#### Transcript Spring-mass oscillators

SPRING-MASS OSCILLATORS AP Physics Unit 8 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 Fr Fapplied Fr kx 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 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? Eo U o K o Eo 1 kA2 0 1 (500)(0.08) 2 2 2 Eo 1.6 J By energy conservation, E=1.6J at every spring position 1 kx2 1 m v2 E 2 2 1 500(0.04) 2 1 (2.0)v 2 1.6 J 2 2 v 1.1m / s 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 2 it follows that 1 2 kA2 1 2 m vmax therefore vmax k A m 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 2R 2A v T T since vmax or k A m T 2A T v 2v m v k 2 m k 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. T 2 m T 2 2.0 k 300 1 f 1.9 Hz T 0.51s 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… 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 Vertical SMOs The net force on the block is now F= k(x+A)- mg But since kx=mg, the force on the block is F= 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! 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? 1 2 1 f 2 f k m 300 2.3Hz 1.5 kx m g 300x (1.5)(9.8) x 0.049m 4.9cm Since A=2.0 cm, the spring is stretched a maximum of 6.9 cm and a minimum of 2.9 cm