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SPRING-MASS OSCILLATORS

SPRING-MASS OSCILLATORS. AP Physics Unit 8. Recall Hooke’s Law. Applied force (F applied ) stretches or compresses spring from its natural length Restoring force (F r ) acts to return spring to lowest energy state. An energy approach to SHM.

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SPRING-MASS OSCILLATORS

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  1. SPRING-MASS OSCILLATORS AP Physics Unit 8

  2. Recall Hooke’s Law • Applied force (Fapplied) stretches or compresses spring from its natural length • Restoring force (Fr) acts to return spring to lowest energy state

  3. An energy approach to SHM • Stretched/compressed spring stores elastic potential energy: Us = ½kx2 • When released, mass oscillates about its equilibrium position as PE  KE  etc • Amplitude of oscillation is xmax • At x=0, Us = 0 so K is maximized • At x=A, Us is maximized, so K=0 Us, max = ½kA2

  4. Example #1 • A 2.0 kg block is attached to an ideal spring with a force constant of 500 N/m. The spring is stretched 8.0 cm and released. • When the block is 4.0 cm from equilibrium • what is the total energy of the system? • what is the velocity of the block? By energy conservation, E=1.6J at every spring position

  5. An energy approach to SHM Since energy is conserved and • at x=A, Us is maximized and K=0 • at x=0, Us = 0 so K is maximized it follows that therefore

  6. SHM and the Reference Circle • Motion of the shadow cast by a particle moving in a vertical circle mimics SHM • Amplitude corresponds to the radius of the circle • Period of the oscillation corresponds to the period of the UCM or since

  7. Example #2 A 2.0kg block is attached to a spring with a force constant of 300 N/m. Determine the period and frequency of the oscillations.

  8. Vertical Spring-Mass Oscillators • As it turns out, the behavior is the same regardless of the orientation, i.e. gravity does not affect the period or frequency of the oscillations. • Sounds improbable? Let’s see why it is not…

  9. Vertical SMOs • Consider a spring with constant k on which a mass m is hung, stretching the spring some distance x • The spring is in equilibrium: Fapplied= Fr or kx=mg • If the spring is further displaced by some amount A, the restoring force increases to k(x+A) while the weight remains mg

  10. Vertical SMOs The net force on the block is now F= k(x+A)- mg But sincekx=mg, the force on the block isF= kA. This is Hooke’s Law! Instead of oscillating about the natural length of the spring as happens with a horizontal SMO, oscillations of a vertical SMO are about the point at which the hanging mass is in equilibrium!

  11. Example #3 • A 1.5 kg block is attached to the end of a vertical spring with a constant of 300 N/m. After the block comes to rest, it is stretched an additional 2.0 cm and released. • What is the frequency of the oscillation? • What are the maximum & minimum amounts of stretch in the spring? Since A=2.0 cm, the spring is stretched a maximum of 6.9 cm and a minimum of 2.9 cm

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