAYS acting on the spring and mass consistently • Only need to use it to calculate the spring constant • F = -ky (using y for vertical rather than x)

  32. An 83 kg student hangs from a bungee cord with spring constant 270 N/m. He is pulled down to a position 5.0 m below the unstretched length of the bungee, then released. • Calculate the equilibrium length of the bungee/student (3 m) • Calculate the Amplitude (2 m) • Calculate the position and velocity 2.0 s later. (1.8 m, -1.6 m/s)

  33. The Pendulum • Pendulums follow SHM only for small angles (<15o) • The restoring force is at a maximum at the top of the swing. q Fr = restoring Force

  34. Remember the circle (360o = 2p rad) q = x L Fr = mgsinq at small angles sinq = q Fr = mgq q L x q mg s

  35. Fr = mgq Fr = mg s (Look’s like Hook’s Law F = -kx) L k = mg L T = 2pm k T = 2p mL mg

  36. T = 2p L g f = 1 = 1 g T 2p L The Period and Frequency of a pendulum depends only on its length

  37. The Pendulum and the Equation of motion F = -mgsinq ma = -mgsinq a = -gsinq d2s = -gsinq dt2 d2s = -gq (small angle approximation) dt2

  38. d2s = -gs (q = s/L) dt2 L s = Acos(wt + f) d2s = -w2Acos(wt + f) dt2 -w2Acos(wt + f) = -g Acos(wt + f) L

  39. Swings and the Pendulum • To go fast, you need a high frequency • Short length (tucking and extending your legs) f = 1 g 2p L decrease the denominator

  40. Consider a grandfather clock with a 1.0 m long pendulum • Calculate the period of? (2.0 s) • Estimate the length of the pendulum of a grandfather clock that ticks once per second (T = 1.0 s). (0.25 m)

  41. A 300 g mass on a 30 cm long string swings at a speed of 0.25 m/s at its lowest point. • Calculate the period • Calculate the angular velocity • Calculate the maximum angle that the pendulum reaches (HINT: Use the triangle and the small angle approximation).

  42. Physical Pendulum • Center of mass is in the middle (sinq q) t = -Mglsinq t = -Mglq (small angle approximation)

  43. = Ia a = d2q dt2 • = I d2q dt2 • = -Mglq I d2q = -Mglq dt2 d2q + Mglq= 0 dt2 I

  44. d2q + Mglq= 0 dt2 I d2x + k x= 0 dt2 m

  45. A student swings his leg which is 0.90 m long. Assume the center of mass is halfway down the leg. • Write the equation for the moment of inertia of the leg (I = ML2/3) • Substitute this into the period formula to calculate the period (1.6 s) • Calculate the frequency (0.64 Hz)

  46. A nonuniform 1.0 kg physical pendulum has a center of mass 42 cm from the pivot. It oscillates with a period of 1.6 s. • Calculate the moment of inertia (0.27 kg m2) • Using the parallel axis theorem, calculate the period if the pendulum were swung at the center of mass. (0.09 kg m2) I = Icm + Md2

  47. A Christmas ball has a radius R and a moment of inertia of 5/3MR2 when hung by a hook. The ball is given a slight tap and rocks back and forth. • Derive the formula for the period of oscillation. Assume the center of mass is at the center of the ball. • Insert a reasonable radius into your equation to estimate the period.

  48. Damped Harmonic Motion • Most SHM systems slowly stop • For car shocks, a fluid “dampens” the motion

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