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Simple Harmonic Motion Simple Harmonic Motion Simple harmonic motion is the repetitive motion of an oscillating object (simple harmonic oscillator). Example: A spring mass system. Like circular motion, simple harmonic motion has a regular period and frequency. In fact, simple harmonic motion and circular motion are linked both mathematically and graphically. Equilibrium The spring mass system shown below is at equilibrium ( x = 0 ). Equilibrium for a spring is its natural rest position. This spring can either be stretched or compressed, displacing the mass from its current equilibrium position. +F 0 To stretch the spring a force, F , must be applied in the +x direction. This new unbalanced force will displace the mass from equilibrium. A force is applied As the mass is displaced the spring generates an increasing force of it own. +F 0 A force is applied −FS 0 As the mass is displaced the spring generates an increasing force of it own. +F A force is applied −FS 0 As the mass is displaced the spring generates an increasing force of it own. +F A force is applied As the mass is displaced the spring generates an increasing force of it own. −FS 0 +F A force is applied As the mass is displaced the spring generates an increasing force of it own. This force is known as a restoring force, as it acts to restore the spring to its original equilibrium position. −FS 0 +F Amplitude When the applied force, F , and the restoring force, FS , are equal and opposite the oscillator is stationary. R = A = xmax The oscillator has reached maximum displacement, xmax . Maximum displacement is known as the amplitude, A . This is equal to the radius, R , of a circle matching the springs displacement. −FS 0 x +F +A = xmax The Oscillator is Released When the applied force, F, is removed, the restoring force moves the mass left. x −FS +F x 0 +A The Oscillator is Released When the applied force, F, is removed, the restoring force moves the mass left. x −FS x 0 +A The Oscillator is Released When the applied force, F, is removed, the restoring force moves the mass left. x −FS x 0 +A The Oscillator is Released When the applied force, F, is removed, the restoring force moves the mass left. x −FS x 0 +A The Oscillator is Released When the applied force, F, is removed, the restoring force moves the mass left. x −FS x 0 +A The Oscillator is Released R = A = xmax When the applied force, F, is removed, the restoring force moves the mass left. ω Δθ x The radius of the circle is equal to the maximum displacement (amplitude) R = A = xmax The radius, R = A , and displacement, x , are two of the sides of a right triangle. While the mass moved to the left the corresponding point on the circle moved with an angular velocity, ω , through an angle, Δθ . −FS x 0 +A Conclusion R = A = xmax ω Δθ x An oscillator’s displacement equals the x-component of the radius of another object that is in uniform circular motion with the same amplitude (R = A = xmax). Cosine solves for x . ( ) x = Acos Dq In uniform circular motion Dq = w t −FS Therefore, ( ) x = Acos w t x 0 +A Conclusion Period and frequency are related to angular velocity. R = A = xmax ω 2p 1 T= = w f Which rearrange to Δθ x 2p w= = 2p f T These can be substituted into the previous equation ( ) x = Acos w t −FS ( x = Acos 2p ft x 0 +A ) Putting all together ( x = Acos 2p ft ) The motion described by this equation looks like… x(t) FS x(t) −A 0 +A Putting all together ( x = Acos 2p ft ) The motion described by this equation looks like… ω Δθ x(t) FS x(t) −A 0 +A Putting all together ( x = Acos 2p ft ω Δθ x(t) FS x(t) −A 0 +A ) The motion described by this equation looks like… Putting all together ( x = Acos 2p ft ω Δθ x(t) FS x(t) −A 0 +A ) The motion described by this equation looks like… Putting all together ( x = Acos 2p ft ω Δθ x(t) x(t) −A 0 +A ) The motion described by this equation looks like… Putting all together ( x = Acos 2p ft ω Δθ x(t) FS x(t) −A 0 +A ) The motion described by this equation looks like… Putting all together ( x = Acos 2p ft ω Δθ x(t) FS x(t) −A 0 +A ) The motion described by this equation looks like… Putting all together ( x = Acos 2p ft ω Δθ x(t) FS x(t) −A 0 +A ) The motion described by this equation looks like… Putting all together ( x = Acos 2p ft ω Δθ x(t) FS x(t) −A 0 +A ) The motion described by this equation looks like… Putting all together ( x = Acos 2p ft ω Δθ x(t) FS x(t) −A 0 +A ) The motion described by this equation looks like… Putting all together ( x = Acos 2p ft ω Δθ x(t) FS x(t) −A 0 +A ) The motion described by this equation looks like… Putting all together ( x = Acos 2p ft ω Δθ x(t) FS x(t) −A 0 +A ) The motion described by this equation looks like… Putting all together ( x = Acos 2p ft ω Δθ x(t) x(t) −A 0 +A ) The motion described by this equation looks like… Putting all together ( x = Acos 2p ft ω Δθ x(t) FS x(t) −A 0 +A ) The motion described by this equation looks like… Putting all together ( x = Acos 2p ft ω Δθ x(t) FS x(t) −A 0 +A ) The motion described by this equation looks like… Putting all together ( x = Acos 2p ft ω Δθ x(t) FS x(t) −A 0 +A ) The motion described by this equation looks like… Putting all together ( x = Acos 2p ft ω Δθ x(t) FS x(t) −A 0 +A ) The motion described by this equation looks like… Graphing Oscillations + A 0 x(t) − A FS x(t) −A 0 +A π 2π Putting all together ω Δθ x(t) + A 0 − A FS x(t) −A 0 +A π 2π Putting all together ω Δθ + A 0 x(t) − A x(t) −A 0 +A π 2π Putting all together ω Δθ + A 0 x(t) − A FS x(t) −A 0 +A π 2π Putting all together ω Δθ + A 0 x(t) − A FS x(t) −A 0 +A π 2π Putting all together ω Δθ + A 0 x(t) − A FS x(t) −A 0 +A π 2π Putting all together ω Δθ + A 0 x(t) − A x(t) −A 0 +A π 2π Putting all together ω Δθ x(t) + A 0 − A FS x(t) −A 0 +A π 2π Putting all together ω Δθ + A 0 x(t) − A FS x(t) −A 0 +A π 2π Analyzing Oscillation Graphs: Amplitude Amplitude is the largest displacement and it can be easily read on the graph. A = 8m Analyzing Oscillation Graphs: Period T = 12s Period is the time of one cycle (1 revolution). One cycle is one complete wave form. This wave form extends from 3 seconds to 15 s. The time of this wave form is 15 − 3 = 12 s Analyzing Oscillation Graphs: Frequency Frequency is the inverse of period. 1 1 f = = = 0.0833s T 12 Analyzing Oscillation Graphs: Displacement Displacement can be solved using the displacement equation and substituting the values determined in the previous slides. ( f = 0.0833s ) A = 8m x = (8) cos ( 2p ( 0.0833) t ) the given time to determine the x = (8) cos ( 0.523t ) Substitute matching displacement. x = Acos 2p ft Analyzing Oscillation Graphs: Maximum Speed Maximum speed can be found using the equation for speed in uniform circular motion. Set the radius equal to amplitude. 2p r v= T () ( ) 2p A 2p 8 v= = = 4.19m s T 12 Analyzing Oscillation Graphs: Qualitatively When is the object moving to the right at maximum speed? This occurs at equilibrium (x = 0) and when object is moving toward a greater displacement value: 9 and 21 seconds When is the object moving to the left at maximum speed? This occurs at equilibrium (x = 0) and when object is moving toward a lesser displacement value: 3 and 15 seconds When is the object instantaneously at rest? This occurs at maximum displacement: 0, 6, 12, 18, and 24 seconds. Circular Motion R Δθ x(t) y(t) ω Oscillation is only concerned with the component of the circular motion equation that is parallel to the oscillation. If a spring is oscillating horizontally ( ) x(t) = Acos w t If a spring is oscillating vertically ( ) y(t) = Asin w t Circular motion involves both equations simultaneously. Amplitude is equal radius R , and Δθ = ω t . x(t) = Rcos Dq y(t) = Rsin Dq