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Oscillations

Oscillations. Consider a spring-mass system undergoing oscillations or vibrations on a frictionless horizontal surface. - x. + x. F net = 0 N. X= 0 (equilibrium position). When the displacement is positive the restoring force is negative. + x. F net. X= 0.

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Oscillations

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  1. Oscillations

  2. Consider a spring-mass system undergoing oscillations or vibrations on a frictionless horizontal surface. -x + x Fnet = 0 N X= 0 (equilibrium position)

  3. When the displacement is positive the restoring force is negative. + x Fnet X= 0

  4. When the displacement is negative the restoring force is positive. - x Fnet X= 0

  5. Amplitude = X The amplitude is equal to the maximum displacement from the equilibrium position.

  6. The period (T) is the length of time it requires to complete one oscillation

  7. The frequency (ƒ)is equal to the number of oscillations in one second or how often an oscillation occurs.

  8. T = 1 / ƒ The unit for period is the second. 1 / s = 1 hertz = 1 Hz Which represents one oscillation per second.

  9. Any repetitive motion such as a mass on a spring or a pendulum can be described as harmonic motion

  10. Simple Harmonic Motion (SHM) SHM occurs if the restoring force is directly proportional to the displacement.

  11. Therefore in SHM the restoring force: F = -kx

  12. Free-body diagram k x  x mg m Gently place a mass on the spring so that it can hang at rest

  13. Apply a light force to extend the spring  x k (x+ x) x mg

  14.  x New equilibrium position x k (x+ x) mg

  15. The force acting on the mass is F = -k (x+ x) + mg = -k (x+ mg) + mg k therefore F = -k x So a good rule of practice … for vertical springs: find the new equilibrium point, measure the displacements from there … and ignore gravity.

  16. Period of a Pendulum Another example of Simple Harmonic Motion

  17. Period as a function of Length

  18. Using radian measure x = L  which represents the arc path of the pendulum or its actual displacement. x = L   Frestore = - mg (sin  ) Frestore= - kx x L mg (sin  ) x = 0

  19. Equating Acceleration (ax) a Pendulum to the Centripetal Acceleration Ac = 4π2r T2 Then T2 = 4π2r Ac

  20. SHM (reference circle) using Newton’s 2nd Lawwhich infers that any object in SHM will have an acceleration that is directly proportional to its displacement. Frestore = - Kx = ma a = -K x m

  21. If an object is moving in a vertical circle with constant speed v0 and radius R. Consider a light source shining from above that casts a shadow below the object v0 R Shadow moves back and forth with properties equal to the horizontal component of the circularly moving object.

  22. Can we show that acceleration of the shadow is proportional to the displacement. Recall from unit circle that: At the ends vhorizontal = 0 x = R cos  ac = v02 cos v0 R R ac = v02 R(cos) = v02 (x) R2 R2 Since v and R are constant, ac is proportional to x +x

  23. Again since v and R are constant, ac is proportional to x. So if the acceleration of the shadow is directly proportional to its displacement, this motion is SHM. Recall from unit circle that: At the ends vhorizontal = 0 x = R cos  ac = v02 cos v0 R R ac = v02 R(cos) = v02 (x) R2 R2 +x

  24. If we reverse our argument, we will find that for every system undergoing SHM, you can visualize a reference circle where the uniform circular motion corresponds to the real system…

  25. Radius (R) Constant tangential speed Period of revolution (T) Amplitude (A) Maximum speed (v0 ) Period of oscillation (T) Correspondence of circle to system Referencecircle Real System 2πA =v0T > T = 2πA v0

  26. Using a spring-mass system, we can compare the energy stored in the spring at full extension to the kinetic energy as it passes through the equilibrium point ES = EK ½ kSA2 = ½ m v02A = m v0 kS

  27. A = m v0 kS Substituting into our equation for the systems period (T): The period depends on the system properties T = 2πA = 2π m v0 kS

  28. To determine the period of a pendulum we can replace kS with mg/L T = 2π m Spring-mass system kS The period of a pendulum is independent of mass T = 2π L Pendulum g This formula we had previously derived from one of your first labs and then through vector analysis of centripetal acceleration experienced during uniform circular motion.

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